chapter 4

Apr 8, 2023Apr 8, 2023 Dr. B. Mishra (Prob & Stat)Dr. B. Mishra (Prob & Stat) 11

Chapter 4

Continuous Distributions


Continuous Random Variables

In many situations, we come across random variables that take all values lying in a certain interval of the x-axis.

(a) The life length X of a bulb is a continuous random variable that can take all non-negative real values.

Examples


(b) The time between two consecutive arrivals in a queuing system is a random variable that can take all non-negative real values.

(c) The distance R of the point (where a dart hits) (from the centre) is a continuous random variable that can take all values in the interval (0,a) where a is the radius of the board.


It is clear that in all such cases, the probability that the random variable takes any one particular value is meaningless (and is zero). We only ask for the probability that the random variable takes values in an interval. For example, when you buy a bulb, you ask the question? What are the chances that it will work for at least 500 hours? (That is we want the probability that the life length X is 500.)


Definition

(Continuous Random Variable)

X is called a continuous random variable if its range is an interval (or intervals) of real numbers and the probability that it assumes any one specific value is zero.


If X is a continuous random variable, the questions about the probability that X takes values in an interval (a, b) are answered by defining a probability density function.

Probability Density function (pdf)


( ) 0 for alla f x x

( ) 1b f x dx

( ) X .b

ac P a b f x dx

A real function f (x) is called the probability density function of X if

for any two real numbers a, b

DefinitionLet X be a continuous random variable.


Condition (a) is needed as probability is always 0.

Condition (b) says that the probability of the certain event is 1.

Condition (c) says to get the probability that X takes a value between a and b, integrate the probability density function f (x) between a and b.


P(a X b)a b

f(x)

.b

af x dx


1. X X 0a

aP a P a a f x dx

2. X X

X X

P a b P a b

P a b P a b

Remarks

Note: it is immaterial whether we include or exclude one or both the end points.


3. XP x x x f x x

Thus in words, the probability that X takes values in the interval [ x, x + x ] is approximately f (x) times the length of the interval, assuming that it is a small interval.

This is proved using Mean value theorem (of Integral Calculus).


Cumulative Distribution function

X Xx

P x P x f t dt

We denote the above by F(x) and call it the cumulative distribution function (cdf) of X.

If X is a continuous r.v. and if f (x) is its density,


F(x) = P( X x)

- x

f (t)

F (x) = Area under f (t) from t = - to t = x

t

y

x

f t dt


(1) 0 1 for all .F x x

1 2 1 2(2) x x F x F x

Properties of cdf

i.e. F(x) is a non-decreasing function of x.


(3) 0; 1.lim limx x

F F x F F x

(4)xd d

F x f t dt f xdx dx

(Thus we can get density function f (x) by differentiating the distribution function F (x)).

(5) ( )P a X b ( ) ( )F b F a


Problem

If the probability density of a random variable X is given by

2 0 1( )

0

k x xf x

elsewhere

find the value of k and the probability that the random variable takes on a value(a) between ¼ and ¾ ; (b) greater than 2/3 .


Problem

With reference to the preceding exercise, find the corresponding distribution function and use it to determine the probabilities that a random variable having this distribution function will take on a value

(a) greater than 0.8; (b) between 0.2 and 0.4.


ProblemIf the probability density of a random variable is given by for 0 1

( ) 2 for 1 2

0 elsewhere

x x

f x x x

find the probabilities that a random variable having this probability density will take on a value (a) between 0.2 and 0.8; (b) between 0.6 and 1.2.


Problem

With reference to the preceding exercise, find the corresponding distribution function and use it to determine the probabilities that a random variable having this distribution function will take on a value

(a) greater than 1.8; (b) between 0.4 and 1.6.


Exercise 4.1(14a)

Decide whether the following function can be the cdf of a continuous rv X. If yes find the corresponding pdf; if not explain what property fails.

0 1

( ) 1 1 0

1 0

x

F x x x

x


Exercise 4.1(14b)

Decide whether the following function can be the cdf of a continuous rv X. If yes find the corresponding pdf; if not explain what property fails.

2

0 0

0 1/ 2( )

/ 2 1 / 2 1

1 1

x

x xF x

x x

x


Problem

The mileage (in thousands of miles) that car owners get with a certain kind of tire is a random variable having the probability density

/ 201for 0

( ) 200 for 0

xe xf x

x

Expectation Expectation

and and

Distribution ParameterDistribution Parameter



The density of a r.v. X is

23 0 1

0 elsewhere

x xf x

Problem

Find its mean and variance.


The density of a r.v. X is

/ 201

0200 elsewhere

xe xf x

Problem

Find its mean and variance.


Moment Generating Function

Let X be a continuous r.v. with density f (x).We define its mgf as

XX ( ) et txM t E e f x dx

Note that the mgf exists (=is defined) only for those t for which the RHS infinite integral converges.


