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STT 459: Actuarial Models
Albert Cohen
Actuarial Sciences ProgramDepartment of Mathematics
Department of Statistics and ProbabilityC336 Wells Hall
Michigan State UniversityEast Lansing MI
[email protected]@stt.msu.edu
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 1 / 193
Copyright Acknowledgement
Some examples and theorem proofs in these slides, and on in class exampreparation slides, are taken from our textbook ”Loss Models: From Datato Decisions” 4th edition ,by Klugman, Panjer, and Wilmott.
Please note that Wiley owns the copyright for that material. Noportion of the Wiley textbook material may be reproduced in anypart or by any means without the permission of the publisher. Weare very thankful to the publisher for allowing posting of thesenotes on our class website.
Also, we will from time-to-time look at problems from releasedprevious Exams C by the SOA. All such questions belong incopyright to the Society of Actuaries, and we make no claim onthem. It is of course an honor to be able to present analysis of suchexamples here.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 2 / 193
Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .
In our previous experience, F = 2Ω.For now, we take this to be the case. We can define, on this ProbabilitySpace (Ω,F ,P) a Random Variable X as any function
X : Ω→ R (1)
Consider now the experiment of tossing 3 fair coins. Define the randomvariable Y = Number of heads that appear. Then
P[Y = 0] = P[(T ,T ,T )]P[Y = 1] = P[(T ,T ,H), (T ,H,T ), (H,T ,T )]P[Y = 2] = P[(T ,H,H), (H,T ,H), (H,H,T )]P[Y = 3] = P[(H,H,H)]
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 3 / 193
Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .In our previous experience, F = 2Ω.
For now, we take this to be the case. We can define, on this ProbabilitySpace (Ω,F ,P) a Random Variable X as any function
X : Ω→ R (1)
Consider now the experiment of tossing 3 fair coins. Define the randomvariable Y = Number of heads that appear. Then
P[Y = 0] = P[(T ,T ,T )]P[Y = 1] = P[(T ,T ,H), (T ,H,T ), (H,T ,T )]P[Y = 2] = P[(T ,H,H), (H,T ,H), (H,H,T )]P[Y = 3] = P[(H,H,H)]
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 3 / 193
Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .In our previous experience, F = 2Ω.For now, we take this to be the case. We can define, on this ProbabilitySpace (Ω,F ,P) a Random Variable X as any function
X : Ω→ R (1)
Consider now the experiment of tossing 3 fair coins. Define the randomvariable Y = Number of heads that appear. Then
P[Y = 0] = P[(T ,T ,T )]P[Y = 1] = P[(T ,T ,H), (T ,H,T ), (H,T ,T )]P[Y = 2] = P[(T ,H,H), (H,T ,H), (H,H,T )]P[Y = 3] = P[(H,H,H)]
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 3 / 193
Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .In our previous experience, F = 2Ω.For now, we take this to be the case. We can define, on this ProbabilitySpace (Ω,F ,P) a Random Variable X as any function
X : Ω→ R (1)
Consider now the experiment of tossing 3 fair coins. Define the randomvariable Y = Number of heads that appear. Then
P[Y = 0] = P[(T ,T ,T )]P[Y = 1] = P[(T ,T ,H), (T ,H,T ), (H,T ,T )]P[Y = 2] = P[(T ,H,H), (H,T ,H), (H,H,T )]P[Y = 3] = P[(H,H,H)]
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 3 / 193
Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .In our previous experience, F = 2Ω.For now, we take this to be the case. We can define, on this ProbabilitySpace (Ω,F ,P) a Random Variable X as any function
X : Ω→ R (1)
Consider now the experiment of tossing 3 fair coins. Define the randomvariable Y = Number of heads that appear. Then
P[Y = 0] = P[(T ,T ,T )]P[Y = 1] = P[(T ,T ,H), (T ,H,T ), (H,T ,T )]P[Y = 2] = P[(T ,H,H), (H,T ,H), (H,H,T )]P[Y = 3] = P[(H,H,H)]
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 3 / 193
Discrete Random Variables
Consider a random variable X that takes on a countable number of values,denoted by the set R ≡ x1, x2, .... For any a ∈ R, we define
p(a) = P[X = a]∞∑i=1
p(xi ) = 1(2)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 4 / 193
Expected Value
For a random variable X that takes on a countable number of values,define
E[X ] =∑xi∈R
xip(xi ) (3)
As an important example, consider an event A in our σ−field and definethe random variable
IA =
1, if A occurs
0, if A does not occur(4)
Then E[IA] = P[A].
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 5 / 193
Expected Value
For a random variable X that takes on a countable number of values,define
E[X ] =∑xi∈R
xip(xi ) (3)
As an important example, consider an event A in our σ−field and definethe random variable
IA =
1, if A occurs
0, if A does not occur(4)
Then E[IA] = P[A].
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 5 / 193
Expected Value
For a random variable X that takes on a countable number of values,define
E[X ] =∑xi∈R
xip(xi ) (3)
As an important example, consider an event A in our σ−field and definethe random variable
IA =
1, if A occurs
0, if A does not occur(4)
Then E[IA] = P[A].
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 5 / 193
Variance
If X is a random variable with mean µ = E[X ], then the variance Var(X )and standard deviation SD(X ) are defined as
Var(X ) = E[(X − µ)2
]
= E[X 2]− E[X ]2
SD(X ) =√
Var(X )
(5)
A useful identity is
Var(a · X + b) = a2 · Var(X ) (6)
Also notice that E[X 2] ≥ (E[X ])2. This is true regardless of thedistribution of X .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 6 / 193
Variance
If X is a random variable with mean µ = E[X ], then the variance Var(X )and standard deviation SD(X ) are defined as
Var(X ) = E[(X − µ)2
]= E
[X 2]− E[X ]2
SD(X ) =√
Var(X )
(5)
A useful identity is
Var(a · X + b) = a2 · Var(X ) (6)
Also notice that E[X 2] ≥ (E[X ])2. This is true regardless of thedistribution of X .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 6 / 193
Variance
If X is a random variable with mean µ = E[X ], then the variance Var(X )and standard deviation SD(X ) are defined as
Var(X ) = E[(X − µ)2
]= E
[X 2]− E[X ]2
SD(X ) =√
Var(X )
(5)
A useful identity is
Var(a · X + b) = a2 · Var(X ) (6)
Also notice that E[X 2] ≥ (E[X ])2. This is true regardless of thedistribution of X .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 6 / 193
Bernoulli and Binomial Random Variables
Define a Bernoulli random variable X with R = 0, 1 and
p = p(1) = P [X = 1]
1− p = p(0) = P [X = 0](7)
This represents the outcome of 1 trial. Now, imagine this experiment iscarried out n times. Let Yn,p denote the outcome of this repeatedexperiment. We now have R = 0, 1, 2, .., n and
p(i) = P[Yn,p = i ] =
(n
i
)· pi · (1− p)n−i
E[Yn,p] = n · pVar(Yn,p) = n · p · (1− p)
(8)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 7 / 193
The Poisson Random Variable
Recall our r.v. X with p(i) = e−λλi
i! . Straigthforward computation showsthat E[X ] = λ = Var(X )
X can in fact approximate Yn,p if λ = n · p, for large n.
As an example, consider an item that is produced by a certain machine.The probability that it will be defective is p = 0.1. The probability that, ina sample of 10, at most one item will be defective, is
P[Y10,0.1 ≤ 1] =
(10
0
)0.10 · 0.910 +
(10
1
)0.11 · 0.99
= 0.7361 ≈ 0.7358 = e−1 + e−1
= e−(10·0.1) · (10 · 0.1)0
1+ e−(10·0.1) · (10 · 0.1)1
1= P[X = 0] + P[X = 1]
(9)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 8 / 193
Other Discrete Probability Distributions
Geometric Random Variable: Number of trials required until success :P[X = n] = (1− p)n−1 · p
E[X ] = p · 1 + (1− p) · E[1 + X ]⇒ E[X ] = 1p
E[X 2] = p · 12 + (1− p) · E[(X + 1)2]⇒ E[X 2] = 2−pp2
Var(X ) = E[X 2]− (E[X ])2 = 1−pp2
Negative Binomial Random Variable: n trials needed for r successes :P[Xr ,p = n] =
(n−1r−1
)pr · (1− p)n−r
E[Xr ,p] = rp
E[X 2r ,p] = r
p ·(
r+1p − 1
)Var(Xr ,p) = E[X 2
r ,p]− (E[Xr ,p])2 = r ·(1−p)p2
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 9 / 193
Other Discrete Probability Distributions
Geometric Random Variable: Number of trials required until success :P[X = n] = (1− p)n−1 · p
E[X ] = p · 1 + (1− p) · E[1 + X ]⇒ E[X ] = 1p
E[X 2] = p · 12 + (1− p) · E[(X + 1)2]⇒ E[X 2] = 2−pp2
Var(X ) = E[X 2]− (E[X ])2 = 1−pp2
Negative Binomial Random Variable: n trials needed for r successes :P[Xr ,p = n] =
(n−1r−1
)pr · (1− p)n−r
E[Xr ,p] = rp
E[X 2r ,p] = r
p ·(
r+1p − 1
)Var(Xr ,p) = E[X 2
r ,p]− (E[Xr ,p])2 = r ·(1−p)p2
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 9 / 193
Other Discrete Probability Distributions
Geometric Random Variable: Number of trials required until success :P[X = n] = (1− p)n−1 · p
E[X ] = p · 1 + (1− p) · E[1 + X ]⇒ E[X ] = 1p
E[X 2] = p · 12 + (1− p) · E[(X + 1)2]⇒ E[X 2] = 2−pp2
Var(X ) = E[X 2]− (E[X ])2 = 1−pp2
Negative Binomial Random Variable: n trials needed for r successes :P[Xr ,p = n] =
(n−1r−1
)pr · (1− p)n−r
E[Xr ,p] = rp
E[X 2r ,p] = r
p ·(
r+1p − 1
)Var(Xr ,p) = E[X 2
r ,p]− (E[Xr ,p])2 = r ·(1−p)p2
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 9 / 193
Other Discrete Probability Distributions
Geometric Random Variable: Number of trials required until success :P[X = n] = (1− p)n−1 · p
E[X ] = p · 1 + (1− p) · E[1 + X ]⇒ E[X ] = 1p
E[X 2] = p · 12 + (1− p) · E[(X + 1)2]⇒ E[X 2] = 2−pp2
Var(X ) = E[X 2]− (E[X ])2 = 1−pp2
Negative Binomial Random Variable: n trials needed for r successes :P[Xr ,p = n] =
(n−1r−1
)pr · (1− p)n−r
E[Xr ,p] = rp
E[X 2r ,p] = r
p ·(
r+1p − 1
)Var(Xr ,p) = E[X 2
r ,p]− (E[Xr ,p])2 = r ·(1−p)p2
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 9 / 193
Other Discrete Probability Distributions
Geometric Random Variable: Number of trials required until success :P[X = n] = (1− p)n−1 · p
E[X ] = p · 1 + (1− p) · E[1 + X ]⇒ E[X ] = 1p
E[X 2] = p · 12 + (1− p) · E[(X + 1)2]⇒ E[X 2] = 2−pp2
Var(X ) = E[X 2]− (E[X ])2 = 1−pp2
Negative Binomial Random Variable: n trials needed for r successes :P[Xr ,p = n] =
(n−1r−1
)pr · (1− p)n−r
E[Xr ,p] = rp
E[X 2r ,p] = r
p ·(
r+1p − 1
)
Var(Xr ,p) = E[X 2r ,p]− (E[Xr ,p])2 = r ·(1−p)
p2
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 9 / 193
Other Discrete Probability Distributions
Geometric Random Variable: Number of trials required until success :P[X = n] = (1− p)n−1 · p
E[X ] = p · 1 + (1− p) · E[1 + X ]⇒ E[X ] = 1p
E[X 2] = p · 12 + (1− p) · E[(X + 1)2]⇒ E[X 2] = 2−pp2
Var(X ) = E[X 2]− (E[X ])2 = 1−pp2
Negative Binomial Random Variable: n trials needed for r successes :P[Xr ,p = n] =
(n−1r−1
)pr · (1− p)n−r
E[Xr ,p] = rp
E[X 2r ,p] = r
p ·(
r+1p − 1
)Var(Xr ,p) = E[X 2
r ,p]− (E[Xr ,p])2 = r ·(1−p)p2
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 9 / 193
Other Discrete Probability Distributions
Hypergeometric Random Variable: sample of size n chosen randomlyfrom population m white balls and N −m black balls. Let Zn,N,m
denote the number of white balls selected :
P[Zn,N,m = i ] =(m
i )(N−mn−i )
(Nn)
for i = 0, 1, .., n
E[Zn,N,m] = n·mN
E[Z 2n,N,m] = n·m
N ·(
(n−1)·(m−1)N−1 + 1
)Var(Zn,N,m) = E[Z 2
n,N,m]− (E[Zn,N,m])2 = n · mN ·(1− m
N
)·(
1− n−1N−1
)Zipf (Pareto) Distribution: Assume α > 0 and that
P[X = k] =
(∑∞l=1
1l1+α
)−1
k1+α . Then E[X ] =?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 10 / 193
Other Discrete Probability Distributions
Hypergeometric Random Variable: sample of size n chosen randomlyfrom population m white balls and N −m black balls. Let Zn,N,m
denote the number of white balls selected :
P[Zn,N,m = i ] =(m
i )(N−mn−i )
(Nn)
for i = 0, 1, .., n
E[Zn,N,m] = n·mN
E[Z 2n,N,m] = n·m
N ·(
(n−1)·(m−1)N−1 + 1
)Var(Zn,N,m) = E[Z 2
n,N,m]− (E[Zn,N,m])2 = n · mN ·(1− m
N
)·(
1− n−1N−1
)
Zipf (Pareto) Distribution: Assume α > 0 and that
P[X = k] =
(∑∞l=1
1l1+α
)−1
k1+α . Then E[X ] =?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 10 / 193
Other Discrete Probability Distributions
Hypergeometric Random Variable: sample of size n chosen randomlyfrom population m white balls and N −m black balls. Let Zn,N,m
denote the number of white balls selected :
P[Zn,N,m = i ] =(m
i )(N−mn−i )
(Nn)
for i = 0, 1, .., n
E[Zn,N,m] = n·mN
E[Z 2n,N,m] = n·m
N ·(
(n−1)·(m−1)N−1 + 1
)Var(Zn,N,m) = E[Z 2
n,N,m]− (E[Zn,N,m])2 = n · mN ·(1− m
N
)·(
1− n−1N−1
)Zipf (Pareto) Distribution: Assume α > 0 and that
P[X = k] =
(∑∞l=1
1l1+α
)−1
k1+α . Then E[X ] =?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 10 / 193
Expectation is Linear
For a collection of random variables Xini=1, we haveE [∑n
i=1 Xi ] =∑n
i=1 E[Xi ].
As an example, if n die are rolled, what is the expected value for the totalnumber of dots?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 11 / 193
Expectation is Linear
For a collection of random variables Xini=1, we haveE [∑n
i=1 Xi ] =∑n
i=1 E[Xi ].
As an example, if n die are rolled, what is the expected value for the totalnumber of dots?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 11 / 193
CDF
Recall that for an r.v. X , we define F (b) := P[X ≤ b]. It follows from thecontinuity property of probability measures that
If a < b then F (a) < F (b)
limb→∞ F (b) = 1
limb→−∞ F (b) = 0
limbn↓b F (bn) = F (b)
As an example,
P[X < b] = P[
limn→∞
X ≤ b − 1
n
]= lim
n→∞P[
X ≤ b − 1
n
]= lim
n→∞F
(b − 1
n
) (10)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 12 / 193
CDF
Recall that for an r.v. X , we define F (b) := P[X ≤ b]. It follows from thecontinuity property of probability measures that
If a < b then F (a) < F (b)
limb→∞ F (b) = 1
limb→−∞ F (b) = 0
limbn↓b F (bn) = F (b)
As an example,
P[X < b] = P[
limn→∞
X ≤ b − 1
n
]= lim
n→∞P[
X ≤ b − 1
n
]= lim
n→∞F
(b − 1
n
) (10)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 12 / 193
Continuous Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .
