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ENGG 2040C: Probability Models and Applications
Andrej Bogdanov
Spring 2013
4. Random variablespart two
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Review
A discrete random variable X assigns a discrete value to every outcome in the sample space.
Probability mass function of X: p(x) = P(X = x)
Expected value of X: E[X] = ∑x x p(x).
E[N]
N: number of headsin two coin flips
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One die
Example from last time
F = face value of fair 6-sided dieE[F] = 1 + 2 + 3 + 4 + 5 + 6 = 3.5
1616 16 16 16 16
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Two dice
S = sum of face values of two fair 6-sided diceSolution 1
2 3 4 5 6 7 89 10 11 12
spS(s) 1
36236
336
436
536
636
536
436
336
236
136
E[S] = 2 + 3 + … + 12
136236 136 =
7
We calculate the p.m.f. of S:
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Two dice again
S = sum of face values of two fair 6-sided dice
F1 F2
S = F1 + F2
F1 = outcome of first dieF2 = outcome of second die
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Sum of random variables
Let X, Y be two random variables.
X + Y is the random variable that assigns value X(w) + Y(w) to outcome w.
X assigns value X(w) to outcome wY assigns value Y(w) to outcome w
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Sum of random variables
F1 F2 S = F1 + F2
11 1 1 212 1 2 3
21 2 1 3
… …
66 6 6 12
… …
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Linearity of expectation
E[X + Y] = E[X] + E[Y]
For every two random variables X and Y
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Two dice again
S = sum of face values of two fair 6-sided diceSolution 2
E[S] = E[F1] + E[F2]
= 3.5 + 3.5 = 7
F1 F2
S = F1 + F2
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Balls
We draw 3 balls without replacement from this urn:
-1
1
11
-1
-1
0
What is the expected sum of values on the 3 balls?
0
-1
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Balls
-1
1
11
-1
-1
00
-1S = B1 + B2
+ B3where Bi is the value of i-th ball.E[S] = E[B1] + E[B2] + E[B3]
p.m.f of B1:
-1 01
xp(x) 4
9 29
39
E[B1] = -1 (4/9) + 0 (2/9) + 1 (3/9) = -1/9same for B2, B3E[S] = 3 (-1/9) = -1/3.
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Three dice
E[N] = E[I1] + E[I2] + E[I3]
Solution I1 I2 I3
N = I1 + I2 + I3
E[I1] = 1 (1/6) + 0(5/6) = 1/6E[I2], E[I3] = 1/6
= 3 (1/6)
= 1/2
Ik =1 if face value of kth die equals
0 if not
N = number of s. Find E[N].
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Problem for you to solve
Five balls are chosen without replacement from an urn with 8 blue balls and 10 red balls.
What is the expected number of blue balls that are chosen?
What if the balls are chosen with replacement?
(a)
(b)
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The indicator (Bernoulli) random variablePerform a trial that succeeds with probability p and fails with probability 1 – p.
01p(x) 1 – p p
x
p = 0.5
p(x)
p = 0.4
p(x)E[X] = p
If X is Bernoulli(p) then
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The binomial random variable
Binomial(n, p): Perform n independent trials, each of which succeeds with probability p.
X = number of successes
ExamplesToss n coins. (# heads) is Binomial(n, ½). Toss n dice. (# s) is Binomial(n, 1/6).
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A less obvious example
Toss n coins. Let C be the number of consecutive changes (HT or TH).
w C(w)
HTHHHHT 3THHHHHTHHHHHHH
20
Examples:
Then C is Binomial(n – 1, ½).
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A non-example
Draw 10 cards from a 52-card deck.Let N = number of aces among the drawn cards
Is N a Binomial(10, 1/13) random variable?
No!
Different trial outcomes are not independent.
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Properties of binomial random variablesIf X is Binomial(n, p), its p.m.f. is
p(k) = P(X = k) = C(n, k) pk (1 - p)n-k
We can write X = I1 + … + In, where Ii is an indicator random variable for the success of the i-th trial E[X] = E[I1] + … + E[In]
= p + … + p = np.
E[X] = np
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Probability mass functionBinomial(10, 0.5) Binomial(50, 0.5)
Binomial(10, 0.3) Binomial(50, 0.3)
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Investments
You have two investment choices:A: put $25 in one stock B: put $½ in each of 50 unrelated stocks Which do you prefer?
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Investments
Probability model
Each stock
doubles in value with probability ½ loses all value with probability ½
Different stocks perform independently
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Investments
NA = amount on choice A
NB = amount on choice B 50 × Bernoulli(½)
A: put $50 in one stock B: put $½ in each of 50 stocks
Binomial(50, ½)
E[NA] E[NB]
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Variance and standard deviation
Let m = E[X] be the expected value of X.