Properties of MGF

1. For any r.v. X, MX(0) = 1.

2. If X is a r.v. with mgf MX(t) and if Y = aX+b, then the mgf of Y is given by MY(t) = ebt MX(at)

3. If X and Y are independent random variables, then

MX+Y(t) = MX(t) MY(t)


Some simple applications

1. The function

2( )

1g t

t

cannot be the mgf of any random variable because its value at t = 0 is 2 (and NOT 1).

2. If the mgf of X is1

( )1XM t

t

Then the mgf of Y = 3X + 2 is 2 1

( )1 3

tYM t e

t


Let X and Y be independent random variables each having Poisson distribution with parameters and respectively.

Hence mgf of X is MX(t) = ( 1)tee

Hence mgf of Y is MY(t) = ( 1)tee

Hence mgf of X+Y is ( )( 1)tee


Hence we can say that sum of two independent Poisson random variables is also a Poisson random variable.

The parameter of the sum, X + Y, is the sum of the parameters of X and Y.


Exercise: 4.2(15)

Consider the random variable X with density

1( ) , 2 4

6f x x x

(a) Find E[X](b) Find E[X2](c) Find 2, and .


Let X denote the amount in pounds of polyurethane cushioning found in a car. The density for X is given by

1 1( ) , 25 50

ln 2f x x

x

Find the mean, variance and standard deviation of X.

Exercise: 4.2(16)


Exercise: 4.2(17)

Let X denote the length in minutes of a long distance telephone conversation. The density of X is given by

101( ) , 0

10

x

f x e x

(a) Find the mgf of X. (b) Using the mgf find the mean, variance

and standard deviation of X.


Let X be a random variable with Cauchy Distribution. The density for X is given by

2

1 1( ) ,

1f x x

x

Show that E [X] does not exist.

Exercise: 4.2(22)


Let X denote the amount of time in hours that a battery on a solar calculator will operate adequately between exposures to light sufficient to recharge the battery. Assume that the density for X is given by

3

50 1( ) , 2 10

6f x x

x

(a) Verify that this is a valid continuous density.

Exercise: 4.2(23)


(b) Find the expression for the cdf for X and use it to find the probability that a randomly selected solar battery will last at most 4 hours before needing to be recharged.

(c) Find the average time that a battery will last before needing to be recharged.

(d) Find E[X2] and use this to find the variance of X.


Assume that the increase in demand for electric power in millions of kilowatt hours over the next 2 years in a particular area is a random variable whose density is given by

31( ) , 0 4

64f x x x

(a) Verify that this is a valid continuous density.

Exercise: 4.2 (24)


(b) Find the expression for the cdf for X and use it to find the probability that the increase in demand will be at most 2 million kilowatt hours.

(c) If the area has only the capacity to generate an additional 3 million kilowatt hours, what is the probability that the demand will exceed supply?

(d) Find the average increase in demand.


Uniform DistributionGamma DistributionExponential DistributionChi-squared Distribution


The Uniform Distribution

A r.v X is said to have uniform distribution over the interval (, ) if its density is given by

1

0 elsewhere

xf x

Thus the graph of the density is a constant over the interval (, ) .


Graph of the uniform density

f (x)


P c X d

If < c < d < ,

and thus is proportional to the length of the interval ( c, d ).

You may verify that the mean of X is

2

XEX

2

E X

(mid point of the interval ( , ))

1d

cdx

d c


The variance of X is 2

2

12

The cumulative distribution function (cdf) is

0

1

x

xF x x

x


Proof:

2

E X

1x dx

21

2

x

2 21

2

2

E X

( )x f x dx


Proof:

2 2

3

2E X

2 1x dx

31

3

x

3 31

3

2

2

12

2 ( )x f x dx


Hence the variance equals

2 2 2E X

22 2

3 2

2 2 2 22

3 4

2 22

12

2

12


The cdf of the uniform density

Case (ii) < x . In this case f (t) = 0 for

Distribution function is dttfxFx

Case (i) x . In this case f (t) = 0 for all t x.

Hence F(x) = 0.

t and f (t) = 1/( - ) for t x. Hence


1( ) .

x xF x d t

0x

0tf

Case (iii) x > . In this case f(t) = 0 for t or t and f (t) = 1/( - ) for < t < . Hence

1( ) 0 0

xF x d t d t d t

= 1.


Thus the cdf of the uniform density is

0

1

x

xf x x

x


MGF of the uniform density

Let X have the uniform density over the interval (, ). Hence the density of X is

1,f x x

Hence the mgf of X is


MX(t) = E[etX] = txe f x dx

1 txe dx

1 t te et

(Note that MX(0) exists and = 1.)