If we take a random variable that can take on any real value, then we callit a continuous random variable if attached to it there is a non-negativefunction f : R→ R+ where
P[X ∈ B] =
∫B
f (x)dx (11)
for any measurable set B. We can consider all practical sets to bemeasurable. Of course, it follows from our definition that we must haveP[X ∈ R] =
∫∞−∞ f (x)dx = 1. For intervals [a, b] and (−∞, b], we have
P[a ≤ X ≤ b] =
∫ b
af (x)dx
P[X ≤ b] =
∫ b
−∞f (x)dx
(12)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 13 / 193
Continuous Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .If we take a random variable that can take on any real value, then we callit a continuous random variable if attached to it there is a non-negativefunction f : R→ R+ where
P[X ∈ B] =
∫B
f (x)dx (11)
for any measurable set B. We can consider all practical sets to bemeasurable. Of course, it follows from our definition that we must haveP[X ∈ R] =
∫∞−∞ f (x)dx = 1. For intervals [a, b] and (−∞, b], we have
P[a ≤ X ≤ b] =
∫ b
af (x)dx
P[X ≤ b] =
∫ b
−∞f (x)dx
(12)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 13 / 193
Continuous Random Variables
Recall our notion of a sample space S = Ω, and attach to it a probabilitymeasure P and the set of possible events F .If we take a random variable that can take on any real value, then we callit a continuous random variable if attached to it there is a non-negativefunction f : R→ R+ where
P[X ∈ B] =
∫B
f (x)dx (11)
for any measurable set B. We can consider all practical sets to bemeasurable. Of course, it follows from our definition that we must haveP[X ∈ R] =
∫∞−∞ f (x)dx = 1. For intervals [a, b] and (−∞, b], we have
P[a ≤ X ≤ b] =
∫ b
af (x)dx
P[X ≤ b] =
∫ b
−∞f (x)dx
(12)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 13 / 193
Example
Suppose that X is c.r.v. with probability density function
f (x) =
C · (2− x), if 0 < x < 2
0, otherwise .(13)
What is C ? Compute P[X > 1]
Answer: Since f is a density, we require
1 =
∫ ∞−∞
f (x)dx = C ·∫ 2
0(2− x)dx = C · 2⇒ C =
1
2
P[X > 1] =
∫ ∞1
f (x)dx =1
2·∫ 2
1(2− x)dx
=1
2·[
(2(2)− 1
2(22))− (2(1)− 1
2(12))
]=
1
4
(14)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 14 / 193
Example
Suppose that X is c.r.v. with probability density function
f (x) =
C · (2− x), if 0 < x < 2
0, otherwise .(13)
What is C ? Compute P[X > 1]Answer: Since f is a density, we require
1 =
∫ ∞−∞
f (x)dx = C ·∫ 2
0(2− x)dx = C · 2
⇒ C =1
2
P[X > 1] =
∫ ∞1
f (x)dx =1
2·∫ 2
1(2− x)dx
=1
2·[
(2(2)− 1
2(22))− (2(1)− 1
2(12))
]=
1
4
(14)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 14 / 193
Example
Suppose that X is c.r.v. with probability density function
f (x) =
C · (2− x), if 0 < x < 2
0, otherwise .(13)
What is C ? Compute P[X > 1]Answer: Since f is a density, we require
1 =
∫ ∞−∞
f (x)dx = C ·∫ 2
0(2− x)dx = C · 2⇒ C =
1
2
P[X > 1] =
∫ ∞1
f (x)dx =1
2·∫ 2
1(2− x)dx
=1
2·[
(2(2)− 1
2(22))− (2(1)− 1
2(12))
]=
1
4
(14)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 14 / 193
Example
Suppose that X is c.r.v. with probability density function
f (x) =
C · (2− x), if 0 < x < 2
0, otherwise .(13)
What is C ? Compute P[X > 1]Answer: Since f is a density, we require
1 =
∫ ∞−∞
f (x)dx = C ·∫ 2
0(2− x)dx = C · 2⇒ C =
1
2
P[X > 1] =
∫ ∞1
f (x)dx =1
2·∫ 2
1(2− x)dx
=1
2·[
(2(2)− 1
2(22))− (2(1)− 1
2(12))
]=
1
4
(14)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 14 / 193
From c.d.f to p.d.f
Recall
F (y) = P[X ≤ y ] (15)
and so
P[
x − ∆x
2≤ X ≤ x +
∆x
2
]=
∫ x+ ∆x2
x−∆x2
f (u)du ≈ f (x)∆x . (16)
It follows that if Y = 2X , then
FY (x) := P[Y ≤ x ] = P[2X ≤ x ] = P[X ≤ 0.5x ]
= FX (0.5x)
fY (x) :=d
dxFY (x) =
d
dxFX (0.5x) = 0.5fX (0.5x).
(17)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 15 / 193
Expectations
Define E[X ] =∫∞−∞ x · f (x)dx for any c.r.v. X with p.d.f. X . For any real
valued function g , we can define the expectation of the transformed c.r.vY = g(X ) as
E[g(X )] =
∫ ∞−∞
g(x) · f (x)dx . (18)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 16 / 193
Expectations
The following Lemma is useful in many computations. The proof can beseen via an interchange (Fubini) of the order of integration.
Lemma
For any nonnegative random variable Y , we have E[Y ] =∫∞
0 P[Y > y ]dy .
Proof. ∫ ∞0
P[Y > y ]dy =
∫ ∞0
(∫ ∞y
fY (x)dx
)dy
=
∫ ∞0
(∫ x
0fY (x)dy
)dx
=
∫ ∞0
x · fY (x)dx = E[X ].
(19)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 17 / 193
Uniform Random Variables
If we define
f (x) :=
1
B − A, if A < x < B
0, otherwise(20)
then the distribution F is comptuted to be
F (x) =
0, if x ≤ A
x − A
B − A, if A < x < B
1, if x ≥ B
(21)
and so
E [X ] =
∫ ∞−∞
xf (x)dx =
∫ B
Ax · 1
B − A=
B + A
2
Var(X ) = E[X 2]−(
B + A
2
)2
=(B − A)2
12.
(22)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 18 / 193
Optimizing Under Uniform Uncertainty
A 1−unit long city block has a mailbox somewhere at a point U that isuniformly distributed over (0, 1). Determine the expected length of thesection of the block that contains an envelope store E , where 0 ≤ E ≤ 1.Also, for what value of E is this expected value maximized ?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 19 / 193
Optimizing Under Uniform Uncertainty
Take LE (U) as the length of the sub-block that contains the envelopestore E and note that
LE (U) =
1− U, if U < E
U, if U > E .(23)
It follows that
E [LE (U)] :=
∫ 1
0LE (u) · f (u)du =
∫ 1
0LE (u)du
=
∫ E
0(1− u)du +
∫ 1
Eudu =
1
2+ E (1− E )
max0≤E≤1
E [LE (U)] =1
2+
1
2
(1− 1
2
)=
3
4
(24)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 20 / 193
Normal Random Variables
We say that X is a Normal Random Variable with parameters µ, σ2 if
fX (x) =1√2πσ
e−(x−µ)2
2σ2 for −∞ < x <∞ (25)
Furthermore, we say that Z is a Standard Normal Random Variable if it isNormal with parameters µ = 0, σ = 1. Consider
z =x − µσ∫ ∞
−∞fX (x)dx =
∫ ∞−∞
1√2πσ
e−(x−µ)2
2σ2 dx
=
∫ ∞−∞
1√2π
e−(z)2
2 dz
=
∫ ∞−∞
fZ (z)dz
(26)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 21 / 193
Normal Random Variables
We user the notation X ∼ N(µ, σ2) and Z ∼ N(0, 1). By ourtransformation above, it can be seen that if Z ∼ N(0, 1), thenX = µ+ σ · Z ∼ N(µ, σ2). We can see this via
Φ(z) := FZ (z) = P[Z ≤ z ]
= P[
X − µσ
≤ x − µσ
]= P[X ≤ x ] =: FX (x)
1
σfZ
(x − µσ
)=
d
dxFZ (z)
=d
dxFX (x) = fX (x)
fZ (z) =1√2π
e−(z)2
2
(27)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 22 / 193
Normal Random Variables
We user the notation X ∼ N(µ, σ2) and Z ∼ N(0, 1). By ourtransformation above, it can be seen that if Z ∼ N(0, 1), thenX = µ+ σ · Z ∼ N(µ, σ2). We can see this via
Φ(z) := FZ (z) = P[Z ≤ z ]
= P[
X − µσ
≤ x − µσ
]= P[X ≤ x ] =: FX (x)
1
σfZ
(x − µσ
)=
d
dxFZ (z)
=d
dxFX (x) = fX (x)
fZ (z) =1√2π
e−(z)2
2
(27)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 22 / 193
Normal Random Variables
We user the notation X ∼ N(µ, σ2) and Z ∼ N(0, 1). By ourtransformation above, it can be seen that if Z ∼ N(0, 1), thenX = µ+ σ · Z ∼ N(µ, σ2). We can see this via
Φ(z) := FZ (z) = P[Z ≤ z ]
= P[
X − µσ
≤ x − µσ
]= P[X ≤ x ] =: FX (x)
1
σfZ
(x − µσ
)=
d
dxFZ (z)
=d
dxFX (x) = fX (x)
fZ (z) =1√2π
e−(z)2
2
(27)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 22 / 193
DeMoivre-Laplace Limit Theorem
Theorem
If Sn is the number of successes that occur when n independent trials,each resulting in a success with probability p, are performed, then for anya < b, we have
limn→∞
P
[a ≤ Sn − n · p√
n · p · (1− p)≤ b
]= P[a ≤ Z ≤ b]
= Φ(b)− Φ(a)
(28)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 23 / 193
Exponential Random Variables
For any λ > 0, define the random variable X with p.d.f.
f (x) =
λe−λx , if x ≥ 0
0, if x < 0.(29)
Then, if x ≥ 0, we have F (x) =∫ x
0 λe−λydy = 1− e−λx .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 24 / 193
Exponential Random Variables
Furthermore,using integration by parts, we also compute
E[X n] =
∫ ∞0
xnλe−λxdx
= 0 +
∫ ∞0
nxn−1e−λxdx
=n
λ
∫ ∞0
xn−1λe−λx
=n
λE[X n−1]
E[X 1] =1
λE[1] =
1
λ
E[X 2] =2
λE[X ] =
2
λ2
Var(X ) =1
λ2
(30)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 25 / 193
Exponential Random Variables
Furthermore,using integration by parts, we also compute
E[X n] =
∫ ∞0
xnλe−λxdx
= 0 +
∫ ∞0
nxn−1e−λxdx
=n
λ
∫ ∞0
xn−1λe−λx
=n
λE[X n−1]
E[X 1] =1
λE[1] =
1
λ
E[X 2] =2
λE[X ] =
2
λ2
Var(X ) =1
λ2
(30)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 25 / 193
Exponential Random Variables
Notice that X is a Memoryless random variable:
P[X > s + t | X > t] =P[X > s + t]
P[X > t]
=1− (1− e−λ(t+s))
1− (1− e−λt)
=e−λ(t+s)
e−λt
= e−λs = P[X > s]
(31)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 26 / 193
Hazard Rates
Define the Hazard Rate for any crv X as
λ(t) :=f (t)
1− F (t)=
ddt F (t)
1− F (t)(32)
It follows, by solution of the above differential equation, that anequivalent definition is
F (t) = 1− e−∫ t
0 λ(s)ds
F (0) = 1(33)
Crv’s may be defined via their Hazard rates. For example:
Rayleigh distributed random variable: λ(t) = a + btExponential distributed random variable: λ(t) = λ.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 27 / 193
Gamma Distribution
Consider a crv X with p.d.f.
f (x) =
λe−λx(λx)α−1
Γ(α), if x ≥ 0
0, if x < 0.
(34)
where the Gamma function Γ(α) is defined as∫∞
0 e−yyα−1dy . Integrationby parts shows that for α > 1
Γ(α) = (α− 1)Γ(α− 1)
Γ(n) = (n − 1)! for n ∈ N(35)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 28 / 193
Gamma Distribution
Consider a crv X with p.d.f.
f (x) =
λe−λx(λx)α−1
Γ(α), if x ≥ 0
0, if x < 0.
(34)
where the Gamma function Γ(α) is defined as∫∞
0 e−yyα−1dy . Integrationby parts shows that for α > 1
Γ(α) = (α− 1)Γ(α− 1)
Γ(n) = (n − 1)! for n ∈ N(35)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 28 / 193
Gamma Distribution
Assume that you are waiting for n events to occur, and that N(t) is thenumber of events that occur up to time t, which is assumed to be Poisson(λt) distributed. Let Tn denote the time at which the nth event occurs.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 29 / 193
Gamma Distribution
It follows that
F (t) = P[Tn ≤ t]
= P[N(t) ≥ n]
=∞∑j=n
P[N(t) = j ]
=∞∑j=n
e−λt(λt)j
j!
f (t) = F ′(t) =∞∑j=n
e−λt · j · (λt)j−1 · λj!
−∞∑j=n
λe−λt(λt)j
j!
=λe−λt(λt)n−1
(n − 1)!
(36)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 30 / 193
Weibull Distribution
Consider a crv X with p.d.f.
f (x) =
0, if x ≤ ν
β
α
(x − να
)β−1
e−( x−να )
β
, if x > ν(37)
The corresponding cdf is
F (x) =
0, if x ≤ ν
1− e−( x−να )
β
, if x > ν(38)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 31 / 193
Cauchy Distribution
Consider a crv θ ∼ U[−π
2 ,π2
], and a transformation to another random
variable X = tan (θ). Then
F (x) = P[X ≤ x ]
= P[tan (θ) ≤ x ]
= P[θ ≤ tan−1 (x)]
=1
π
(tan−1 (x)−
(−π
2
))=
1
2+
1
πtan−1 (x)
⇒ f (x) = F ′(x) =1
π
1
1 + x2
(39)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 32 / 193
Beta Distribution
Consider a crv X with p.d.f.
f (x) =
xa−1(1− x)b−1
B(a, b), if 0 < x < 1
0, otherwise.
(40)
where B(a, b) is defined as∫ 1
0 xa−1(1− x)b−1dx . It can be shown that
B(a, b) =Γ(a)Γ(b)
Γ(a + b)
E[X ] =a
a + b
Var(X ) =ab
(a + b)2(a + b + 1)
(41)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 33 / 193
Beta Distribution
Consider a crv X with p.d.f.
f (x) =
xa−1(1− x)b−1
B(a, b), if 0 < x < 1
0, otherwise.
(40)
where B(a, b) is defined as∫ 1
0 xa−1(1− x)b−1dx . It can be shown that
B(a, b) =Γ(a)Γ(b)
Γ(a + b)
E[X ] =a
a + b
Var(X ) =ab
(a + b)2(a + b + 1)
(41)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 33 / 193
Distribution of a Function of a Random Variable
Theorem
Let X be a continuous random variable with p.d.f. fX . Suppose that g(x)is a strictly monotonic, differentiable function of x. Then the randomvariable Y = g(X ) has a p.d.f. defined by
fY (y) =
fX[g−1(y)
]| d
dyg−1(y) |, if, for some x, y = g(x)
0, if, for all x, y 6= g(x)
(42)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 34 / 193
Distribution of a Function of a Random Variable
Proof.
Suppose that y = g(x) for some x . Then, with Y = g(X ),
fY (y) =d
dyFY (y)
=d
dyP[g(X ) ≤ y ]
=d
dyP[X ≤ g−1(y)]
=d
dyFX
(g−1(y)
)= fX
(g−1(y)
) d
dyg−1(y)
(43)
If y 6= g(x) for any x , then F (y) is either 0 or 1 ⇒ fY (y) = 0.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 35 / 193
Examples (Can we use the previous theorem?)
Let X have p.d.f fX , and define Y = X 2, Z =| X |. Then
fY (y) =d
dyFY (y)
=d
dyP[X 2 ≤ y ]
=d
dyP[−√y ≤ X ≤ √y ]
=d
dy(FX (
√y)− FX (−√y))
=1
2√
y[fX (√
y) + fX (−√y)]
fZ (z) =d
dzP[−z ≤ X ≤ z ]
= fX (z) + fX (−z)
(44)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 36 / 193
Mixed Distributions
There are random variables X such that, for a countable number of pointsxi ,
P[X = xi ] = pi > 0∞∑i=0
pi ≤ 1.(45)
As an example, consider X such that
P[X = 0] =1
2
fX (x) =1
2e−x for x > 0.