The variance of X is the quantityVar[X] = E[(X –
m)2]The standard deviation of X is s = √Var[X]
They measure how close X and m are typically.
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Calculating variancem = E[NA]
y 252q(y) 1
p.m.f of (NA – m)2
Var[NA] = E[(NA – m)2]
x 050p(x) ½½ p.m.f of NA
= 252
s = std. dev. of NA = 25
m – s m + s
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Another formula for variance
Var[X] = E[(X – m)2]
= E[X2 – 2mX + m2]= E[X2] + E[–2mX] + E[m2]= E[X2] – 2m E[X] + m2
= E[X2] – 2m m + m2
= E[X2] – m2
for constant c, E[cX] = cE[X]
for constant c, E[c] = c
Var[X] = E[X2] – E[X]2
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Variance of binomial random variableSuppose X is Binomial(n, p).Then X = I1 + … + In, where Ii =
1, if trial i succeeds0, if trial i fails
Var[X] = E[X2] – m2 = E[X2] – (np)2
m = E[X] = np
E[X2]= E[(I1 + … + In)2]= E[I1
2 + … + In2 + I1I2 + … + In-1In]
= E[I12] + … + E[In
2] + E[I1I2] + … + E[In-
1In]
E[Ii2] = E[Ii] = p E[Ii Ij] = P(Ii = 1 and Ij = 1)
= P(Ii = 1) P(Ij = 1) = p2
= n p = n(n-1) p2
= np + n(n-1) p2 – (np)2
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Variance of binomial random variable
Suppose X is Binomial(n, p).
Var[X] = np + n(n-1) p2 – (np)2
m = E[X] = np
= np – p2 = np(1-p)
Var[X] = np(1-p)
s = √np(1-p).
The standard deviation of X is
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Investments
NA = amount on choice A
NB = amount on choice B 50 × Bernoulli(½)
A: put $50 in one stock B: put $½ in each of 50 stocks
Binomial(50, ½)m m
s = 25 s = √50 ½ ½ = 3.536…
m – s m + s
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Apples
About 10% of the apples on your farm are rotten.You sell 10 apples. How many are rotten?
Probability modelNumber of rotten apples you sold isBinomial(n = 10, p = 1/10).
E[N] = np = 1
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Apples
You improve productivity; now only 5% apples rot.You can now sell 20 apples and only one will be rotten on average.N is now Binomial(20, 1/20).
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Binomial(10, 1/10).349
.387
.194
.001
10-
10
Binomial(20, 1/20).354
.377
.189
.002
10-8 10-
26
.367
.367
.183
.003
10-7 10-
19
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The Poisson random variable
A Poisson(m) random variable has this p.m.f.:
p(k) = e-m mk/k!
k = 0, 1, 2, 3, …
Poisson random variables do not occur “naturally” in the sample spaces we have seen.They approximate Binomial(n, p) random variables when m = np is fixed and n is large (so p is small)
pPoisson(m)(k) = limn → ∞ pBinomial(n, m/n)(k)
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Rain is falling on your head at an average speed of 2.8 drops/second.
Divide the second evenly in n intervals of length 1/n.
Raindrops
Let Ei be the event “raindrop hits during interval i.”Assuming E1, …, En are independent, the number of drops in the second N is a Binomial(n, p) r.v.Since E[N] = 2.8, and E[N] = np, p must equal 2.8/n.
0 1
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Raindrops
Number of drops N is Binomial(n, 2.8/n)
0 1
As n gets larger, the number of drops within the second “approaches” a Poisson(2.8) random variable:
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Expectation and variance of Poisson
If X is Binomial(n, p) then E[X] = np Var[X] = np(1-
p)When p = m/n, we get
E[X] = m Var[X] = m(1-m/n)
As n → ∞, E[X] → m and Var[X] → m. This suggests
When X is Poisson(m), E[X] = m and Var[X] = m.
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Problem for you to solve
Rain falls on you at an average rate of 3 drops/sec.
You walk for 30 sec from MTR to bus stop.
When 100 drops hit you, your hair gets wet.
What is the probability your hair got wet?
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Problem for you to solve
SolutionOn average, 90 drops fall in 30 seconds.
So we model the number of drops N you receive as a Poisson(90) random variable.
Using the online Poisson calculator at or the poissonpmf(n, L) function in 13L07.py we get
P(N > 100) = 1 - ∑i = 0 P(N = i) ≈ 13.49%99