In certain experiments, the error X made in determining the solubility of a substance is a random variable having the uniform density with = - 0.025 and = 0.025. What are the probabilities that such an error will be

Problem

(b) between –0.012 and 0.012?

(a) between 0.010 and 0.015?


Exercise: 4.1(6)

If a pair of coils were placed around a homing pigeon and a magnetic field was applied that reverses the earth’s field, it is thought that the bird would become disoriented. Under these circumstances it is just as likely to fly in one direction as in any other. Let denote the direction in radians of the bird’s initial flight. is uniformly distributed over the interval [0, 2π].


(a) Find the density for .

HomePigeon

= direction of the initial flight of a homing pigeon measured in radians


(b) Sketch the graph of the density. The uniform distribution is sometimes called the “rectangular” distribution. Do you see why?

(c) Shade the area corresponding to the probability that a bird will come within π/4 radians of home, and find this area using plane geometry.

(d) Find the probability that a bird will orient within π/4 radians of home by integrating the density over the appropriate region(s)


and compare your answer with that obtained in part (c).

(e) If 10 birds are released independently and at least seven orient within π/4 radians of home, would you suspect that perhaps the coils are not disorienting the birds to the extent expected? Explain, based on the probability of this occurring.


From experience, Mr. Harris has found that the low bid on a construction job can be regarded as a r.v. X having uniform density

Problem

where C is his own estimate of the cost of the job. What percentage should Mr. Harris add to his cost estimate when submitting bids to

3 2

24 30 elsewhere

Cx C

f x C


maximize his expected profit?


Gamma Function

1

0

t xe t dt

converges to a finite real number which we denote by (x) (Capital gamma of x).

x

x

If x > 0, it is shown that the improper integral


Thus for all real no x > 0, we define

1

0

( ) t xx e t dt

Properties of Gamma Function

1. 1x x x

2. 1 1

3. 2 1 1 1, 3 2 2 2 1 2!


1 ! for all + ve integers .n n n

More generally

14.

2 .

decreases in the interval (0,1)

,2

a minimum somewhere between 1 and 2.

5. x

and increases in the interval and has

Graph of the Gamma function in the next slide.


The Gamma Distribution

Let , be positive real numbers. A r.v. X is said to have a Gamma Distribution with parameters > 0, > 0 if its density is

. 11

0

0 elsewhere

x

e x xf x


It can be shown that

Mean of X E X

2 2Variance of X .

1MGF of X (1 ) ,t t


Proof : Mean =

Mean of X E X

1 /

0

1

( )xx x e dx

Put /x u

0( )ue u du

( 1)( )

as (+1) = ().


Proof: Variance = 2

2E X 2 1 /

0

1

( )xx x e dx

Put /x u

21

0( )ue u du

2

( 2)( )

2( 1)

Hence Variance = E[X2]-(E[X])2 = 2 .

2 2 2

as (+2) = ( +1) ().

2 2( [ ])E X


Derivation of the MGF:

X(t)=E tXM e 1 /

0

1

( )tx xe x e dx

Put

/x u

(1 ) 1

0

1

( )u te u du

Put 1u t v

1

0

1

( )(1 )ve v dv

t

Assume t < 1/

1(1 ) , ( )t t


Example

In a certain city, the daily consumption of electric power (in millions of kilowatt-hours) can be treated as a random variable X having a Gamma distribution with parameters = 3 and = 2. If the power plant of this city has a daily capacity of 12 million kilowatt-hours, what is the probability that the power supply will be inadequate on any given day?


Problem

If n salespeople are employed in a door-to-door selling campaign, the gross sales volume in thousands of dollars may be regarded as a random variable having the Gamma distribution with = 100 and = ½. If the sales costs are $5,000 per salesperson, how many salespeople should be employed to maximize the expected profit?

n


The Exponential Distribution

If, in the gamma distribution, = 1, we say X has exponential distribution. Thus X has an exponential distribution (with parameter > 0) if its density is

/1

0

0 elsewhere

xe xf x


Graph of the Exponential Density

0

1/f (x)


We also see that

1. Mean of X E X

2 22. Variance of X .

13. MGF of X , 1/

1t

t


4. The cumulative distribution function of X is

/1 0

0 elsewhere

xe xF x


4. X has the memoryless property:

X | X X , , 0P s t s P t s t

Proof of (4):

X 1 XP s P s

/1 sF s e by (3).


X | XP s t s Now

X X

X

P s t s

P s

X

X

P s t

P s

( ) /

/

s t

s

e

e

/te P X t


Example

The amount of time that a surveillance camera will run without having to be reset is a random variable X having exponential distribution with = 50 days. Find the probability that such a camera will

(a) have to be reset in less than 20 days;

(b) will not have to be reset in at least 60 days.