(46)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 37 / 193
HW
Please work through
2.2
2.3, 2.4, 2.5
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 38 / 193
Raw and Central Moments
Given a random variable X , we define the kth Raw Moment via,
µ′k := E[X k ] (47)
and the kth Central Moment via,
µk := E[(X − µ′1)k ]. (48)
HW: Compute the first 5 raw and central moments for a random variableX with fX (x) = λe−λx for all x ≥ 0.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 39 / 193
Variance, Skewness, and Kurtosis
Given a random variable X , we define
µ′1 := µ = Mean
µ2 = E[(X − µ′1)2] := σ2 = Varianceσ
µ:= Coefficient of Variation
γ1 :=µ3
σ3= Skewness
γ2 :=µ4
σ4= Kurtosis
(49)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 40 / 193
Excess Loss
Given a random variable X , a value u ∈ R, and a value d ∈ R withP[X > d ] > 0 we define
Y P := X − d = Excess Loss Variable
eX (d) := E[Y P | Y P > 0] = Mean Excess Loss Function
ekX (d) := E[(Y P)k | Y P > 0]
Y L := (X − d)+ = Left Censored and Shifted Variable
Y := X ∧ u = Limited Loss Variable
E[Y ] := Limited Expected Value.
(50)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 41 / 193
HW
Please work through
3.2
3.3, 3.4, 3.5
3.6, 3.7, 3.11
3.14, 3.15, 3.16
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 42 / 193
Percentiles and Moment Generating Functions
Define (any number) πp as the 100pth percentile if it satisfies
F (π−p ) ≤ p ≤ F (πp). (51)
The 50th percentile is also called the median. For all z ∈ R, we define
MX (z) := E[ezX ] = Moment Generating Function
PX (z) := E[zX ] = Probability Generating Function(52)
for which these expectations exist. Note that MX (z) = PX (ez).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 43 / 193
Moment Generating Functions
Theorem
Let Sk = X1 + ...+ Xk , where the X are independent. Then, provided thefollowing exist:
MSk (z) = Πkj=1MXj
(z)
PSk (z) = Πkj=1PXj
(z)(53)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 44 / 193
Example
Consider a loss model where at time t > 0, the liability with a randomlyselected individual is distributed as
P[Xt ≥ x ] = e−xt for x ≥ 0, t > 0
P[X0 = 0] = 1.(54)
It follows that
MXt (z) = E[ezXt ] =
∫ ∞0
ezx fXt (x)dx =
∫ ∞0
ezx · − d
dx
(e−
xt
)dx
=
∫ ∞0
ezx1
te−
xt dx =
1
t
∫ ∞0
e−( 1t−z)xdx =
1
t
11t − z
=1
1− zt.
(55)
HW: For Sk(t) :=∑k
j=1 X jt , compute MSk (t)(z).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 45 / 193
Example
Consider a loss model where at time t > 0, the liability with a randomlyselected individual is distributed as
P[Xt ≥ x ] = e−xt for x ≥ 0, t > 0
P[X0 = 0] = 1.(54)
It follows that
MXt (z) = E[ezXt ] =
∫ ∞0
ezx fXt (x)dx =
∫ ∞0
ezx · − d
dx
(e−
xt
)dx
=
∫ ∞0
ezx1
te−
xt dx =
1
t
∫ ∞0
e−( 1t−z)xdx =
1
t
11t − z
=1
1− zt.
(55)
HW: For Sk(t) :=∑k
j=1 X jt , compute MSk (t)(z).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 45 / 193
Example
Consider a loss model where at time t > 0, the liability with a randomlyselected individual is distributed as
P[Xt ≥ x ] = e−xt for x ≥ 0, t > 0
P[X0 = 0] = 1.(54)
It follows that
MXt (z) = E[ezXt ] =
∫ ∞0
ezx fXt (x)dx =
∫ ∞0
ezx · − d
dx
(e−
xt
)dx
=
∫ ∞0
ezx1
te−
xt dx =
1
t
∫ ∞0
e−( 1t−z)xdx =
1
t
11t − z
=1
1− zt.
(55)
HW: For Sk(t) :=∑k
j=1 X jt , compute MSk (t)(z).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 45 / 193
Example
For a standard Normal random variable Z ∼ N(0, 1), we have
MZ (t) = E[etZ ] =
∫ ∞−∞
etz fZ (z)dx =
∫ ∞−∞
etz1√2π
e−12z2
dz
=
∫ ∞−∞
e12t2 1√
2πe−
12
(z−t)2dz = e
12t2.
(56)
Therefore, for Y = µ+ σZ ∼ N(µ, σ2), it follows that
MY (t) = E[et(µ+σZ)] = etµE[e(tσZ)] = etµe12
(tσ)2= etµ+ 1
2t2σ2
. (57)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 46 / 193
Example
Consider an i.i.d sequence Xini=1 such that all the Xi ∼ X , where for aparameter θ > 0,
fX (x) =1
θe−
xθ (58)
Define Sn := X1+...+Xnn , and so
E[Sn] = θ
Var[Sn] =θ2
n.
(59)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 47 / 193
Example
It follows that
MSn(t) = E[etSn ] =
[E[e
tnX]]n
=[MX
( t
n
)]n=
[1
1− tθn
]n
∴ limn→∞
MSn(t) = lim
n→∞
[1
1− tθn
]n= etθ
∴ PSn⇒ PS := N(θ, 0).
(60)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 48 / 193
Example
Furthermore, define
Zn :=Sn − E[Sn]√
Var[Sn]=√
n · Sn − θθ
. (61)
It follows that
MZn(t) = E[etZn ] = E[et√n· Sn−θ
θ
]= e−t
√nE[et√n· Sn
θ
]= e−t
√nMSn
(t√
n
θ
)= e−t
√n
[1
1− t√n
]n→ e
12t2
∴ PZn ⇒ PZ := N(0, 1).
(62)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 49 / 193
HW
Please work through
3.21
3.23
3.24
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 50 / 193
Heavy Tails
Consider a pair of random variables (X1,X2) with E[X1] = E[X2], andrecall the notation
Si (x) := P[Xi > x ] = 1− Fi (x). (63)
It can be shown via L’Hopital’s rule that
limx→∞
S1(x)
S2(x)= lim
x→∞
f1(x)
f2(x). (64)
If limx→∞S1(x)S2(x) =∞, then we say that X1 has a heavier tail than X2,
based on limiting tail behavior.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 51 / 193
Heavy Tails
Individually, for a random variable X if
d
du(eX (u)) > 0 (65)
we say that X has a heavy tail based on mean excess loss function.
Often, the standard for comparison is the exponential distribution todetermine a heavy tail: X is heavier tailed than any exponentialdistribution if for all λ > 0,
limx→∞
eλxP[X > x ] =∞. (66)
HW: Please work through
3.25
3.29
3.30
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 52 / 193
Parametric Distributions
A Parametric Distribution is determined by a set of parameters. Asubset of set of parametric distributions, known as Scale Distributions, isdetermined by scale parameters. In this case, the r.v. X that is linked tothe parametric distribution FX (x) can be transformed by scalarmultiplication into Y = cX , and the resulting distribution FY (y) = FX
( yc
)is in the same class of distributions.
For example, if X ∼ exp(λ) then FX = 1− e−λX . Then,
FY (y) = 1− e−λcy and so Y ∼ exp(λc ).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 53 / 193
Discrete Mixture Distributions
A random variable X is k-point mixture of (X1, ..,Xk) if
FX (x) = a1FX1(x) + ...+ akFXk(x)
1 =k∑
n=1
an.(67)
For example, let Θ be another random variable, with
ai = P[Θ = i ]
1 = P[X = Xi | Θ = i ]
FXi(x) = P[Xi ≤ x ] = P[X ≤ x | Θ = i ].
(68)
It follows that
P[X ≤ x ] =k∑
i=1
P[X ≤ x | Θ = i ]P[Θ = i ] =k∑
i=1
aiFXi(x). (69)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 54 / 193
Discrete Mixture Distributions
A random variable X is variable-component mixture of (X1, ..,XN) if Ncan be any positive integer, with
FX (x) = a1FX1(x) + ...+ aNFXN(x)
1 =N∑
n=1
an.(70)
For example, let Θ, Θ be random variables, with
p = P[Θ = 0] = 1− P[Θ = 1]
pi = P[Θ = i ].(71)
It follows that
P[X ≤ x ] = pFX0(x) + (1− p)FX1(x)
P[X ≤ x ] =N∑i=1
aiFXi(x).
(72)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 55 / 193
Discrete Mixture Distributions
A Data-Dependent Distribution is at least as complex as the datathat produced it. As the number of data points increases, so does thenumber of parameters.
An Empirical Model is associate with a discrete model that assignsequal probability 1
n to each of the n data points.
HW: 4.1, 4.2, 4.5, 4.7, 4.10.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 56 / 193
Raising To A Power
Let X be a random variable, and Y = X γ its transformation. Then wedefine
Y =
Transformed : γ > 0Inverse : γ = −1Inverse Transformed : γ < 0, γ 6= −1.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 57 / 193
Gamma and Transformed Gamma Distribution
Recall our definitions
Γ(α) =
∫ ∞0
tα−1e−tdt
Γ(α; x) =
∫ x0 tα−1e−tdt
Γ(α).
(73)
If X ∼ Gamma(α), the distribution we use is Γ(x ;α) to transform X intoY = X γ .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 58 / 193
Transformed Exponential Distribution
If SX (x) = e−x , then assuming γ > 0
Y =
Transformed : SY (y) = e−y
γ
Inverse Transformed : SY (y) = e−y−γ
Weibull : SY (y) = e−( θy
)γ
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 59 / 193
Mixing
Define X | Λ as a conditional random variable, andSX |Λ(x | λ) = 1− FX |Λ(x | λ) as its conditional survival (probability)function.
Theorem
Define fΛ(λ) as the density of our parameter Λ, and fX (x) as theunconditional density for our random variable X . Then
fX (x) =
∫R
fX |Λ(x | λ)fΛ(λ)dλ
FX (x) =
∫R
FX |Λ(x | λ)fΛ(λ)dλ
µ′k = E[X k ] = E[E[X k | Λ]
]µ2 = Var[X ] = E [Var[X | Λ]] + Var [E[X | Λ]] .
(74)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 60 / 193
Mixing
Theorem
If X | Λ ∼ exp(Λ) and Λ ∼ Gamma, then X ∼ Pareto.
Proof.
By our previous theorem,
fX (x) =
∫R
fX |Λ(x | λ)fΛ(λ)dλ
=
∫ ∞λ=0
λe−λxλα−1e−θλθα
Γ(α)dλ
=θα
Γ(α)
∫ ∞λ=0
λαe−λ(x+θ)dλ =θα
Γ(α)
Γ(α + 1)
(x + θ)α+1.
(75)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 61 / 193
Mixing
Theorem
If X | Λ ∼ N (Θ, σ) and Θ ∼ N(µ, α), then X ∼ N(µ, σ + α).
Proof.
By our previous theorem, and by definining σ = σασ+α and µ = σµ+αx
σ+α ,
fX (x) =
∫R
fX |Θ(x | θ)fΘ(θ)dθ =
∫ ∞θ=−∞
1√2πσ
e−(x−θ)2
2σ1√2πα
e−(θ−µ)2
2α dθ
=1√
2π(σ + α)e− (x−µ)2
2(σ+α)
∫ ∞θ=−∞
1√2πσ
e−(θ−µ)2
2σ dθ
=1√
2π(σ + α)e− (x−µ)2
2(σ+α) .
(76)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 62 / 193
Frailty
A Fraility random variable Λ > 0 is one where the hazard rate is, for aknown function a(x),
µX |Λ = Λ · a(x). (77)
It follows that for A(x) :=∫ x
0 a(s)ds,
SX |Λ(x | λ) = e−λ·A(x)
SX (x) = E[e−Λ·A(x)
]= MΛ[−A(x)]
(78)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 63 / 193
Frailty
For example, consider an exponential mixture (i.e. A(x) = x) and that ourparameter is uniformly distributed: Λ ∼ U[0, L].Then
SX (x) =
∫ ∞0
SX |Λ(x | λ) · fΛ(λ)dλ
=
∫ L
0e−λx · 1
Ldλ
=1
Lx· (1− e−Lx).
(79)
HW: 5.7.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 64 / 193
Splicing
A k-component spliced distribution has density
fX (x) =k∑
n=1
ak fk(x)1(cn−1,cn) (80)
HW: Can you splice together an exponential and a Pareto distribution?Give an example if you can!
HW:
5.1,5.3,5.6,5.9,5.12
5.13,5.14,5.20
Limiting Distributions: 5.10
5.21,5.22,5.26
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 65 / 193
Linear Exponential Family
A random variable X has a linear exponential density if
f (x ; θ) =p(x)er(θ)x
q(θ). (81)
For example, if X ∼ N(θ, σ), then
f (x ; θ) =1√
2πσ2e−
12σ2 (x−θ)2
=
[1√
2πσ2e−
x2
2σ2
] [eθσ2 x]
eθ2
2σ2
.
(82)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 66 / 193
Linear Exponential Family
∴ ln [f (x ; θ)] = ln [p(x)] + r(θ)x − ln q(θ)
⇒ ∂f
∂θ=[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)
⇒ 0 =∂
∂θ[1] =
∂
∂θ
[ ∫ x=∞
x=−∞f (x ; θ)dx
]=
∫ x=∞
x=−∞
∂f
∂θdx =
∫ x=∞
x=−∞
[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)dx
= r ′(θ)E[X ]− q′(θ)
q(θ).
⇒ E[X ] =q′(θ)
r ′(θ)q(θ)=: µ(θ)
⇒ Var[X ] =µ′(θ)
r ′(θ)(Proof: Left as an exercise!).
(83)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 67 / 193
Linear Exponential Family
∴ ln [f (x ; θ)] = ln [p(x)] + r(θ)x − ln q(θ)
⇒ ∂f
∂θ=[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)
⇒ 0 =∂
∂θ[1] =
∂
∂θ
[ ∫ x=∞
x=−∞f (x ; θ)dx
]=
∫ x=∞
x=−∞
∂f
∂θdx =
∫ x=∞
x=−∞
[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)dx
= r ′(θ)E[X ]− q′(θ)
q(θ).
⇒ E[X ] =q′(θ)
r ′(θ)q(θ)=: µ(θ)
⇒ Var[X ] =µ′(θ)
r ′(θ)(Proof: Left as an exercise!).
(83)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 67 / 193
Linear Exponential Family
∴ ln [f (x ; θ)] = ln [p(x)] + r(θ)x − ln q(θ)
⇒ ∂f
∂θ=[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)
⇒ 0 =∂
∂θ[1] =
∂
∂θ
[ ∫ x=∞
x=−∞f (x ; θ)dx
]=
∫ x=∞
x=−∞
∂f
∂θdx =
∫ x=∞
x=−∞
[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)dx
= r ′(θ)E[X ]− q′(θ)
q(θ).
⇒ E[X ] =q′(θ)
r ′(θ)q(θ)=: µ(θ)
⇒ Var[X ] =µ′(θ)
r ′(θ)(Proof: Left as an exercise!).
(83)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 67 / 193
Linear Exponential Family
∴ ln [f (x ; θ)] = ln [p(x)] + r(θ)x − ln q(θ)
⇒ ∂f
∂θ=[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)
⇒ 0 =∂
∂θ[1] =
∂
∂θ
[ ∫ x=∞
x=−∞f (x ; θ)dx
]=
∫ x=∞
x=−∞
∂f
∂θdx =
∫ x=∞
x=−∞
[r ′(θ)x − q′(θ)
q(θ)
]f (x ; θ)dx
= r ′(θ)E[X ]− q′(θ)
q(θ).
⇒ E[X ] =q′(θ)
r ′(θ)q(θ)=: µ(θ)
⇒ Var[X ] =µ′(θ)
r ′(θ)(Proof: Left as an exercise!).
(83)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 67 / 193
Counting Distributions
Let N be a discrete random variable with
P[N = k] := pk ≥ 0. (84)
It follows that
P(z) = PN(z) = E[zN ] :=∞∑k=0
pkzk
P(m)(z) =dm
dzmE[zN ] = E
[ dm
dzm(zN)
]= E
[N(N − 1) · .. · (N −m + 1)zN−m
]=∞∑
k=m
k(k − 1) · ... · (k −m + 1)zk−mpk .