ProblemGiven a Poisson process with on the average arrivals per unit time, find the probability that there will be no arrivals during a time interval of length t, namely, the probability that the waiting time between successive arrivals will be at least of length t.

Let Xt = Number of arrivals during a time interval of length t

Solution


We know that Xt is a random variable having Poisson distribution with parameter t.

Thus, the probability that there will be no arrivals during a time interval of length t

X 0tP te


Given that the switchboard of a consultant’s office receives on an average 0.6 calls per minute (in a Poisson process), find the probabilities that the time between successive calls arriving at the switchboard of the consultant’s office will be

Problem

(a) less than ½ minute;

(b) more than 3 minutes.


The occurrences of blemishes on a continuously produced bolt of cloth follows a Poisson process with an average of 0.01 per meter. Find the probabilities that the length of cloth between two consecutive blemishes is

(a) less than 10 meters.

(b) more than 50 meters.

Problem


Exercise: 4.3 (34)

A particular nuclear plant releases a detectable amount of radioactive gases twice a month on the average. Find the probability that at least 3 months will elapse before the release of the first detectable emission. What is the average time one must wait to observe the first emission?


Problem

Suppose that telephone calls arrive at a switchboard in a Poisson fashion with an average rate of 5 calls per minute. What is the probability that up to a minute will elapse until 2 calls have come in to the switchboard?

Solution

Unit of measurement is minute.Average number of arrivals per unit of measurement (= minute) = = 5.


Hence the time T until 2 arrivals is a random variable having a Gamma distribution with parameters 2 and 1/ i.e. 2 and 1/5.

Hence the density of T is

5 2 1

2

1, 0

2 (1/ 5)tf t e t t

5. . 25 , 0ti e f t te t


Chi-Square Distribution

Definition

Let X be a random variable having Gamma Distribution with parameters = /2, = 2, where is a positive integer. We then say X is a random variable having 2 distribution with degrees of freedom.


The density of a 2 random variable with degrees of freedom is given by

12 2

2

10

22

0 elsewhere

x

e x x

f x



Mean of X E X

2Variance of X 2 .

2

1 1MGF of X ,

2(1 2 )

t

t


2

P (2 > 2 ) =

Graph of 2 Density


If X = 2 is a random variable having 2 distribution with degrees of freedom, then we define

2 (= ) to be that unique number such that

2,

P (2 > 2 ) =

That is 2 is that unique point on the

horizontal axis so that the are under the density curve to the right of it is . We also sometimes write it as to indicate the degrees of freedom.

2ν,αχ


In your book, 1-2 has been tabulated for

various degrees of freedom and for various in Table IV, Appendix A.

For example

0.952 (where = 12) is 5.23

The intersection of row = 12 and (4th) column where 1 – 0.95 = 0.05

0.052 (where = 18) is 28.9


Note that for > 30, 2 is to be approximated

by n + z .

Hence for = 50,

0.052 50 +10 z0.05 = 66.45

2n

Hence for = 60,

0.012 60 + z0.01120 = 60 +

10.9542.33= 85.523


Exercise: 4.3(38)

Consider a chi-squared random variable with 15 degrees of freedom.

(a) What is the mean of What is its variance?

215 ?

Answer: Mean = = 15

Variance = 2 = 30


(b) What is the expression for the density for215 ?

(c) What is the expression for the mgf for215 ?


(d) Use Table IV of Appendix A to find each of the following:

P( 5.23)215

Answer

0.010

P( 30.6)215 0.010

P(6.26 27.5)215 0.975 – 0.025 = 0.950

2 20.05 15,0.05 25.0

2 20.95 15,0.95 7.26


Normal Distribution

A random variable X is said to have a normal distribution (also called Gaussian distribution) with parameters and (a positive number) if its probability density is given by

2

2

( )

2 21( ; , ) ,

2

x

f x e x


Normal Density Curve with mean 10 and s.d. 2

0

0.05

0.1

0.15

0.2

0.25

0 5 10 15 20

x

De

ns

ity


Properties of the Normal DistributionProperties of the Normal Distribution

The total area under the curve is equal to 1The total area under the curve is equal to 1The mean is The mean is and the variance is and the variance is 22

The distribution is bell shaped and The distribution is bell shaped and symmetrical about its mean symmetrical about its mean The curve extends to infinity in both The curve extends to infinity in both directionsdirectionsThe distribution is centered at the meanThe distribution is centered at the meanIt has two points of inflection, namely It has two points of inflection, namely xx = = - - and and xx = = + + . .