∴ P(m)(0) = m!pm.
(85)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 68 / 193
Poisson Distributions
For example, consider the Poisson distribution with pk = e−λλk
k! for allk ≥ 0. We can compute
P(z) = PN(z) = E[zN ] :=∞∑k=0
e−λλk
k!zk
=∞∑k=0
e−λ(λz)k
k!
= e−λ∞∑k=0
(λz)k
k!= e−λeλz
= eλ(z−1).
(86)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 69 / 193
Poisson Distributions
Theorem
If Njnj=1 is a sequence of i.i.d. Poisson random variables with respective
parameters λjnj=1, then
N := N1 + ...+ Nn ∼ Poisson(λ1 + ...+ λn). (87)
Proof.
PN(z) = Πnj=1PNj
(z) = Πnj=1eλj (z−1) = e(
∑nj=1 λj)(z−1). (88)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 70 / 193
Poisson Distributions
Theorem
If Njnj=1 is a sequence of i.i.d. Poisson random variables with respective
parameters λjnj=1, then
N := N1 + ...+ Nn ∼ Poisson(λ1 + ...+ λn). (87)
Proof.
PN(z) = Πnj=1PNj
(z) = Πnj=1eλj (z−1) = e(
∑nj=1 λj)(z−1). (88)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 70 / 193
Poisson Distributions : Parameter Uncertainty
Theorem
Assume Λ has a density fΛ(λ), and P[N = k | Λ = λ] = e−λλk
k! .Then
pk = P[N = k] =
∫ ∞0
e−λλk
k!fΛ(λ)dλ =
1
k!E[Λke−Λ]. (89)
For example, if fΛ(λ) = λα−1e−λθ
θαΓ(α) , then
pk =1
k!
∫ ∞0
e−λλkλα−1e−
λθ
θαΓ(α)dλ =
1
k!
1
θαΓ(α)
∫ ∞0
e−λ(1+ 1θ
)λk+α−1dλ
=Γ(k + α)
k!Γ(α)
θk
(1 + θ)k+α.
(90)
HW: Compute pk if Λ ∼ Exp(λ).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 71 / 193
Poisson Distributions : Parameter Uncertainty
Theorem
Assume Λ has a density fΛ(λ), and P[N = k | Λ = λ] = e−λλk
k! .Then
pk = P[N = k] =
∫ ∞0
e−λλk
k!fΛ(λ)dλ =
1
k!E[Λke−Λ]. (89)
For example, if fΛ(λ) = λα−1e−λθ
θαΓ(α) , then
pk =1
k!
∫ ∞0
e−λλkλα−1e−
λθ
θαΓ(α)dλ =
1
k!
1
θαΓ(α)
∫ ∞0
e−λ(1+ 1θ
)λk+α−1dλ
=Γ(k + α)
k!Γ(α)
θk
(1 + θ)k+α.
(90)HW: Compute pk if Λ ∼ Exp(λ).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 71 / 193
Binomial Distributions Revisited
Define X and N such that for a sequence Xjnj=1 with each of theXj ∼ X and
p = P[X = 1] = 1− P[X = 0]
N = N1 + ...+ Nn.(91)
It follows that
PX (z) = (1− p)z0 + pz1 = 1− p + pz
⇒ PN(z) = [1− p + pz ]n
⇒ P[N = k] =m!
(m − k)!k!pk(1− p)m−k .
(92)
We can also show that
E[N] = np
Var[N] = np(1− p).(93)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 72 / 193
Binomial Distributions Revisited
Define X and N such that for a sequence Xjnj=1 with each of theXj ∼ X and
p = P[X = 1] = 1− P[X = 0]
N = N1 + ...+ Nn.(91)
It follows that
PX (z) = (1− p)z0 + pz1 = 1− p + pz
⇒ PN(z) = [1− p + pz ]n
⇒ P[N = k] =m!
(m − k)!k!pk(1− p)m−k .
(92)
We can also show that
E[N] = np
Var[N] = np(1− p).(93)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 72 / 193
Binomial Distributions Revisited
Define X and N such that for a sequence Xjnj=1 with each of theXj ∼ X and
p = P[X = 1] = 1− P[X = 0]
N = N1 + ...+ Nn.(91)
It follows that
PX (z) = (1− p)z0 + pz1 = 1− p + pz
⇒ PN(z) = [1− p + pz ]n
⇒ P[N = k] =m!
(m − k)!k!pk(1− p)m−k .
(92)
We can also show that
E[N] = np
Var[N] = np(1− p).(93)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 72 / 193
(a, b, 0) class
A discrete random variable N is a member of the (a, b, 0) class if , forpk = P[N = k] and all k ∈ 1, 2, 3...,
pk
pk−1= a +
b
k. (94)
It follows that
kpk = akpk−1 + bpk−1 = a(k − 1)pk−1 + (a + b)pk−1
⇒∞∑k=1
kpk = a∞∑k=1
(k − 1)pk−1 + (a + b)∞∑k=1
pk−1
∴ E[N] = aE[N] + (a + b) · 1 =a + b
1− a.
(95)
HW: Ex. 6.1, 6.2.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 73 / 193
(a, b, 0) class
A discrete random variable N is a member of the (a, b, 0) class if , forpk = P[N = k] and all k ∈ 1, 2, 3...,
pk
pk−1= a +
b
k. (94)
It follows that
kpk = akpk−1 + bpk−1 = a(k − 1)pk−1 + (a + b)pk−1
⇒∞∑k=1
kpk = a∞∑k=1
(k − 1)pk−1 + (a + b)∞∑k=1
pk−1
∴ E[N] = aE[N] + (a + b) · 1 =a + b
1− a.
(95)
HW: Ex. 6.1, 6.2.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 73 / 193
(a, b, 0) class
A discrete random variable N is a member of the (a, b, 0) class if , forpk = P[N = k] and all k ∈ 1, 2, 3...,
pk
pk−1= a +
b
k. (94)
It follows that
kpk = akpk−1 + bpk−1 = a(k − 1)pk−1 + (a + b)pk−1
⇒∞∑k=1
kpk = a∞∑k=1
(k − 1)pk−1 + (a + b)∞∑k=1
pk−1
∴ E[N] = aE[N] + (a + b) · 1 =a + b
1− a.
(95)
HW: Ex. 6.1, 6.2.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 73 / 193
(a, b, 0) class: Fitting to data
By definition, if we take succesive probabilities (frequencies) of observeddata, and we know that the data is drawn from an (a, b, 0)-classdistribution, then it must be that for yk := k pk
pk−1,
yk = ak + b. (96)
For example, if we observe f total data points, and fk are the total numberof observed points with property k , then
pk =fkf⇒ yk = k
fkfk−1
. (97)
To fit, plot yk against k, and fit for a and b.HW: Read on the (a, b, 1) class, read pages 90− 94, and complete Ex.6.4, 6.6.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 74 / 193
(a, b, 0) class: Fitting to data
According to results by Sundt and Jewell, and extended in Hess,Liewald, and Schmidt, the three distributions in this class are
Distribution a b p0 pk
Poisson 0 λ e−λ e−λ λk
k!
Negative Binomial 1− p (1− p)(r − 1) pr Γ(r+k)k!Γ(r) pr (1− p)k
Binomial − p1−p (n + 1) p
1−p (1− p)n(nk
)pk(1− p)n−k
We can use this table to match empirically observed data with anappropriate distribution.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 75 / 193
(a, b, 0) class: Finan - Example 28.1
Let N be a member of C (a, b, 0) satisfying the recursive probabilities
kpk
pk−1=
3
4k + 3. (98)
Identify the distribution N.
Answer:As a > 0, b > 0, it follows that N is a negative binomial distribution.Furthermore,
3
4= a = 1− p
3 = b = (1− p)(r − 1) =3
4(r − 1)
⇒ (p, r) = (0.25, 5).
(99)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 76 / 193
(a, b, 0) class: Finan - Example 28.1
Let N be a member of C (a, b, 0) satisfying the recursive probabilities
kpk
pk−1=
3
4k + 3. (98)
Identify the distribution N.
Answer:As a > 0, b > 0, it follows that N is a negative binomial distribution.Furthermore,
3
4= a = 1− p
3 = b = (1− p)(r − 1) =3
4(r − 1)
⇒ (p, r) = (0.25, 5).
(99)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 76 / 193
(a, b, 0) class: Finan - Example 28.2
The distribution of accidents for 84 randomly selected policies is as follows:
Number of Accidents Number of Policies
0 32
1 26
2 12
3 7
4 4
5 2
6 1
Identify the frequency model that best represents these data
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 77 / 193
(a, b, 0) class: Finan - Example 28.2
The distribution of accidents for 84 randomly selected policies, and theiradjusted frequency ratios, are as follows:
Number of Accidents Number of Policies k pkpk−1
1 26 0.8125
2 12 0.9231
3 7 1.75
4 4 2.2857
5 2 2.5
6 1 3
Best fit line is 0.462969k + 0.25816, hence a negative binomialdistribution. Check out Wolfram Alpha’s computation by entering ”linearregression (1, 0.8125), (2, 0.9231), (3, 1.75), (4, 2.2857), (5, 2.5), (6, 3)” inthe search bar.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 78 / 193
(a, b, 0) class: Finan - Example 28.2
The distribution of accidents for 84 randomly selected policies, and theiradjusted frequency ratios, are as follows:
Number of Accidents Number of Policies k pkpk−1
1 26 0.8125
2 12 0.9231
3 7 1.75
4 4 2.2857
5 2 2.5
6 1 3
Best fit line is 0.462969k + 0.25816, hence a negative binomialdistribution. Check out Wolfram Alpha’s computation by entering ”linearregression (1, 0.8125), (2, 0.9231), (3, 1.75), (4, 2.2857), (5, 2.5), (6, 3)” inthe search bar.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 78 / 193
(a, b, 0) class: Interpretation
Consider now the distribution of accidents for randomly selected policies isas follows:
Number of Accidents Number of Policies
0 400
1 200
2 50
3 8
4 1
Identify the frequency model that best represents these data
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 79 / 193
(a, b, 0) class: Interpretation
The distribution of accidents for 84 randomly selected policies, and theiradjusted frequency ratios, are as follows:
Number of Accidents Number of Policies k pkpk−1
1 200 1 · 200400
2 50 2 · 50200
3 8 3 · 850
4 1 4 · 18
Best fit line is −0.002k + 0.5. Is this a Binomial or Poisson distribution?Check out Wolfram Alpha’s computation by entering ”linear regression(1, 1 · 200
400 ), (2, 2 · 50200 ), (3, 3 · 8
50 ), (4, 4 · 18 )” in the search bar.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 80 / 193
(a, b, 0) class: Interpretation
The distribution of accidents for 84 randomly selected policies, and theiradjusted frequency ratios, are as follows:
Number of Accidents Number of Policies k pkpk−1
1 200 1 · 200400
2 50 2 · 50200
3 8 3 · 850
4 1 4 · 18
Best fit line is −0.002k + 0.5. Is this a Binomial or Poisson distribution?Check out Wolfram Alpha’s computation by entering ”linear regression(1, 1 · 200
400 ), (2, 2 · 50200 ), (3, 3 · 8
50 ), (4, 4 · 18 )” in the search bar.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 80 / 193
Multiple Exposures to Risk
A bank, insurance company, hedge fund, or pension sponsor among otherentities is charged with navitgating the risks borne by their endeavor. Inorder to mitigate and price such risks, one must first set out to identifytheir various forms.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 81 / 193
Multiple Exposures to Risk
A few known variants include
Market Risk.
Interest Rate Risk.
FX Risk.
Liquidity Risk.
Credit Risk.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 82 / 193
Market Risk
Exposure to capital markets.
Can affect both assets and liabilities.
Examples include equities and other financial instruments.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 83 / 193
Interest Rate Risk
Exposure to changes in short rate.
Underestimating volatility of short rate evolution (shocks to economy.)
Exposure to changes in yield curve and term structure.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 84 / 193
FX Risk
Inability to predict or affect foreign government domestic policy(prime rate) and foreign policy (devaluation.)
Can lead to inability to properly hedge financial instruments held inforeign countries.
Foreign interest rate fluctuation can lead to possible shortfalls thatbanks may wish to immunize their portfolio against. Currencyfluctuation also leads to daily operational risk, and should factor intoplanning.
For example, should a band tour Europe or America this summer ?Click here for an insightful article by Neil Shah in the Wall StreetJournal TM, with comment from the manager of a very prominentrock band
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 85 / 193
Liquidity Risk
Exposure to depressed prices in return for expedited sale (liquidation)of assets.
Also affects strategies that hedge long term contracts with short termassets.
Banks are also prone to this in the form of bank runs.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 86 / 193
Credit Risk
This also falls under Default Risk, where one is exposed to(counterparty) default on contracts that exchange cash flows.
Correlated with Market Risk, although different opinions on how so.
For life insurance companies, a life (x) can be seen as a default risk ifthat person outlives the present value of their retirement annuity.Also known as Longevity (Mortality) Risk.
This happens, for example, when pension sponsors overestimate themortality of the insured population. Reinsurance can be bought, butsuch reinsurers are also susceptible to the previously defined risks,depending on their asset and liability portfolio.As an example, please refer to the paper by Tsai, Tzeng, and Wang onHedging Longevity Risk When Interest Rates Are Uncertain
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 87 / 193
Credit Risk
This also falls under Default Risk, where one is exposed to(counterparty) default on contracts that exchange cash flows.
Correlated with Market Risk, although different opinions on how so.
For life insurance companies, a life (x) can be seen as a default risk ifthat person outlives the present value of their retirement annuity.Also known as Longevity (Mortality) Risk.
This happens, for example, when pension sponsors overestimate themortality of the insured population. Reinsurance can be bought, butsuch reinsurers are also susceptible to the previously defined risks,depending on their asset and liability portfolio.As an example, please refer to the paper by Tsai, Tzeng, and Wang onHedging Longevity Risk When Interest Rates Are Uncertain
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 87 / 193
Credit Risk
Complicated to measure and perhaps even define:
Risk = P[Default]?Risk = E[Loss | Default]?or...Risk = E[Loss] = E[Loss | Default]× P[Default]?However, does E[Loss | Default] relate in some fashion to P[Default] ?In an adverse economy mired in recession, perhaps so. In fact,Frye inhis paper ”Collateral Damage” (2000) argues that the same factorsthat increase default rates can also decrease the value of loan collateral.Click here for a review of Structural Models for Credit Riskpublishedby the SOA.
Clearly, regulators, counterparties, clearinghouses, and shareholders wouldlike a precise definition of the risk involved with an entity’s business. Onepossible definition is the capital a business entity should hold. So what todo?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 88 / 193
Credit Risk
Complicated to measure and perhaps even define:
Risk = P[Default]?Risk = E[Loss | Default]?
or...Risk = E[Loss] = E[Loss | Default]× P[Default]?However, does E[Loss | Default] relate in some fashion to P[Default] ?In an adverse economy mired in recession, perhaps so. In fact,Frye inhis paper ”Collateral Damage” (2000) argues that the same factorsthat increase default rates can also decrease the value of loan collateral.Click here for a review of Structural Models for Credit Riskpublishedby the SOA.
Clearly, regulators, counterparties, clearinghouses, and shareholders wouldlike a precise definition of the risk involved with an entity’s business. Onepossible definition is the capital a business entity should hold. So what todo?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 88 / 193
Credit Risk
Complicated to measure and perhaps even define:
Risk = P[Default]?Risk = E[Loss | Default]?or...Risk = E[Loss] = E[Loss | Default]× P[Default]?
However, does E[Loss | Default] relate in some fashion to P[Default] ?In an adverse economy mired in recession, perhaps so. In fact,Frye inhis paper ”Collateral Damage” (2000) argues that the same factorsthat increase default rates can also decrease the value of loan collateral.Click here for a review of Structural Models for Credit Riskpublishedby the SOA.
Clearly, regulators, counterparties, clearinghouses, and shareholders wouldlike a precise definition of the risk involved with an entity’s business. Onepossible definition is the capital a business entity should hold. So what todo?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 88 / 193
Credit Risk
Complicated to measure and perhaps even define:
Risk = P[Default]?Risk = E[Loss | Default]?or...Risk = E[Loss] = E[Loss | Default]× P[Default]?However, does E[Loss | Default] relate in some fashion to P[Default] ?