Proof : Mean

Put2

xt

21

21( )

2

x

E X x e dx

21Hence (X) ( 2 ) 2

2tE t e dt

2 22

0t te dt t e dt


Proof: variance

2(X )E

Put2

xt

21

2 21( )

2

x

x e dx


Hence 2(X )E

2 22

t tt e e dt

2

20

22 2 12 2

2tt e dt


MGF of the Normal Distribution

XX ( ) et txM t E e f x dx

21

21

2

xt xe e dx

Put2

xy


2( 2 ) 12

2t y ye e dy

Now

2( 2 )1 t y tye e dy

2 2 2 21 1( 2 ) ( )

22y ty y t t

22 2 11 ( )221 y tt t

e e dy


X ( )M t

22 2 11 ( )221 y tt t

e e dy

2 21

21 t te

2 212

X ( ) (all )t t

M t e t

Hence

Thus


A random variable having normal distribution with mean = 0 and s.d. = 1 is said to have a standard normal distribution and usually is denoted by the symbol Z.

Standard Normal Distribution


Standard Normal curve


Let Z have standard normal distribution. Then its cdf,

2 / 21

2

zte dt

cdf of the standard normal distribution

is tabulated for various values of z in Table V.

F(z) = P(Z z)


For example, F(1.96) = P( Z 1.96) = 0.9750

F(2.33) =

0.01

Can you relate F(z) and F(-z) ?

Answer: F(-z) = 1 – F(z).

Also P ( a Z b) = F(b) – F(a)

0.95

0.025

0.99 F(1.645) =

F(-1.96) = F(-2.33) =

Reason: By symmetry of the standard normal curve.


Thus all probability questions about Z can be answered using F (z).

Now we ask:

What is that z for which F(z) = a given value?

For example find z such that F(z) = 0.8438.

Answer: z = 1.1

In particularP ( - a Z a) = F(a) – F(-a) = 2F(a) –1


Notation: Given , we denote by z that unique z such that P( Z > z ) = . (That is for which P( Z z ) = 1- .)

Thus

It is important to note that z1- = - z

We can now relate the probabilities that a normal random variable X will have in terms of F(z).

z0.05 = 1.645,

z0.01 = 2.33, z0.95 = - 1.645


We note that if X is a random variable having a normal distribution with mean and s.d. , then the random variable

has a standard normal distribution.

XZ

Z is called the random variable obtained by “standardizing” X.


Thus P ( a X b)

.b a

F F

Za b

P


Given a random variable having the normal distribution with mean = 16.2 and variance 2 = 1.5625, find the probabilities that it will take on a value

(c) between 13.6 and 18.8;

Problem

(b) less than 14.9;

(d) between 16.5 and 16.7.

(a) greater than 16.8;

Exercise: 4.4 (41)Exercise: 4.4 (41)Most galaxies take the form of a flattened disc, with Most galaxies take the form of a flattened disc, with the major part of the light coming from this very thin the major part of the light coming from this very thin fundamental plane. The degree of flattening differs fundamental plane. The degree of flattening differs from galaxy to galaxy. In the milky way Galaxy from galaxy to galaxy. In the milky way Galaxy most gases are concentrated near the centre of the most gases are concentrated near the centre of the fundamental plane. X is normally distributed with fundamental plane. X is normally distributed with mean 0 and standard deviation 100 parsecs ( A mean 0 and standard deviation 100 parsecs ( A parsec is equal to approximately 19.2 trillion miles).parsec is equal to approximately 19.2 trillion miles).

112112Dr. B. Mishra (Prob & Stat)Dr. B. Mishra (Prob & Stat)Apr 8, 2023Apr 8, 2023

(a)(a) Sketch a graph of the density for X. Indicate on this Sketch a graph of the density for X. Indicate on this graph the probability that a gaseous mass is located graph the probability that a gaseous mass is located within 200 parsecs of the centre of the fundamental within 200 parsecs of the centre of the fundamental plane. Find this probability.plane. Find this probability.

(b)(b) Approximately, what percentage of the gaseous Approximately, what percentage of the gaseous masses are located more than 250 parsecs from the masses are located more than 250 parsecs from the centre of the plane?centre of the plane?

(c)(c) What distance has the property that 20% of the What distance has the property that 20% of the gaseous masses are at least this far from the gaseous masses are at least this far from the fundamental plane?fundamental plane?

(d)(d) What is the moment generating function for X.What is the moment generating function for X.

-100 -50 50 100

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

0.0007


Exercise: 4.4(43)Exercise: 4.4(43)Let X denote the time in hours needed to locate Let X denote the time in hours needed to locate and correct a problem in the software that governs and correct a problem in the software that governs the timing of traffic lights in the downtown area of the timing of traffic lights in the downtown area of a large city. Assume X is normally distributed a large city. Assume X is normally distributed with mean 10 hours and variance 9.with mean 10 hours and variance 9.

(a) Find the probability that the next problem will (a) Find the probability that the next problem will requir at most 15 hours to find and correct.requir at most 15 hours to find and correct.