In an adverse economy mired in recession, perhaps so. In fact,Frye inhis paper ”Collateral Damage” (2000) argues that the same factorsthat increase default rates can also decrease the value of loan collateral.Click here for a review of Structural Models for Credit Riskpublishedby the SOA.
Clearly, regulators, counterparties, clearinghouses, and shareholders wouldlike a precise definition of the risk involved with an entity’s business. Onepossible definition is the capital a business entity should hold. So what todo?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 88 / 193
Credit Risk
Complicated to measure and perhaps even define:
Risk = P[Default]?Risk = E[Loss | Default]?or...Risk = E[Loss] = E[Loss | Default]× P[Default]?However, does E[Loss | Default] relate in some fashion to P[Default] ?In an adverse economy mired in recession, perhaps so. In fact,Frye inhis paper ”Collateral Damage” (2000) argues that the same factorsthat increase default rates can also decrease the value of loan collateral.Click here for a review of Structural Models for Credit Riskpublishedby the SOA.
Clearly, regulators, counterparties, clearinghouses, and shareholders wouldlike a precise definition of the risk involved with an entity’s business. Onepossible definition is the capital a business entity should hold. So what todo?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 88 / 193
Credit Risk
Complicated to measure and perhaps even define:
Risk = P[Default]?Risk = E[Loss | Default]?or...Risk = E[Loss] = E[Loss | Default]× P[Default]?However, does E[Loss | Default] relate in some fashion to P[Default] ?In an adverse economy mired in recession, perhaps so. In fact,Frye inhis paper ”Collateral Damage” (2000) argues that the same factorsthat increase default rates can also decrease the value of loan collateral.Click here for a review of Structural Models for Credit Riskpublishedby the SOA.
Clearly, regulators, counterparties, clearinghouses, and shareholders wouldlike a precise definition of the risk involved with an entity’s business. Onepossible definition is the capital a business entity should hold. So what todo?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 88 / 193
Stochastic Actuarial Reserving (Ko and Shiu)
Fix a probability space(
Ω,F ,P)
and a standard Brownian motion W
that lives on this space.
Consider now a term contact with term T and let α denote themanagement charges factor along with β representing the policyholder’sparticipation factor.
Furthermore, assume mean and standard deviation parameters (µ, σ)respectively and the corresponding Geometric Brownian Mutual Fund Asset
St = S0eµt+σWt . (100)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 89 / 193
Stochastic Actuarial Reserving (Ko and Shiu)
Using this as the model of the asset returns upon which premiums areinvested, the policyholder wishes to purchase a contract that pays amaturity benefit credited at a rate of return which is the greater of
the customer’s risk discount rate r , where r < µ or
the participation rate of the stock index returns of S .
Symbolically, for a current premium P invested in the , the contractpayout value at maturity is
V (T ) = (1− α)P max
erT , 1 + β
(ST
S0− 1)
. (101)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 90 / 193
Stochastic Actuarial Reserving (Ko and Shiu)
Assume that the policyholder is able to fully participate in the returns fromthe fund (i.e. β = 1.)Then
V (T ) = (1− α)PST
S0+ (1− α)P max
erT −
(ST
S0
), 0
:= V1(T ) + V2(T ).
(102)
Here, V1(T ) is the net premium, or payoff, for investing in the index fundand V2(T ) is the guaranteed option payoff if the index fundunder-performs relative to the risk discount rate r .
How does one reserve to meet the obligations of V2(T ).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 91 / 193
Stochastic Actuarial Reserving (Ko and Shiu)
One can see that the probability of a payout, that V2(T ) 6= 0 is for large T
P[V2(T ) 6= 0] = P[rT > µT + σWT ] = Φ( r − µ
σ
√T)≈ 0. (103)
Since it is a low probability event that we have to prepare for a payoutV2(T ) and since we can directly replicate the payoff V1(T ) by initially
purchasing (1−α)PS0
units of the index fund, an actuary may be tempted tonot reserve for the uncertain portion of the guarantee, V2(T ), if thecontract has a relatively long term T .
Is this a wise decision?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 92 / 193
Stochastic Actuarial Reserving (Ko and Shiu)
One can see that the probability of a payout, that V2(T ) 6= 0 is for large T
P[V2(T ) 6= 0] = P[rT > µT + σWT ] = Φ( r − µ
σ
√T)≈ 0. (103)
Since it is a low probability event that we have to prepare for a payoutV2(T ) and since we can directly replicate the payoff V1(T ) by initially
purchasing (1−α)PS0
units of the index fund, an actuary may be tempted tonot reserve for the uncertain portion of the guarantee, V2(T ), if thecontract has a relatively long term T .
Is this a wise decision?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 92 / 193
Stochastic Actuarial Reserving (Ko and Shiu)
One can see that the probability of a payout, that V2(T ) 6= 0 is for large T
P[V2(T ) 6= 0] = P[rT > µT + σWT ] = Φ( r − µ
σ
√T)≈ 0. (103)
Since it is a low probability event that we have to prepare for a payoutV2(T ) and since we can directly replicate the payoff V1(T ) by initially
purchasing (1−α)PS0
units of the index fund, an actuary may be tempted tonot reserve for the uncertain portion of the guarantee, V2(T ), if thecontract has a relatively long term T .
Is this a wise decision?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 92 / 193
Return Measures
A bank, insurance company, hedge fund, or pension sponsor can have theirefficiency measured by how they return profits over a time horizon.For example, if the value of a portfolio at time t is denoted by X (t), thenreturns are measured at times t0, t1, t2, ..., tn and the Loss in period(tk−1, tk) is
L[X ](tk) := − (X (tk)− X (tk−1)) (104)
and the Loss Distribution is
r [X ](tk) := − lnX (tk)
X (tk−1)(105)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 93 / 193
Return Over Benchmark
Comparing to a benchmark rB(t) for returns, a manager can be rated via
RoB :=1
tn − t0
n∑k=1
(r [X ](tk)− rB(tk)) (106)
as well as the deviation away from the benchmark returns:
Dev :=
√√√√ 1
tn − t0
n∑k=1
(r [X ](tk)− rB(tk))2 (107)
Notice that RoB is a linear measure.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 94 / 193
Sharpe Ratio
Recall from (financial) economics that there is a measure of return vs risk,known as the Sharpe Ratio. This measures the reward gained versus therisk ventured to return the gains won.
If α is the realized return on anasset over a risk free rate r , then for a standard deviation σ of the returnson the asset we have the informational measure
Sharpe Ratio :=α− r
σ(108)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 95 / 193
Sharpe Ratio
Recall from (financial) economics that there is a measure of return vs risk,known as the Sharpe Ratio. This measures the reward gained versus therisk ventured to return the gains won. If α is the realized return on anasset over a risk free rate r , then for a standard deviation σ of the returnson the asset we have the informational measure
Sharpe Ratio :=α− r
σ(108)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 95 / 193
Sharpe Ratio
Going back to our periodic returns, we can enact a similar measure ofperformance via
S :=1
tn−t0
∑nk=1 (r [X ](tk)− rB(tk))√
1tn−t0
∑nk=1 (r [X ](tk)− rB(tk))2
(109)
The numerical values of S can be used to rank manager performance, butas with any models there are criticisms. First off is the usefulness of thedenominator as a risk measure.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 96 / 193
Risk Measures
Define a probability space (Ω,F ,P). Define the set A ⊂ L0(Ω,F ,P) ofpossible risks, where
X ,Y ∈ A ⇒ X + Y ∈ A.
X ∈ A ⇒ αX ∈ A for all α ∈ [0, 1].
X ∈ A ⇒ X + a ∈ A for all a ∈ R.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 97 / 193
Risk Measures
A risk measure ρ : A → R ∪ ∞ may be required to satisfy some of thefollowing properties:
Law Invariance: ∀X ,Y ∈ A, x ∈ R,P[X ≤ x ] = P[Y ≤ x ]⇒ ρ[X ] = ρ[Y ] .
Monotonicity: ∀X ,Y ∈ A, P[X ≤ Y ] = 1⇒ ρ[X ] ≤ ρ[Y ].
Subadditivity: ∀X ,Y ∈ A, ρ[X + Y ] ≤ ρ[X ] + ρ[Y ].
Positive Homogeneity: ∀X ∈ A, α ≥ 0, ρ[αX ] = αρ[X ].
Translation Invariance: ∀X ∈ A, a ∈ R, ρ[X + a] = ρ[X ] + a.
Convexity: ∀X ,Y ∈ A, λ ∈ [0, 1],ρ[λX + (1− λ)Y ] ≤ λρ[X ] + (1− λ)ρ[Y ]
A risk measure that satisfies Monotonicity, Subadditivity, PositiveHomogeneity, and Translation Invariance is said to be Coherent.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 98 / 193
Risk Measures
A risk measure ρ : A → R ∪ ∞ may be required to satisfy some of thefollowing properties:
Law Invariance: ∀X ,Y ∈ A, x ∈ R,P[X ≤ x ] = P[Y ≤ x ]⇒ ρ[X ] = ρ[Y ] .
Monotonicity: ∀X ,Y ∈ A, P[X ≤ Y ] = 1⇒ ρ[X ] ≤ ρ[Y ].
Subadditivity: ∀X ,Y ∈ A, ρ[X + Y ] ≤ ρ[X ] + ρ[Y ].
Positive Homogeneity: ∀X ∈ A, α ≥ 0, ρ[αX ] = αρ[X ].
Translation Invariance: ∀X ∈ A, a ∈ R, ρ[X + a] = ρ[X ] + a.
Convexity: ∀X ,Y ∈ A, λ ∈ [0, 1],ρ[λX + (1− λ)Y ] ≤ λρ[X ] + (1− λ)ρ[Y ]
A risk measure that satisfies Monotonicity, Subadditivity, PositiveHomogeneity, and Translation Invariance is said to be Coherent.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 98 / 193
Risk Measures
Note that Translation Invariance ⇒ ρ[X − ρ[X ]
]= 0.
In the insurance field, ρ[X ] is seen as the solvency capital requiredby a regulating agency for a company exposed to loss (risk) X .
The capital is a requirement imposed on the firm so that the surplusof assets over liabilities for the company is at least ρ[X ].
ρ[X ] should be chosen to ensure that X > ρ[X ] is a sufficiently rareevent.
Regulators should, and do, take into consideration the costs incurredby a company that is required to hold excessive solvency capital.Hence, a balance must be sought between solvency and onerouscapital requirements. The choice of ρ should be decided upon withthese thoughts in mind.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 99 / 193
Premium Principles vs Risk Measures
Companies or individuals who wish to exchange a future liability foran up-front payment seek out willing partners. They find them in theinsurance industry.
However, insurance firms pricing the risk of being liable for a futurepayment is coupled must also meet the demands of regulators toremain solvent.
One of the most common methods is the Expected Value PremiumPrinciple (EPP) which returns ρ[X ] = E[X ].
However, such a principle doesn’t take into account possible (large)deviations from the average loss. Some corrections:
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 100 / 193
Premium Principles vs Risk Measures
Companies or individuals who wish to exchange a future liability foran up-front payment seek out willing partners. They find them in theinsurance industry.
However, insurance firms pricing the risk of being liable for a futurepayment is coupled must also meet the demands of regulators toremain solvent.
One of the most common methods is the Expected Value PremiumPrinciple (EPP) which returns ρ[X ] = E[X ].
However, such a principle doesn’t take into account possible (large)deviations from the average loss. Some corrections:
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 100 / 193
Premium Principles vs Risk Measures
EPP with loading: ρ[X ] = (1 + α)E[X ] for some α > 0.
Standard Deviation Premium Principle: ρ[X ] = E[X ] + α√
V [X ]for some α ≥ 0.
Variance Premium Principle: ρ[X ] = E[X ] + αV [X ] for someα ≥ 0.
Dutch Premium Principle: ρ[X ] = E[X ] + θE[(X − αE [X ])+] forsome α ≥ 1, θ ∈ [0, 1].
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 101 / 193
Value at Risk
Let X ∈ L0(Ω,F ,P) and α ∈ [0, 1]. Define
VaRα[X ] := Qα[X ] = inf x | P[X ≤ x ] ≥ α (110)
The VaR is the amount of cash a company requires to set the probabilityof the company becoming insolvent equal to α. The Basel Committee onBanking has in the past set VaR as a measure to set capital requirements.VaR is also used in the Portfolio Premium Pricing Principle in actuarialsettings.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 102 / 193
Example: Value at Risk
Consider now the distribution of Losses for accident claims from randomlyselected policies is as follows:
Claim Amount L Number of Policies Probability
0 400 0.607
100 200 0.303
200 50 0.076
500 8 0.012
1000 1 0.002
Compute the Value at Risk for the 90%, 95% and 99% levels.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 103 / 193
Example: Value at Risk
Consider now the distribution of Losses for accident claims from randomlyselected policies is as follows:
Claim Amount L P[L ≤ x ]
1000 1
500 0.998
200 0.986
100 0.910
0 0.607
It follows that
(VaR0.90[L],VaR0.95[L],VaR0.99[L]) = (100, 200, 500). (111)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 104 / 193
Why Choose VaR as a Risk Measure?
Consider that a firm is allowed to penalize for the cost of capital by apre-determined factor 0 < ε ≤ 1.
Define the (penalized) Expected Shortfall via
ESε(ρ[X ]) := E[(X − ρ[X ])+
]+ ερ[X ] (112)
Question: What capital u = ρ[X ] level solves
minu∈R+
ESε(u)? (113)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 105 / 193
Why Choose VaR as a Risk Measure?
Consider that a firm is allowed to penalize for the cost of capital by apre-determined factor 0 < ε ≤ 1.
Define the (penalized) Expected Shortfall via
ESε(ρ[X ]) := E[(X − ρ[X ])+
]+ ερ[X ] (112)
Question: What capital u = ρ[X ] level solves
minu∈R+
ESε(u)? (113)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 105 / 193
Why Choose VaR as a Risk Measure?
Consider that a firm is allowed to penalize for the cost of capital by apre-determined factor 0 < ε ≤ 1.
Define the (penalized) Expected Shortfall via
ESε(ρ[X ]) := E[(X − ρ[X ])+
]+ ερ[X ] (112)
Question: What capital u = ρ[X ] level solves
minu∈R+
ESε(u)? (113)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 105 / 193
VaR as Minimal Capital Requirement (DGK [2003])
Theorem
The smallest capital ρ[X ] that solves
minu∈R+
ESε(u) (114)
is given by ρ[X ] = Q1−ε[X ].
There is a nice geometric proof for this, given by the authors.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 106 / 193
VaR as a Minimal Capital Requirement (DGK [2003])
Proof.
(Sketch:) Assume that X has a smooth distribution. The critical point isobtained via
0 =d
du
(E[(X − u)+] + εu
)=
d
du
(∫ ∞0
P[(X − u)+ > x ]dx + εu)
=d
du
(∫ ∞u
P[X > x ]dx + εu)
=d
du
(∫ ∞u
(1− FX (x))dx + εu)
= FX (u)− 1 + ε
→ ρ[X ] = u = F−1X (1− ε) = Q1−ε[X ].
(115)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 107 / 193
VaR as a Minimal Capital Requirement (DGK [2003])
Proof.
(Sketch:) Assume that X has a smooth distribution. The critical point isobtained via
0 =d
du
(E[(X − u)+] + εu
)=
d
du
(∫ ∞0
P[(X − u)+ > x ]dx + εu)
=d
du
(∫ ∞u
P[X > x ]dx + εu)
=d
du
(∫ ∞u
(1− FX (x))dx + εu)
= FX (u)− 1 + ε
→ ρ[X ] = u = F−1X (1− ε) = Q1−ε[X ].
(115)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 107 / 193
Value at Risk
Note that VaRα[X − VaRα[X ]
]= 0
This implies that the risk of holding an asset with uncertain outcomeX can be neutralized by holding a reserve amount VaRα[X ].
But VaR doesn’t account for how big the loss for the company in theevent of insolvency.