(b) The fastest 5% of repairs take at most how (b) The fastest 5% of repairs take at most how many hours to complete?many hours to complete?


Exercise: 4.4(44)Exercise: 4.4(44)Assuming that during seasons of normal rainfall Assuming that during seasons of normal rainfall the water level in feet at a particular lake follows the water level in feet at a particular lake follows a normal distribution with mean of 1876 feet and a normal distribution with mean of 1876 feet and standard deviation of 6 inches.standard deviation of 6 inches.

(a) During such a season, would it be unusual to (a) During such a season, would it be unusual to observe a water level of at most 1875 feet.observe a water level of at most 1875 feet.

(b) Suppose that the water will crest the spillway (b) Suppose that the water will crest the spillway if the water level exceeds 1878 feet. What is the if the water level exceeds 1878 feet. What is the probability that this will occur during a season of probability that this will occur during a season of normal rainfall?normal rainfall?



Exercise: 4.5 (48)

The number of Btu’s of petroleum and petroleum products used per person in the US in 1975 was normally distributed with mean 153 million Btu’s and standard deviation of 25 million Btu’s. Approximately, what percentage of the population used between 128 and 178 million Btu’s during that year? Approximately what percentage of the population used in excess of 228 million Btu’s?

Log-Normal DistributionLog-Normal Distribution

The log-normal distribution is the The log-normal distribution is the distribution of a random variable whose distribution of a random variable whose natural logarithm follows a normal natural logarithm follows a normal distribution. Thus if X is a normal random distribution. Thus if X is a normal random variable, then Y=evariable, then Y=eXX follows a log-normal follows a log-normal distribution.distribution.


Exercise: 4.4 (45)Exercise: 4.4 (45)

Let X be normal with mean Let X be normal with mean and variance and variance 22. Let . Let G denote the cumulative distribution function for G denote the cumulative distribution function for Y = eY = eX X and let F denote the cumulative distribution and let F denote the cumulative distribution for X. for X.

a)a) Show that G(y) = F(lny)Show that G(y) = F(lny)

b)b) Show that G’(y) =F’(lny)/y Show that G’(y) =F’(lny)/y

c)c) Show that the density for Y is given byShow that the density for Y is given by

.0,);,(,2

1)(

2

2

2

ln

yey

ygy


Now, G(y) = P(YNow, G(y) = P(Yy) = P(ey) = P(eX X y) = P(X y) = P(X lny) lny)

.0);,(,,2

1)ln()(

ln2 2

2

xdxeyXPyGy x

x

y y

eywhere

yydeyGor

.0,);,(),(ln2

1)(,

0

2

ln2

2

2

2

ln

2

0

1, ( ) (ln ) (ln )

2

yy

or G y e d y F y


Hence, show that the density for Y is given Hence, show that the density for Y is given

g(y) = G’(y)g(y) = G’(y)

.0,);,(,2

1)(

2

2

2

ln

yey

ygy

)(ln)(ln2

1)(

2

2

2

ln

yFydeyGy y



In a photographic process, the developing time of prints may be looked upon as a random variable having normal distribution with mean 16.28 seconds and s.d. 0.12 second. Find the probability that it will take

Problem

(a) anywhere from 16.00 to 16.50 seconds to develop one of the prints;

(b) at least 16.20 seconds to develop one of the prints;


(c) at most 16.35 seconds to develop one of the prints.


With reference to the preceding exercise, for which value is the probability 0.95 that it will be exceeded by the time it takes to develop one of the prints?

i.e. Find c such that P ( X > c) = 0.95

Problem

Normal Approximation to the Binomial Normal Approximation to the Binomial DistributionDistribution

Let X be binomial with parameters Let X be binomial with parameters n and p. For large n, X is n and p. For large n, X is approximately normal with mean approximately normal with mean np and variance np(1-p);np and variance np(1-p);

i.e.i.e. )1(,~ pnpnpNX


For most practical purposes the For most practical purposes the approximations is acceptable for approximations is acceptable for values of n and p such that either values of n and p such that either

p≤.5 and n.p>5p≤.5 and n.p>5

oror p>.5 and n.(1-p)>5p>.5 and n.(1-p)>5


Correction for continuity:Correction for continuity:1.P(a < X ≤ b)=P(a-0.5 < X ≤ b+0.5)1.P(a < X ≤ b)=P(a-0.5 < X ≤ b+0.5)

2. P(X ≤ b)= P(-∞ < X ≤ b) = P(-∞ < X ≤ b+0.5)2. P(X ≤ b)= P(-∞ < X ≤ b) = P(-∞ < X ≤ b+0.5)

= P(X ≤ b+0.5)= P(X ≤ b+0.5)

3. P(X ≥ a)= P(a ≤ X < ∞)= P(a-0.5 ≤ X < ∞)3. P(X ≥ a)= P(a ≤ X < ∞)= P(a-0.5 ≤ X < ∞)

=P( X ≥a-.5)=P( X ≥a-.5)

4. P(X = a)=P(a-0.5 < X < a+0.5)4. P(X = a)=P(a-0.5 < X < a+0.5)

The number 0.5 is called the half-unit The number 0.5 is called the half-unit correction factor for continuity.correction factor for continuity.