And regarding diversification..See Example 3.14 in our textbook andexample on page 19 of Hardy Study Note
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 108 / 193
Value at Risk and Solvency II
From the European Commission FAQ on Solvency II:
”The aim of a solvency regime is to ensure the financial soundness ofinsurance undertakings, and in particular to ensure that they cansurvive difficult periods. This is to protect policyholders (consumers,businesses) and the stability of the financial system as a whole. ”
” ..insurers must have available resources sufficient to cover both aMinimum Capital Requirement (MCR) and a Solvency CapitalRequirement (SCR). The SCR is based on a Value-at-Risk measurecalibrated to a 99.5% confidence level over a 1-year time horizon.The SCR covers all risks that an insurer faces (e.g. insurance, market,credit and operational risk) and will take full account of any riskmitigation techniques applied by the insurer (e.g. reinsurance andsecuritisation).. ”
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 109 / 193
Some other (related) Risk Measures
The quantile measure VaR doesn’t give us information on the tail behaviorof X beyond Qp[X ]. One way to account for this is to define
TVaRp[X ] :=1
1− p
∫ 1
pQq[X ]dq. (116)
Furthermore, in the tail region, we can also determine how much the lossX exceeds Qp[X ], on average, by
CTEp[X ] := E[X | X > Qp[X ]]. (117)
Finally, regulators are interested in the average shortfall
ESFp[X ] := E[(X − Qp[X ])+]. (118)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 110 / 193
Equivalence of Risk Measures [DGK (2006)]
Theorem
TVaRp[X ] = Qp[X ] +1
1− pESFp[X ] (119)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 111 / 193
Equivalence of Risk Measures [DGK (2006)]
Proof.
Under the transformation x → q, where x = F−1X (q) = Qq[X ] and
(q, dq) =(
FX (x), fX (x)dx)
ESFp[X ] =
∫ ∞−∞
(x − Qp[X ])+fX (x)dx
=
∫ 1
0(Qq[X ]− Qp[X ])+dq
=
∫ 1
pQq[X ]dq − (1− p)Qp[X ]
⇒ TVaRp[X ] =1
1− p
∫ 1
pQq[X ]dq =
ESFp[X ]
1− p+ Qp[X ].
(120)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 112 / 193
Equivalence of Risk Measures [DGK (2006)]
Proof.
Under the transformation x → q, where x = F−1X (q) = Qq[X ] and
(q, dq) =(
FX (x), fX (x)dx)
ESFp[X ] =
∫ ∞−∞
(x − Qp[X ])+fX (x)dx
=
∫ 1
0(Qq[X ]− Qp[X ])+dq
=
∫ 1
pQq[X ]dq − (1− p)Qp[X ]
⇒ TVaRp[X ] =1
1− p
∫ 1
pQq[X ]dq =
ESFp[X ]
1− p+ Qp[X ].
(120)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 112 / 193
Equivalence of Risk Measures [DGK (2006)]
Proof.
Under the transformation x → q, where x = F−1X (q) = Qq[X ] and
(q, dq) =(
FX (x), fX (x)dx)
ESFp[X ] =
∫ ∞−∞
(x − Qp[X ])+fX (x)dx
=
∫ 1
0(Qq[X ]− Qp[X ])+dq
=
∫ 1
pQq[X ]dq − (1− p)Qp[X ]
⇒ TVaRp[X ] =1
1− p
∫ 1
pQq[X ]dq =
ESFp[X ]
1− p+ Qp[X ].
(120)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 112 / 193
Equivalence of Risk Measures: Some Points
Note that if E[X ] <∞ and if FX is continuous, then
limp→0
TVaRp[X ] = E[X ]
TVaRp[X ] = CTEp[X ](121)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 113 / 193
Normally Distributed Risk
For an X ∼ N(µ, σ2), it can be shown for 0 < p < 1 that
Qp[X ] = µ+ σΦ−1(p)
CTEp[X ] = E[X | X > Qp[X ]] =1
1− p
∫ ∞Qp
1√2πσ
ye−(y−µ)2
2σ dy
=1
1− p
∫ ∞Qp−µσ
µ+ σz√2πσ
e−z2
2 dz
=µ
1− p
∫ ∞Φ−1(p)
e−z2
2
√2πσ
dz +σ
1− p
∫ ∞Φ−1(p)
ze−z2
2
√2πσ
dz
= µ+σ
1− pφ(Φ−1(p))
E[(X − ρ[X ])+] = σφ(ρ[X ]− µ
σ
)− (ρ[X ]− µ)
[1− Φ
(ρ[X ]− µσ
)]ESFp[X ] = σφ
(Φ−1(p)
)− σ(1− p)Φ−1(p)
(122)Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 114 / 193
Lognormally Distributed Risk
For a ln X ∼ N(µ, σ2), it can be shown for 0 < p < 1 that
Qp[X ] = eµ+σΦ−1(p)
E[(X − ρ[X ])+] = eµ+σ2
2 Φ
(σ +
µ− ln ρ[X ]
σ
)− ρ[X ]Φ
(µ− ln ρ[X ]
σ
)
ESFp[X ] = eµ+σ2
2 Φ
(σ − Φ−1(p)
)− eµ+σΦ−1(p)(1− p)
CTEp[X ] = eµ+σ2
2
Φ
(σ − Φ−1(p)
)1− p
(123)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 115 / 193
Pareto Distributed Risk
If SX (x) =(
θθ+x
)α, then
Qp[X ] = θ[(1− p)−1α − 1]
CTEp[X ] = Qp[X ] +Qp[X ] + θ
α− 1.
(124)
HW:
Compute ESFp[X ].
3.31, 3.32, 3.35, 3.36.
Show that CTEp[X ] is, in general, not subadditive.
How about ESFp[X ] - is it subadditive ?
How about ρ[X ] := E[X ] - is it subadditive ?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 116 / 193
HW: CTEp[X ]
Consider again the empirically observed distribution of Losses for accidentclaims from randomly selected policies is as follows:
Claim Amount L Probability
0 0.607
100 0.303
200 0.076
500 0.012
1000 0.002
Compute the CTEp[X ] for p ∈ 0.90, 0.95, 0.99. Consult the HardyStudy Note for cases where the VaR levels Qp = Qp+ε for some ε > 0.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 117 / 193
Thoughts on CTEp[X ]
CTEp[X ] ≥ Qp[X ], making it more preferable in the eyes of aregulator who always desires higher solvency capital.
CTEp[X ] is used for equity-linked life contingent insurance in Canada.
Minimum Capital Test For Federally Regulated Property and CasualtyInsurance Companies report from the Government of Canada:”..(u)nder the MCT, regulatory capital requirements for various risksare set directly at a pre-determined target confidence level. OSFI haselected 99% of the expected shortfall (conditional tail expectation orCTE 99%) over a one-year time horizon as a target confidence level”,with the footnote ”As an alternative, a value at risk (VaR) at99.5% confidence level or expert judgement was used when it was notpractical to use the CTE approach.”
CTEp is not the only non-subadditive measure, apparently..
What to do? Is there a ”better” property than sub-additivity?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 118 / 193
Distortion Risk Measures
Recalling actuarial terminology, for the loss distributionF : R+ → [0, 1], define the Survival Function
S(x) = 1− F (x). (125)
Define the Distortion Function g : [0, 1]→ [0, 1] withg(0) = 0, g(1) = 1, and g increasing.Define the Distortion Risk Measure that inflates the probability oflarge losses via
ρg [X ] =
∫ ∞0
g(S(x))dx
= Eg [X ].
(126)
Theorem
DVTGKV (2004) If g is also concave, then for any losses X ,Y ∈ A,
P[(X ,Y ) ∈ R2+] = 1⇒ ρg [X + Y ] ≤ ρg [X ] + ρg [Y ]. (127)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 119 / 193
Distortion Risk Measures
HW Are CTEp[X ],ESFp[X ],TTVaRp[X ] and Qp[X ] Distortion RiskMeasures ? Proof or counterexamples..
Hint: Tryg(x) = 1x≥1−p
g(x) = min
x
1− p, 1
(128)
and use the relationships obtained for the above risk measures.What about probability masses?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 120 / 193
Proportional Hazard Transform
For α ≥ 1, define
g(s) = s1α (129)
If X ∼ Pareto(γ, θ), then E[X ] = θγ−1 and (assuming γ > α,)
S(x) = P[X > x ] =( θ
θ + x
)γg(S(x)) =
( θ
θ + x
) γα
⇒ ρg [X ] =
∫ ∞0
( θ
θ + x
) γα
dx
=θ
γα − 1
(130)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 121 / 193
Proportional Hazard Transform
For α ≥ 1, define
g(s) = s1α (131)
If X ∼ exp(λ), then E[X ] = 1λ and (assuming γ > α,)
S(x) = P[X > x ] = e−λx
g(S(x)) = e−λαx
⇒ ρg [X ] =
∫ ∞0
e−λαxdx
=α
λ
(132)
HW: How about if X ∼ N(µ.σ2)?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 122 / 193
Dual Power Transform
Define g(s) = 1− (1− s)N . Note that if N ∈ N, then for N i.i.d. copies ofX : i.e. X1,X2, ..,XN it follows that
g(S(x)) = 1− (1− S(x))N = P[max X1,X2, ..,XN > x ]
∴ ρg [X ] =
∫ ∞0
g(S(x))dx = E[max X1,X2, ..,XN].(133)
HW: Compute ρg [X ] for X ∼ exp(λ). Use (λ,N) = (0.1, 10) andwhatever numerical tools you feel comfortable with.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 123 / 193
The Wang Transform
Wang, in A Universal Framework for Pricing Financial and Insurance(2002) presents the following idea: Define λ ∈ R and a distortion functiong such that
g(S(x)) = Φ(
Φ−1(S(x)) + λ). (134)
Then there is a universal pricing method based on this transform forfinancial assets or liabilities over a time horizon [0,T ]. In this distortionmeasure, λ is the market price of risk, reflecting the level of systematicrisk.
HW: Consider an X ∼ N(µ, σ2). How does the Wang Transform acton X ? Specifically, compute ρg [X ]!
HW: Consider an X ∼ LN(µ, σ2). How does the Wang Transformact on X ? Specifically, compute ρg [X ]. How does ρg [X ] relate to thequantile measure Qp[X ] in this case?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 124 / 193
Spectral Risk Measure
Assume a non-negative, non-increasing, right-continuous, integrablefunction γ : [0, 1]→ R+ such that∫ 1
0γ(p)dp = 1 (135)
and the corresponding risk measure
ργ [X ] :=
∫ 1
0γ(p)Qp[X ]dp. (136)
Then ργ [X ] is a Spectral Risk Measure.
HW: Consider X ∼ exp(λ), and γ(p) = 2p. Compute ργ [X ].
Is Expected Shortfall a Spectral Risk Measure? How about Value atRisk? How about just Expected Value?
Are Distortion and Spectral Risk Measures related? Check outTheorem 3.1 here for a nice connection.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 125 / 193
Regulators, Economic Capital, and Expected Shortfall
In the case of choosing an adequate capital requirement, a regulator ischarged with making the event X > ρ[X ] ”unlikely.” A selection processcan entail first choosing the a shortfall measure M, where M satisfies
ρ1[X ] ≤ ρ2[X ]⇒ M
[(X − ρ1[X ])+
]≥ M
[(X − ρ2[X ])+
]. (137)
This reflects the view of the view of the regulator that an increase ofsolvency capital implies a reduction of the shortfall risk that M measures.
A choice of a monotonic M is enough to guarantee the above condition.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 126 / 193
Regulators, Economic Capital, and Expected Shortfall
Note that this condition relates the view of the regulator thatincreasing regulatory capital decreases the likelihood of insolvency.
As mentioned earlier, this view must compete with the cost of holdingexcess capital.
Given the choice of regulatory measure M, the optimal solutionu = ρ[X ] solves
minu∈R+
(E[M[(X − u)+]
]+ εu
). (138)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 127 / 193
Regulators, Economic Capital, and Expected Shortfall
It was also shown earlier that for M[x ] = x , the optimal solvencycapital is ρ[X ] = Q1−ε[X ].
It follows from our Equivalence Of Risk Measures Theorem that theminimized expected penalized shortfall using this capital is also themeasure
minu∈R+
(E[(X − u)+
]+ εu
)= εTVaR1−ε[X ]. (139)
From the view of a company that must self-determine the optimalsolvency capital, the problem mentioned above is also known asdetermining the economic capital required.
Please refer to Economic Capital Allocation Derived From RiskMeasures [DGK (2003)] for more on economic capital from anactuarial perspective.
Great overall review paper is Risk Measures and Comonotonicity: AReview [DGK (2006)]
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 128 / 193
Representation Theorem for Coherent Measures
The original authors Artzner, P., Delbaen, F., Eber, J.-M., Heath, D.(1999) Coherent measures of risk, Mathematical Finance 9(3), 203-228.state and prove the foundational theorem:
Theorem
For a finite set of outcomes, ρ is a coherent risk measure if and only if ∃ afamily P of probability measures on the set of outcomes such that
ρ[X ] = supEQ[X ] : Q ∈ P
. (140)
To see how this relates to the margin system SPAN (Standard PortfolioAnalysis of Risk) developed by the Chicago Mercantile Exchange, pleaselook through Section 3.1 here and the references within.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 129 / 193
Introduction
An estimator is a quantity, rule, or formula that estimates properties of aset of values. For example, if our set is A = x1, x2, ..., xn, then acommon estimator is
x :=
∑nj=1 xj
n. (141)
A single estimate that results in a value for a an unknown parameter θ iscalled a point estimator θ. Such an estimator is called unbiased if
bias(θ) := E[θ − θ] = 0. (142)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 130 / 193
Example 46.1 (Finan)
A population consists of the values 1, 3, 5, and 9. We want to estimate themean of the population µ. A random sample of two values from thispopulation is taken without replacement, and the mean of the sample µ isused as an estimator of the population mean µ.
Find the probability distribution of µ
Is µ an unbiased estimator?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 131 / 193
Example 46.1 (Finan)
We find that there are six possible outcomes for this experiment,consisting of 6 pairs of values drawn:
S = 1, 3 , 1, 5 , 1, 9 , 3, 5 , 3, 9 , 5, 9 . (143)
Each of these pairs is equally likely, and so we assign probability 16 to each
of the possible values of µ:
µ ∈ 2, 3, 4, 5, 6, 7 . (144)
and so it follows that
µ =1 + 3 + 5 + 9
4= 4.5
E[µ] =1
6(2 + 3 + 4 + 5 + 6 + 7) = 4.5
∴ bias(θ) = E[µ− µ] = 0.
(145)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 132 / 193
Asymptotically Unbiased and Consistent Estimators
If, as the sample size increases, the bias vanishes we say that the estimatorθn is asymptotically unbiased if forall θ :
limn→∞
E[θn − θ] = 0. (146)
If, for all δ > 0,
limn→∞
P[| θn − θ |> δ] = 0 (147)
then θn is a consistent estimator.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 133 / 193
Theorem
Theorem
If limn→∞ E[θn − θ] = 0 = limn→∞Var[θn], then θn is consistent.
Proof.
Fix a δ > 0. By Markov’s Inequality,
0 ≤ P[|θn − θ| > δ] ≤ E[(θn − θ)2]
δ2=
E[(
(θn − E[θn]) + (E[θn]− θ))2]
δ2
=Var[θn]− 2E[θn − θ] · E[θn − E[θn]] + E[(E[θn]− θ)2]
δ2→ 0.
(148)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 134 / 193
HW: Asymptotically Unbiased and Consistent Estimators
Assume that X1, ...,Xn are i.i.d., and for all i ∈ 1, 2, ..n, Xi ∼ exp(λ).
Are there any values λ > 0 such that
λn = max X1, ..,Xnλn = min X1, ..,Xn
(149)
as estimators of E[X ] = 1λ are Asymptotically Unbiased? Consistent?
If not, can we multiply them by scaling functions of n to get thesedesired properties?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 135 / 193
Mean-Squared Error (MSE)
The mean-squared error associated with an estimator θ is
MSE(θ) = E[(θ − θ)2] = Var[θ] + bias2(θ). (150)
If two unbiased estimators θ1, θ2 are such that MSE (θ1) < MSE (θ2),then we say that θ1 is more efficient than θ2.
If θ1 is more efficient that any other unbiased estimator θ, then wesay that θ1 is a uniformly minimum variance unbiased estimator.