A manufacturer knows that, on average, 2% of the electric toasters that he makes will require repairs within 90 days after they are sold. Use normal approximation to the binomial distribution to determine the probability that among 1200 of these toasters at least 30 will require repairs within the first 90 days after they are sold?

Problem


A safety engineer feels that 30% of all industrial accidents in her plant are caused by failure of employees to follow instructions. Find approximately the probability that among 84 industrial accidents in this plant anywhere from 20 to 30 (inclusive) will be due to failure of employees to follow instructions.

Problem

(Normal Approximation to the Poisson Distribution)(Normal Approximation to the Poisson Distribution)

Let X be Poisson with parameter Let X be Poisson with parameter s.Then s.Then for large values of for large values of s, X is approximately s, X is approximately normal with mean normal with mean s and variance s and variance s.s.

,.......3,2,1,0,!

)(

xx

xsse

xf


Exercise:4.6 (56)Exercise:4.6 (56)

Let X be a poisson random variable with Let X be a poisson random variable with parameter parameter s = 15. Find P[X ≤ 12] from the s = 15. Find P[X ≤ 12] from the table. Approximate this probability using a table. Approximate this probability using a normal curve. Be sure to employ the half-normal curve. Be sure to employ the half-unit correction factor. unit correction factor.


Exercise:4.6 (57)Exercise:4.6 (57)The average number of jets either The average number of jets either arriving at or departing from O’Hare arriving at or departing from O’Hare Airport is one every 40 seconds. What Airport is one every 40 seconds. What is the approximate probability that at is the approximate probability that at least 75 such flights will occur during a least 75 such flights will occur during a randomly select hour? What is the randomly select hour? What is the probability that fewer than 100 flights probability that fewer than 100 flights will take place in an hour?will take place in an hour?


Normal Probability Rule

Let X be normally distributed with parameters mean µ and standard deviation .Then

P[- < X-µ < ]=0.68

P[-2 < X-µ < 2 ]=0.95

P[-3 < X-µ < 3 ]=0.997


Let X be a r.v. having mean and s.d. . Then for any positive number k, the probability of getting a value of X which deviates from by at least k is at most 1/k2.

In symbols

2

1| |P X k

k

Chebyshev’s Inequality

Apr 8, 2023Apr 8, 2023 137137Dr. B. Mishra (Prob & Stat)Dr. B. Mishra (Prob & Stat)

Proof

-k +k

R1R2

R3


We divide the range of X into three regions

R1, where X -k

R2, where -k < X < +k

R3, where X +k

We assume that X is a continuous random variable with density f (x).


Variance of X =2 2( ) ( )x f x dx

2 2

2

( ) ( ) ( ) ( )

( ) ( )

k k

k

k

x f x dx x f x dx

x f x dx


Neglecting the middle term which is 0, we get

In the region R1,i.e. (-, - k )

x -k or x- -k and so (x - )2 k22 In the region R3, i.e. (-, + k ),

x +k or x- k and so (x - )2 k22

2 2 2( ) ( ) ( ) ( )k

k

x f x dx x f x dx


Hence

Or

Now

and


2 2 2 2 2( ) ( )k

k

k f x dx k f x dx

21 ( ) ( )k

k

k f x dx f x dx

( ) ( )k

P X k f x dx

( ) ( )k

P X k f x dx

Hence we get

Or


21 ( ) ( )k P X k P X k

2

1( )P X k

k

2k P X k


If X is a discrete random variable, the above proof holds by replacing integral by summation:

i.e. replace by etc.k

k

The event { |X - | < k } is complementary to the event { |X - | k }. Hence we can also say that

Putting k = , we can also say that

2

1| | 1P X k

k

2

2 2

Var| |

XP X



Normal Distribution and Chebyshev’s inequality

Let X be a random variable with mean and s.d. . Then by Chebyshev’s equality,

( 3 )P X 11 0.8999

9

But from normal tables, we find

( 3 )P X (| | 3)P Z

2 (3) 1F 0.9974


Then by Chebyshev’s inequality,

( 2 )P X 11 0.75

4


( 2 )P X (| | 2)P Z 0.9544

( )P X 11 0.00

1


( )P X (| | 1)P Z 0.6826

By Chebyshev’s inequality,

In 1 out of 6 cases, material for bulletproof vests fail to meet puncture standards. If 405 specimens are tested, what does Chebyshev’s inequality tell us about the probability of getting at most 30 or at least 105 cases that do not meet puncture standards?