HW:
Find MSE(λ) for the previous HW problem.
HW: Practice problems 46.1− 46.4, 46.7 in Finan.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 136 / 193
Example: Estimator for Uniform Random Variable
Assume that X1, ...,Xn are i.i.d., and for all i ∈ 1, 2, ..n, Xi ∼ U(0, L).Define Ln as an estimator of L such that
Ln = max X1, ..,Xn . (151)
Is Ln Asymptotically Unbiased? Consistent? Asympotically zero -MSE?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 137 / 193
Example: Estimator for Uniform Random Variable
By definition, if we set FLn(y) := P[Ln ≤ y ], then
FLn(y) =
(y
L
)nfLn(y) =
nyn−1
Ln.
(152)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 138 / 193
Example: Estimator for Uniform Random Variable
It follows that
Var[Ln] + bias2(Ln) = MSE(Ln)
= E[(Ln − L)2] =
∫ L
0(y − L)2 nyn−1
Lndy
=n
Ln
∫ L
0(y 2 − 2Ly + L2)yn−1dy
=3L2
(n + 1)(n + 2)→ 0.
(153)
So not only is Ln asympotically zero -MSE, but also Var[Ln]→ 0 andbias(Ln)→ 0 and so Ln is consistent by our previous theorem.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 139 / 193
Method of Moments and Matching Percentile
For a random variable X with parameters θ1, ..., θp, defineθ = (θ1, ..., θp) and
F (x | θ) = E[1X≤x | θ]
µ′k(θ) = E[X k | θ]
µ′k =1
n
n∑i=1
(xi )k
(154)
for a sequence of realizations X1 = x1 , ..., Xn = xn.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 140 / 193
Method of Moments and Matching Percentile
A method-of-moments estimate of the vector θ is any solution to thesystem of p equations
µ′k(θ) = µ′k . (155)
The smoothed empirical estimate of the 100g th percentile is obtained byarranging the observed values x1, ..., xn from smallest to largest:
x1, ..., xn →
x(1), ..., x(n)
(156)
and finding the integer p such that pn+1 ≤ g ≤ p+1
n+1 .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 141 / 193
Method of Moments and Matching Percentile
From this, we define
ˆVaRg [X ] := [p + 1− (n + 1)g ]x(p) + [(n + 1)g − p]x(p+1). (157)
and the method of matching percentiles corresponds to matching
F ( ˆVaRgk [X ] | θ) = gk ∀k ∈ 1, ..., p (158)
and a set of arbitrarily chosen percentiles g1, ..., gp.
See Examples 58.5 and 58.8 in Finan for an application of matchingpercentiles.
See Examples 58.7 and 58.10 in Finan for an application of method ofmoments.
HW: Practice problems 58.6− 58.10 in Finan.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 142 / 193
SOA Exam C Practice Question #1
You are given:
Losses X follow a Losses follow a loglogistic distribution withcumulative distribution function characterized by parameters (θ, γ):
FX (x) =
(xθ
)γ1 +
(xθ
)γ , ∀x ≥ 1 (159)
A random sample of losses is
Sample Losses = 10, 35, 80, 86, 90, 120, 158, 180, 200, 210, 1500 .(160)
Calculate the estimate of θ by percentile matching, using the 40th and80th empirically smoothed percentile estimates.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 143 / 193
SOA Exam C Practice Question #1
The 40th percentile is the .4(12) = 4.8th smallest observation. Byinterpolation, it is .2(86) + .8(90) = 89.2. The 80th percentile is the.8(12) = 9.6th smallest observation. By interpolation it is.4(200) + .6(210) = 206. The equations to solve are
0.4 =
(89.2θ
)γ1 +
(89.2θ
)γ0.8 =
(206θ
)γ1 +
(206θ
)γ⇒ (γ, θ) = (2.1407, 107.08).
(161)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 144 / 193
SOA Exam C Practice Question #19
You are given:
A sample x1, x2, , x10 is drawn from a distribution with probabilitydensity function (with θ > σ):
fX (x) =1
2
(1
θe−
xθ +
1
σe−
xσ
), ∀x ∈ (0,∞) (162)
Summed values10∑i=1
xi = 150
10∑i=1
x2i = 5000.
(163)
Estimate θ by matching the first two sample moments to thecorresponding population quantities.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 145 / 193
SOA Exam C Practice Question #19
By definition of exponential distributions, we obtain the pair of equations:
150
10= E[X ] =
1
2(θ + σ)
5000
10= E[X 2] =
1
2(2θ2 + 2σ2)
⇒ (θ, σ) = (20, 10).
(164)
The solution can be found via
30 = θ + σ ⇒ σ = 30− θ500 = θ2 + σ2 = θ2 + (30− θ)2 = 2θ2 − 60θ + 900
⇒ θ ∈ 10, 20 .(165)
We choose θ = 20 to fix σ = 10.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 146 / 193
Two-sided Uniform Interval Estimation
Define the empirically observed quantities
X :=1
n
n∑k=1
xk
X 2 :=1
n
n∑k=1
x2k
(166)
for a sequence of n observed values X1 = x1, ...,Xn = xn . Assume thatthe Xini=1 are i.i.d. and drawn from a population X ∼ U[a, b]. Use
method of moments to determine a, b in terms of X ,X 2.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 147 / 193
Two-sided Uniform Interval Estimation
By definition,
E[X ] =a + b
2
E[X 2] =a2 + b2 + ab
3.
(167)
Estimation via matching of the first two moments:
X =a + b
2
X 2 =a2 + b2 + ab
3.
(168)
Algebra leads to
(a, b) =
(X −
√3
√X 2 − X
2,X +
√3
√X 2 − X
2). (169)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 148 / 193
Maximum Likelihood Estimation for Complete Data
Assume again a probability space (Ω,F ,P), where a subset Aini=1 of Fis contained within Ω, and
∪ni=1Ai ⊆ Ω. (170)
Here, the Ai represent observations obtained from an experimentconsisting of a sequence of i.i.d. random variables Xini=1, pulled from acommon distribution F (· | θ).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 149 / 193
Maximum Likelihood Estimation for Complete Data
The idea now is to use information obtained by the outcomes observedXi ∈ Ai. In fact, we determined that the likelihood function
L(θ) := Πni=1P[Xi ∈ Ai | θ] (171)
should be maximized, due to our observation that this independentsequence of events has occurred!
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 150 / 193
Maximum Likelihood Estimation for Complete Data
For ease of computation, we may wish to maximize the log-likelihoodfunction
l(θ) := ln [L(θ)] =n∑
i=1
ln (P[Xi ∈ Ai | θ]). (172)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 151 / 193
SOA Exam C Practice Question #4
You are given:
Losses X follow a Single-parameter Pareto distribution with densityfunction characterized by a parameter α ∈ (0,∞)
fX (x) =α
xα+1, ∀x > 1 (173)
A random sample of size five produced three losses with values 3, 6and 14, and two losses exceeding 25.
Determine the maximum likelihood estimate of α.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 152 / 193
SOA Exam C Practice Question #4
By definition, SX (x) = 1xα and so
L(α) = fX (3)fX (6)fX (14)SX (25)2 =α
3α+1
α
6α+1
α
14α+1
1
252α
=1
252
α3
157500α
l(α) = ln [L(α)] = 3 lnα− α ln (157500)− ln (252)
⇒ l ′(α) =3
α− ln (157500)
⇒ α =3
ln (157500)= 0.2507.
(174)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 153 / 193
SOA Exam C Practice Question #44
You are given:
Losses X follow an exponential distribution with mean θ
A random sample of 20 losses is distributed as follows:
Loss Range Frequency
[0,1000] 7
(1000,2000] 6
(2000,∞) 7
Determine the maximum likelihood estimate of θ.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 154 / 193
SOA Exam C Practice Question #44
Our loss function is, for p := e−1000θ ,
L(θ) = F (1000)7[F (2000)− F (1000)]6[1− F (2000)]7
=(
1− e−1000θ
)7 (e−
1000θ − e−
2000θ
)6 (e−
2000θ
)7
L(p) = (1− p)7(p − p2)6(p2)7 = p20(1− p)13
l(p) = 20 ln p + 13 ln (1− p)
l ′(p) =20
p− 13
1− p= 0
⇒ p =20
33= 1− e−
10001996.90
⇒ θ = 1996.90.
(175)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 155 / 193
Maximum Likelihood Estimation for Complete Data
See Examples 59.1− 59.5 in Finan for an application of maximumlikelohood estimators to loss models.
HW: Practice problems 59.2− 59.7 in Finan.
HW: Read Section 60 in Finan on Maximum Likelihood Estimationfor Incomplete Data.
HW: Read Section 67 in Finan on Estimation of Class (a, b, 0).
HW: Practice problems 67.2.2− 67.4 in Finan.
Note: Take-home final exam may have problems very similar in scopeto those above!
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 156 / 193
Introduction
Assume now that there is a history of claims X1, ...,Xn, all with acommon risk parameter Θ such that the Xk | Θnk=1 are i.i.d.
Furthermore, based on the history we have observed, we would like to useit to estimate the (common) expected loss
µ(Θ) := E[Xn+1 | Θ]. (176)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 157 / 193
Introduction
The question to ask is how Credible the data X1, ...,Xn is. As a riskmanager, there is the option to charge the book or manual premium
µ := E[µ(Θ)] (177)
or charge the weighted average premium for some sequence of weightsβ1, ..., βn:
X :=n∑
k=1
βkXk . (178)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 158 / 193
Introduction
As in the section on risk measures, we seek a metric Z for the credibilityof the observed claims X1, ...,Xn so far.
This value Z should depend on everything, and measure the likelihoodthat the observed data reflects reality, and should be used for premiumsover the assumed average µ:
Z : (Θ,X1, ...,Xn)→ [0, 1]. (179)
The balance reached between history and assumed µ results in an optimalpremium at time n + 1:
Pn+1 = (1− Z )µ+ Z X . (180)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 159 / 193
Limited Fluctuation Credibility
A risk manager may be tempted to give full credibility, Z = 1, toobserved data. In order to justify this, there must be some stability to theobserved data.
By assigning equal weight to the observed claims, αk := 1n , we seek to
determine whether, by the Central Limit Theorem
p ≤ P[∣∣∣∣ X − µµ
∣∣∣∣ ≤ r
]= P
[∣∣∣∣∣ X − µσ√n
∣∣∣∣∣ ≤ rµ√
n
σ
]≈ 2Φ
(rµ√
n
σ
)−1 (181)
for company-determined values of r and p, and assumed parameters(µ, σ2) = (E[Xj ],Var[Xj ]).
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 160 / 193
Limited Fluctuation Credibility
Hence, the question of full-credibility now reduces to a large enoughnumber of exposure units:
rµ√
n
σ≥ Φ−1
(1 + p
2
)=: yp. (182)
Rewritten in terms of n, the inequality is now
n ≥ λ0
(σ
µ
)2
λ0 :=(yp
r
)2.
(183)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 161 / 193
Limited Fluctuation Credibility: Aggregate Claims
Consider now the joint sequence (Ni ,Xi )ni=1, where Ni is the totalnumber of claims in year i , and Xi is the total loss in year i .
Assume that the Xi and Ni are independent of each other, and that theNi ∼ Poisson(λ) are i.i.d. as well as Xi independent with commondistribution FX .
Let Yij denote the j th claim in year i . Assume that the Yij are i.i.d. withmean θY and variance σ2
Y . It follows that
Xi =
Ni∑j=1
Yij . (184)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 162 / 193
Limited Fluctuation Credibility: Aggregate Claims
Based on this set-up, the independence of the claim frequency and claimsize leads to
E[Xi ] = E[Ni ] · E[Yij ] = λθY
Var[Xi ] = Var[Ni ] · E[Yij ]2 + E[Ni ] ·Var[Yij ]
= λ(θ2Y + σ2
Y ).
(185)
It follows that for full credibility, we need
n ≥ λ0
σ2Xi
E[Xi ]2= λ0
θ2Y + σ2
Y
λθ2Y
(186)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 163 / 193
SOA Exam C Practice Question #2
You are given:
The number of claims has a Poisson distribution.
Claim sizes Yij have a Pareto distribution with common values(θY , σ
2Y ) = (0.10, 0.015).
The number of claims and claim sizes are independent.
The observed pure premium should be within 2% of the expectedpure premium 90% of the time.
Determine the expected number of claims needed for full credibility.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 164 / 193
SOA Exam C Practice Question #2
The expected number of claims is equal to λn. From the informationprovided, r = 0.02 and y0.90 = 1.645, and so for full credibility, we require
λn ≥ λ ·
(λ0
σ2Xi
E[Xi ]2
)
= λ0θ2Y + σ2
Y
θ2Y
=
(1.645
0.02
)2(1 +
0.0152
0.102
)= 16913.
(187)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 165 / 193
Greatest Accuracy Credibility
One way of determining Z ∈ (0, 1) is by optimizing over all linear functionsof past data:
minα1,...αn
E
(µ(Θ)−
[µ(1−
n∑k=1
αk) +n∑
k=1
αkXk
])2 . (188)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 166 / 193
Greatest Accuracy Credibility
Using Hilbert space theory, Shiu and Sing show the optimal weights to be
αk =
E[(µ−µ(Θ))2]E[(Xk−µ(Θ))2]
1 +∑n
j=1E[(µ−µ(Θ))2]E[(Xj−µ(Θ))2]
=
Var[µ(Θ)]E[Var[Xk |Θ]]
1 +∑n
j=1Var[µ(Θ)]
E[Var[Xj |Θ]]
1− Z = 1−n∑
k=1
αk =1
1 +∑n
j=1E[(µ−µ(Θ))2]E[(Xj−µ(Θ))2]
=1
1 +∑n
j=1Var[µ(Θ)]
E[Var[Xj |Θ]]
(189)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 167 / 193
Buhlmanns k−factor
From the previous slide, we have
Z =
∑nj=1
Var[µ(Θ)]E[Var[Xj |Θ]]
1 +∑n
j=1Var[µ(Θ)]
E[Var[Xj |Θ]]
. (190)
If, however, we have a common value k such that, for all j ∈ 1, ..., n,
1
k:=
Var[µ(Θ)]
E[Var[Xj | Θ]](191)
then
Z =n 1k
1 + n 1k
=n
n + k. (192)
This common value k is known as Buhlmanns k−factor. Notice that ifn→∞, then Z → 1.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 168 / 193
Buhlmanns k−factor
Furthermore, we have that for constant β := E[Var[Xk | Θ]],
βk :=
1E[Var[Xk |Θ]]∑nj=1
1E[Var[Xj |Θ]]
=1
n(193)
and so we can write
Pn+1 = (1− Z )µ+ Z
(X1 + ...+ Xn
n
). (194)
In this setting, we can extend to number of claims or other observedvariables V1, ...,Vn and say that our Buhlmann Credible Estimate forVn+1 is
Vn+1 = (1− Z )µV + Z
(V1 + ...+ Vn
n
). (195)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 169 / 193
Compound Poisson Process
Assume the existence of a claim count random variable such thatN ∼ Poisson(λ).
Assume that the claim sizes X1,X2, .. to be i.i.d. copies of arandom variable X .
Assume that X , X1,X2, .. are independent of N.
Based on these assumptions, define the aggregate claim Y | N such that
Y | N :=N∑
k=1
Xk . (196)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 170 / 193
Compound Poisson Process
By the Tower Property of Conditional Expectations, and the fact that N isPoisson-distributed, we have
E[Y ] = E[E[Y | N]] = E[NE[X ]] = E[N]E[X ].
Var[Y ] = E[Var[Y | N]] + Var [E[Y | N]]
= E[NVar[X ]] + Var [NE[X ]]
= E[N]Var[X ] + Var [N]E[X ]2
= E[N]Var[X ] + E [N]E[X ]2
= E[N]E[X 2]
(197)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 171 / 193
SOA Exam C Practice Question #8
You are given:
Claim counts follow a Poisson distribution with mean θ.
Claim sizes follow an exponential distribution with mean 10θ .
Claim counts and claim sizes are independent, given θ.