Problem


Show that for 1 million flips of a balanced coin, the probability is at least 0.99 that the proportion of heads will fall between 0.495 and 0.505.

Problem


How many times do we have to flip a balanced coin to be able to assert with a probability of at most 0.01 that the difference between the proportion of tails and 0.50 will be at least 0.04?

Problem



Weibull Distribution

A r.v X is said to have a Weibull distribution with parameters >0, > 0, and if its density is given by

1 ( )( ) ,xf x x e x



1/Mean of X E X (1 1/ )

2Variance of X

22/ (1 2 / ) (1 1/ )


Proof: Mean =

Mean of X E X

1 ( )( ) xx x e dx

Put

( )x u

1/

0

u ue du

1/ (1 1/ )

1/ (1 1/ )


Variance:

2E X 12 ( )xx x e dx

Put

( 1)x u

Hence Variance

2/ 1/2

0

2 uu ue du

2/ 1/ 2(1 2 / ) 2 (1 1/ )

22/ (1 2 / ) (1 1/ )

Transformation of VariablesTransformation of VariablesLet X be a continuous random variable with Let X be a continuous random variable with density . Let , where is strictly density . Let , where is strictly monotonic and differentiable. The density for monotonic and differentiable. The density for Y Y is denoted by and is given byis denoted by and is given by


Monotonic functionsMonotonic functionsMonotonic functions are functions that tend to Monotonic functions are functions that tend to move in only one direction as x increases.move in only one direction as x increases.A monotonic increasing function always increases A monotonic increasing function always increases as x increases i.e. f(a) > f(b), for all a > b.as x increases i.e. f(a) > f(b), for all a > b.A monotonic decreasing function always A monotonic decreasing function always decreases as x increases i.e. f(a)<f(b), for all a>b.decreases as x increases i.e. f(a)<f(b), for all a>b.In calculus speak, In calculus speak, A monotonic decreasing function’s derivative is A monotonic decreasing function’s derivative is always negative.always negative.A monotonic increasing function’s derivative is A monotonic increasing function’s derivative is always positive.always positive.


ExamplesExamples

(1) Let X be a random variable with density (1) Let X be a random variable with density

Let Let . Find the density for Y.. Find the density for Y.

(2) Let X be uniformly distributed over (0,4) and (2) Let X be uniformly distributed over (0,4) and let .Find the density for let .Find the density for Y.Y.


Problem-70Problem-70

Let X be a random variable with densityLet X be a random variable with density

Let Let . .

(a)(a)Find E[X] and use the rules for Find E[X] and use the rules for expectation to find E[Y].expectation to find E[Y].

(b)(b)Find the density for Find the density for YY..

(c)(c)Use the density for Y to find E[Y].Use the density for Y to find E[Y].



Let Let XX be a random variable with density be a random variable with density

Let Let . Find the density for . Find the density for Y.Y.



Let Let XX be a random variable with density be a random variable with density

Let Let . Find the density for . Find the density for YY..


Problem-74Problem-74Let X denote the velocity of a random gas Let X denote the velocity of a random gas molecule. According to the Maxwell-Boltzmann molecule. According to the Maxwell-Boltzmann law, the density for X is given bylaw, the density for X is given by

Here Here c c is a constant that depends on the gas is a constant that depends on the gas involved, and involved, and ββ is a constant whose value depends is a constant whose value depends on the mass of the molecule and its absolute on the mass of the molecule and its absolute temperature. The kinetic energy of the molecule, temperature. The kinetic energy of the molecule, YY, is given by , is given by YY = (1/2)m = (1/2)mXX22 , where m > 0. Find , where m > 0. Find the density for the density for YY..



Exercise:4.8(75)

Let X be a continuous random variable with density fX (x). Let Y = X2. Then the density of Y is given by

1( ) ( ) ( ) , 0

2Y X Xf y f y f y y

y


ProofLet y > 0.

( X )y y

Then the cdf of Y = G(y) = P( Y y)

(F (x) is the cdf of X).

( ) ( )F y F y

Hence the density of Y = g (y)

( )d

G ydy

1( ) ( ) , 0

2X Xf y f y y

y


Exercise:4.8(76d)

Let Z have standard normal distribution. Then the random variable Y = Z2 has a chi-square distribution with one degree of freedom.

Problem-77Problem-77Let X be normally distributed with mean µ Let X be normally distributed with mean µ and variance and variance σσ22. Let Y = e. Let Y = eXX. Show that Y . Show that Y follows the log-normal distribution.follows the log-normal distribution.



Exercise:4.8(78)

Let Z be a standard normal variable and let Y = 2Z2 – 1. Find the density of Y.

chapter 4

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