The prior distribution has probability density function
fΘ(θ) =5
θ6for θ > 1. (198)
Calculate Buhlmanns k−factor for aggregate losses.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 172 / 193
SOA Exam C Practice Question #8
Define N as the Poisson claim count variable, X as the claim size variable,and Y as the aggregate loss variable. It follows that
µ(θ) = E[Y | Θ = θ] = E[N | Θ = θ] · E[X | Θ = θ] = θ · 10θ = 10θ2.
v(θ) = Var[Y | Θ = θ] = E[N | Θ = θ] · E[X 2 | Θ = θ] = 200θ3.(199)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 173 / 193
SOA Exam C Practice Question #8
To compute k , we need
µ = E[µ(Θ)] =
∫ ∞1
10θ2 · 5
θ6dθ =
50
3
Var[µ(Θ)] = Var[10Θ2] =
∫ ∞1
(10θ2 − 50
3
)2
fΘ(θ)dθ
=11111
50= 222.22
E[v(Θ)] =
∫ ∞1
200θ3 · fΘ(θ)dθ = 500
⇒ k =E[v(Θ)]
Var[µ(Θ)]=
500
222.22= 2.25.
(200)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 174 / 193
Bayesian Parameter Estimation
Another way to use previously observed data to estimate future claims andlosses is via Bayesian Parameter Estimation. In this approach, weassume that we wish to estimate X using observed data by firstconditioning the parameter distribution on observed data :
E[Xn+1 | X1, ...,Xn] =
∫θ∈R
E[Xn+1 | θ]fΘ(θ | X1, ...,Xn)dθ. (201)
Question: Is there a connection between Bayesian Parameter Estimationand Credibility Theory?
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 175 / 193
SOA Exam C Practice Question #45
You are given:
The size X of a claim is uniformly distributed on the interval [0, θ]
The prior distribution has probability density function
fΘ(θ) =500
θ2for θ > 500. (202)
Two claims, x1 = 400 and x2 = 600, are observed. You calculate theposterior distribution as:
fΘ(θ | X1 = 400,X2 = 600) = 3
(6003
θ4
)for θ > 600. (203)
Calculate the Bayesian premium E[X3 | X1 = x1,X2 = x2].
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 176 / 193
SOA Exam C Practice Question #45
As X ∼ U[0, θ], we have E[X | Θ] = 12 Θ. It follows that
E[X3 | X1 = 400,X2 = 600] =
∫θ∈R
E[X3 | θ]fΘ(θ | X1 = 400,X2 = 600)dθ
=
∫ ∞600
1
2θ · 3
(6003
θ4
)dθ
= 36003
2
∫ ∞600
1
θ3dθ
= 36003
2
1
2 · 6002= 450.
(204)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 177 / 193
Introduction
Recall that if we take a sequence of i.i.d. (independent, identicallydistributed) random variables X1,X2, ...,Xn, called a random sample,with common mean and variance:
E[Xi ] = µ
Var[Xi ] = σ2(205)
then
E
[X1+X2+...+Xn
n − µσ√n
]= 0
Var
(X1+X2+...+Xn
n − µσ√n
)= 1
(206)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 178 / 193
Introduction
Recall that if we take a sequence of i.i.d. (independent, identicallydistributed) random variables X1,X2, ...,Xn, called a random sample,with common mean and variance:
E[Xi ] = µ
Var[Xi ] = σ2(205)
then
E
[X1+X2+...+Xn
n − µσ√n
]= 0
Var
(X1+X2+...+Xn
n − µσ√n
)= 1
(206)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 178 / 193
Introduction
If the random sample comes from a Normal population, then by theprevious slide, we know
X1+X2+...+Xnn − µσ√n
∼ N(0, 1) (207)
This implies we can use our Observed Sample Mean X1+X2+...+Xnn to
estimate the Population Sample Mean µ as long as we know σ a priori:
P
[−zα
2≤
X1+X2+...+Xnn − µσ√n
≤ zα2
]= P
[−zα
2≤ Z ≤ zα
2
]= Φ
(zα
2
)− Φ
(−zα
2
)= 1− α
(208)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 179 / 193
Introduction
If the random sample comes from a Normal population, then by theprevious slide, we know
X1+X2+...+Xnn − µσ√n
∼ N(0, 1) (207)
This implies we can use our Observed Sample Mean X1+X2+...+Xnn to
estimate the Population Sample Mean µ as long as we know σ a priori:
P
[−zα
2≤
X1+X2+...+Xnn − µσ√n
≤ zα2
]= P
[−zα
2≤ Z ≤ zα
2
]= Φ
(zα
2
)− Φ
(−zα
2
)= 1− α
(208)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 179 / 193
Introduction
In fact, we can tune the accuracy of our prediction by how small wechoose α > 0 and how large n of a sample we take. Symbolically, if weobserve our sample mean to be x , then we are 100 · (1− α)% that
µ ∈(
x − zα2
σ√n, x + zα
2
σ√n
)
Some values:
100 · (1− α)% α2 zα
2
90% 0.05 1.645
95% 0.025 1.960
99% 0.005 2.576
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 180 / 193
Introduction
In fact, we can tune the accuracy of our prediction by how small wechoose α > 0 and how large n of a sample we take. Symbolically, if weobserve our sample mean to be x , then we are 100 · (1− α)% that
µ ∈(
x − zα2
σ√n, x + zα
2
σ√n
)Some values:
100 · (1− α)% α2 zα
2
90% 0.05 1.645
95% 0.025 1.960
99% 0.005 2.576
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 180 / 193
Interval Width
In general, a confidence interval with width n observations and100 · (1− α)% certainty has a total width
w = 2zα2· σ√
n. (209)
Inverting this for n, we derive the number n observations required for agiven width w , confidence level 100 · (1− α)%, and known deviation σ tobe
n =(
2zα2· σ
w
)2. (210)
For example, if α = 0.05, σ = 25,w = 10, then n =(2 · 1.96 · 25
10
)2= 96.
Notice that n ∝ 1w2 .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 181 / 193
Interval Width
In general, a confidence interval with width n observations and100 · (1− α)% certainty has a total width
w = 2zα2· σ√
n. (209)
Inverting this for n, we derive the number n observations required for agiven width w , confidence level 100 · (1− α)%, and known deviation σ tobe
n =(
2zα2· σ
w
)2. (210)
For example, if α = 0.05, σ = 25,w = 10, then n =(2 · 1.96 · 25
10
)2= 96.
Notice that n ∝ 1w2 .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 181 / 193
Interval Width
In general, a confidence interval with width n observations and100 · (1− α)% certainty has a total width
w = 2zα2· σ√
n. (209)
Inverting this for n, we derive the number n observations required for agiven width w , confidence level 100 · (1− α)%, and known deviation σ tobe
n =(
2zα2· σ
w
)2. (210)
For example, if α = 0.05, σ = 25,w = 10, then n =(2 · 1.96 · 25
10
)2= 96.
Notice that n ∝ 1w2 .
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 181 / 193
Unknown σ
In most cases, one can safely assume that we are also estimating theVariance of the population along with the population mean. By the CLT,we know that for any random sample with mean µ and variance σ2,
P
[−zα
2≤
X1+X2+...+Xnn − µσ√n
≤ zα2
]→ 1− α. (211)
If we work only from observed data, then we recall that our SampleMean-Variance pair (X ,S2) has the form
S2 =n∑
i=1
(Xi − X
)2
n − 1
X =X1 + X2 + ...Xn
n
(212)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 182 / 193
Unknown σ
In most cases, one can safely assume that we are also estimating theVariance of the population along with the population mean. By the CLT,we know that for any random sample with mean µ and variance σ2,
P
[−zα
2≤
X1+X2+...+Xnn − µσ√n
≤ zα2
]→ 1− α. (211)
If we work only from observed data, then we recall that our SampleMean-Variance pair (X ,S2) has the form
S2 =n∑
i=1
(Xi − X
)2
n − 1
X =X1 + X2 + ...Xn
n
(212)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 182 / 193
If we work only from observed data, we can use our modified version ofthe CLT:
Theorem
P
[−zα
2≤ X − µ
S√n
≤ zα2
]→ 1− α. (213)
The speed of convergence depends on the underlying populationdistribution, of course. As a general rule of thumb, when using onlyobserved data, one can consider the above limit achieved if n ≥ 40.
As a computational consideration, recall that
S2 =n∑
i=1
(Xi − X
)2
n − 1
=1
n − 1
n∑i=1
X 2i −
1
n · (n − 1)
(n∑
i=1
Xi
)2
(214)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 183 / 193
If we work only from observed data, we can use our modified version ofthe CLT:
Theorem
P
[−zα
2≤ X − µ
S√n
≤ zα2
]→ 1− α. (213)
The speed of convergence depends on the underlying populationdistribution, of course. As a general rule of thumb, when using onlyobserved data, one can consider the above limit achieved if n ≥ 40.
As a computational consideration, recall that
S2 =n∑
i=1
(Xi − X
)2
n − 1=
1
n − 1
n∑i=1
X 2i −
1
n · (n − 1)
(n∑
i=1
Xi
)2
(214)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 183 / 193
Bionomial Distribution - Polling ?
Imagine now that we are polling a group of people whose answer is eitheryes or no. Assign the value 1 for an answer of yes, 0 for the answer no.Define Xi to be the numerical value of the i th respondent’s answer, and
p :=
∑ni=1 Xi
n(215)
the sample proportion of positive respondents. The question remains : howgood of an estimation of the total population positive proportion p is p ?.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 184 / 193
Bionomial Distribution - Polling ?
Luckily, we have seen this Binomial random variable before, and we knowthat given a population proportion p of a yes answer, we have
E[p] := E
[1
n
n∑i=1
Xi
]= p
Var(p) := Var
(1
n
n∑i=1
Xi
)=
p · (1− p)
n
= σ2p
= E[S2] ............... Really ?
(216)
It follows that for large n , we can make the approximation
p − p
σp∼ N(0, 1). (217)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 185 / 193
Bionomial Distribution - Polling ?
Luckily, we have seen this Binomial random variable before, and we knowthat given a population proportion p of a yes answer, we have
E[p] := E
[1
n
n∑i=1
Xi
]= p
Var(p) := Var
(1
n
n∑i=1
Xi
)=
p · (1− p)
n
= σ2p = E[S2]
............... Really ?
(216)
It follows that for large n , we can make the approximation
p − p
σp∼ N(0, 1). (217)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 185 / 193
Bionomial Distribution - Polling ?
Luckily, we have seen this Binomial random variable before, and we knowthat given a population proportion p of a yes answer, we have
E[p] := E
[1
n
n∑i=1
Xi
]= p
Var(p) := Var
(1
n
n∑i=1
Xi
)=
p · (1− p)
n
= σ2p = E[S2] ............... Really ?
(216)
It follows that for large n , we can make the approximation
p − p
σp∼ N(0, 1). (217)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 185 / 193
Bionomial Distribution - Polling ?
Luckily, we have seen this Binomial random variable before, and we knowthat given a population proportion p of a yes answer, we have
E[p] := E
[1
n
n∑i=1
Xi
]= p
Var(p) := Var
(1
n
n∑i=1
Xi
)=
p · (1− p)
n
= σ2p = E[S2] ............... Really ?
(216)
It follows that for large n , we can make the approximation
p − p
σp∼ N(0, 1). (217)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 185 / 193
Bionomial Distribution - Polling ?
Formally, the modified CLT tells us that
P
−zα2≤ p − p√
p·(1−p)√n
≤ zα2
→ 1− α. (218)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 186 / 193
Bionomial Distribution - Polling ?
This can be inverted to say that for large n,
P [p− ≤ p ≤ p+] ≈ 1− α
p− =p +
z2α2n − zα
2
√pqn +
z2α2
4n2
1 +z2α2n
p+ =p +
z2α2n + zα
2
√pqn +
z2α2
4n2
1 +z2α2n
(219)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 187 / 193
One sided Confidence Bounds
Since
P
[X1+X2+...+Xn
n − µS√n
≤ zα
]→ 1− α
P
[X1+X2+...+Xn
n − µS√n
≥ −zα
]→ 1− α
(220)
we say that
µ < x + zαs√n
is a Large Sample Upper Confidence Bound for µ
µ > x − zαs√n
is a Large Sample Lower Confidence Bound for µ.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 188 / 193
T-distributions
If the population of interest is normal, then X1,X2, ...,Xn constitutes anormal random sample with both µ, σ unknown.
Since we need σ to estimate µ, we do the next best thing and substitute Sfor σ, as before.
Tn−1 :=
(1n
∑ni=1 Xi
)− µ
S√n
fTn−1(x) =1√
(n − 1)π
Γ(n2 )
Γ(n−12 )
1(1 + x2
n−1
) n2
→ 1√2π
e−x2
2 =: fZ (x)
E[Tn−1] = 0
Var(Tn−1) =n − 1
n − 3
(221)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 189 / 193
T-distributions
If the population of interest is normal, then X1,X2, ...,Xn constitutes anormal random sample with both µ, σ unknown.Since we need σ to estimate µ, we do the next best thing and substitute Sfor σ, as before.
Tn−1 :=
(1n
∑ni=1 Xi
)− µ
S√n
fTn−1(x) =1√
(n − 1)π
Γ(n2 )
Γ(n−12 )
1(1 + x2
n−1
) n2
→ 1√2π
e−x2
2 =: fZ (x)
E[Tn−1] = 0
Var(Tn−1) =n − 1
n − 3
(221)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 189 / 193
T-distributions
If the population of interest is normal, then X1,X2, ...,Xn constitutes anormal random sample with both µ, σ unknown.Since we need σ to estimate µ, we do the next best thing and substitute Sfor σ, as before.
Tn−1 :=
(1n
∑ni=1 Xi
)− µ
S√n
fTn−1(x) =1√
(n − 1)π
Γ(n2 )
Γ(n−12 )
1(1 + x2
n−1
) n2
→ 1√2π
e−x2
2 =: fZ (x)
E[Tn−1] = 0
Var(Tn−1) =n − 1
n − 3
(221)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 189 / 193
T-distributions
If the population of interest is normal, then X1,X2, ...,Xn constitutes anormal random sample with both µ, σ unknown.Since we need σ to estimate µ, we do the next best thing and substitute Sfor σ, as before.
Tn−1 :=
(1n
∑ni=1 Xi
)− µ
S√n
fTn−1(x) =1√
(n − 1)π
Γ(n2 )
Γ(n−12 )
1(1 + x2
n−1
) n2
→ 1√2π
e−x2
2 =: fZ (x)
E[Tn−1] = 0
Var(Tn−1) =n − 1
n − 3
(221)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 189 / 193
T-distributions
Define tα,n−1 to satisfy
P[Tn−1 ≥ tα,n−1] = α (222)
It follows that
P[−tα2,n−1 ≤ Tn−1 ≤ tα
2,n−1] = 1− α (223)
and the corresponding two-sided and one-sided confidence intervals can bederived.
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 190 / 193
Prediction Intervals
Given a normal random sample X1,X2, ...,Xn, we want to predict the next,independent value Xn+1. One very tempting predictor is the historical, orsample, average X := 1
n
∑ni=1 of the previous n values. The residual error
attached with this prediction is X − Xn+1, and we can compute
E[X − Xn+1] = µ− µ = 0
Var(X − Xn+1
)= Var
(X)
+ Var(Xn+1)
=σ2
n+ σ2 = σ2 ·
(1 +
1
n
) (224)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 191 / 193
Prediction Intervals
With the above information, it follows that
X − Xn+1√(σ2 ·
(1 + 1
n
)) ∼ N(0, 1)
X − Xn+1√(S2 ·
(1 + 1
n
)) ∼ Tn−1
(225)
and we say that if the sample mean is x , and sample deviation is s, thenthe prediction interval for a single observation from a normal populationis
P
[∣∣∣Xn+1 − x∣∣∣ ≤ tα
2,n−1 · s ·
√1 +
1
n
]= 1− α (226)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 192 / 193
Confidence Intervals for Variance
Given a normal random sample X1,X2, ...,Xn, we see that
n − 1
σ2S2 =
∑ni=1(xi − X )2
σ2(227)
has a chi-squared (χ2n−1) distribution with n − 1 degrees of freedom.
Because S2 is always positive, and because the chi-squared distribution isasymmetric, we compute the confidence interval
P[χ2
1−α2,n−1 <
n − 1
σ2S2 < χ2
α2,n−1
]= 1− α
where P[χ2n−1 > χ2
α,n−1]]
= α
(228)
Albert Cohen (MSU) STT 459: Actuarial Models MSU Spring 2015 193 / 193