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Chapter 4. Continuous Probability Distributions

4.1 The Uniform Distribution4.2 The Exponential Distribution4.3 The Gamma Distribution4.4 The Weibull Distribution4.5 The Beta Distribution

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4.1 The Uniform Distribution4.1.1 Definition of the Uniform Distribution(1/2)

• It has a flat pdf over a region.– if X takes on values between a and b, and

– Mean and Variance

( , ),X U a b

1( ) ,

( ) ,

f x for a x bb a

d cP c X d

b a

for a c <d b

2

( )2

( )( )

12

a bE X

b aV X

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Figure 4.1 Probability density function of aU(a, b) distribution

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4.2 The Exponential Distribution4.2.1 Definition of the Exponential Distribution

• Pdf:

• Cdf:

• Mean and variance:

( ) 0xf x e for x ,

( ) 1 0xF x e for x ,

0, otherwise

2

1 1( ) ( )E X and V X

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Figure 4.3

Probability density function of an exponential distribution with parameter = 1

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• The exponential distribution often arises, in practice, as being of the amount of time until some specific event occurs.

For example, the amount of time until an earthquake occurs, the amount of time until a new war breaks out, or the amount of time until a telephone call you receive turns

out to be a wrong number, etc.

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4.2.2 The memoryless property of the Exponential Distribution

• For any non-negative x and y

• The exponential distribution is the only continuous distribution that has the memoryless property

( | ) ( )P X x y X x P X y

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• Memoryless property of exponential random variable:

This is equivalent to

When X is an exponential random variable,

The memoryless condition is satisfied since

{ | } { }P X x y X x P X y

{ , } { } { } { }P X x y X x P X x y P X x P X y

{ } xP X x e

( )x y x ye e e

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• Example: Suppose that a number of miles that a car run before its battery wears out is exponentially distributed with an average value of 10,000 miles. If a person desires to take a 5,000 mile trip, what is the probability that he will be able to complete his trip without to replace the battery?

(Sol) Let X be a random variable including the remaining lifetime

(in thousand miles) of the battery. Then,

What if X is not exponential random variable?

That is, an additional information of t should be known.

[ ] 1/ 10E X 1/10 5 1/ 2{ 5} 1 (5) 0.604P X F e e

1 ( 5){ 5 | }

1 ( )

F tP X t X t

F t

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• Proposition: If are independent exponential random variables having respective parameters , then is the exponential random variable with

parameter

(proof)

1 2, , , nX X X1 2, , , n

1 2min( , , , )nX X X

1

n

ii

1 2 1 2

1

1

1

{min( , , , )} { , , , }

{ }

exp( )

exp( )

n n

n

ii

n

ii

n

ii

P X X X P X x X x X x

P X x

x

x

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• Example: A series system is one that all of its components to function in order for the system itself to be functional. For an n component series system in which the component lifetimes are independent exponential random variables with respective parameters .

What is the probability the system serves for a time t? (Sol) Let X be a random variable indicating the system lifetime. Then, X is an exponential random variable with parameter Hence,

1 2, , , n

1

n

ii

1

{ } exp( )n

ii

P X t t

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4.2.3 The Poisson process

• A stochastic process is a sequence of random events

• A Poisson process with parameter is a stochastic process where the time (or space) intervals between event-

occurrences follow the Exponential distribution with parameter .

• If X is the number of events occurring within a fixed time (or space) interval of length t, then

( )X P t

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Figure 4.7 A Poisson process. The number of events occurring in a time interval of length t has a Poisson distribution with mean t

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• The Poisson Process Suppose that events are occurring at random time points

and let N(t) denote the number of points that occur in time

interval [0,t]. Then, A Poisson process having rate is defined if (a) N(0)=0, (b) the number of events that occur in disjoint time

intervals are independent, (c) the distribution of N(t) depends only on the length of

interval, (d) and (e)

( 0)

{ ( ) 1}lim ,h

P N h

h

{ ( ) 2}lim 0.h

P N h

h

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• Let us break the interval [0,t] into n non-overlapping subintervals each of length t/n. Now, there will be k events in [0,t] if either

(1) N(t) equals to k and there is at most one event in each subinterval, or

(2) N(t) equals to k and at least one of subintervals contain 2 or more events. Then,

Since the number of events in the different subintervals are

independent, it follows that

{ \1\ \ \ \ } \t

P exactly event in a subin andn

{0 \ \ \ \ } 1 .t

P event in a subinn

{ \ \ \ \ \1\ \ \

\ \ \ \ 0 \ } 1 .k n k

P k of subin contain exactly event and

n t tthe other n k contain event

k n n

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Hence, as n approach infinity,

That is, the number of events in any interval of length t has a Poisson distribution with mean

For a Poisson process, let denote the time of the first event and for n>1, denote the elapsed time between the (n-1)st and nth event. Then, the sequence

is called the sequence of inter-arrival times.

( ){ ( ) } , \ 0,1,

!

kt t

P N t k e kk

.t

1XnX

{ , 1,2, }nX n

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The distribution of The event takes place if and only if no events of the Poisson process occur in [0,t] and thus,

Likewise,

Repeating the same argument yields are independent exponential random variables with mean

:nX

1{ }X t

1{ } { ( ) 0} tP X t P N t e

2 1 1{ | } {0 \ \ \ ( , ] | }

{0 \ \ \ ( , ]}

.t

P X t X s P event in s s t X s

P event in s s t

e

1 2, ,X X 1/ .

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• Example 32 (Steel Girder Fractures, p.209)– 42 fractures on average on a 10m long girder between-fracture length: 10/43=0.23m on average

– If the between-fracture length (X) follows an exponential distribution, how would you define the gap?

– How would you define the number of fractures (Y) per 1m steel girder?

– How are the Exponential distribution and the Poisson distribution related?

X E( ) with =43/10

( )Y P

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Figure 4.9 Poisson process modeling fracturelocations on a steel girder

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• P(the length of a gap is less than 10cm)=?

• P(a 25-cm segment of a girder contains at least two fractures)=?

4.3 0.1{ 0.1} 1 0.35P X e

1.075 0 1.075 1

0.25 4.3 0.25 1.075,

{ 2} 1 { 0} { 1}

1.075 1.0751

0! 1!1 0.341 0.367 0.292

P Y P Y P Y

e e

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Figure 4.10 The number of fractures in a 25-cm segment of the steel girder has a Poisson distribution with mean 1.075

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4.3 The Gamma Distribution4.3.1 Definition of the Gamma distribution

• Useful for reliability theory and life-testing and has several important sister distributions

• The Gamma function:

• The Gamma pdf with parameters k>0 and >0:

• Mean and variance:

1

0

( ) 0k xk x e dx for k

1( )

( )( )

k xx ef x

k

for x 0

2( ) / ( ) /E X k and V X k

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Curves of Gamma pdf

1

( )( )

k k xx ef x

k

for x 0

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• Calculation of Gamma function

When k is an integer,

When k=1, the gamma distribution is reduced to the exponential

with mean

1 1

0 0

( ) ( ) \ \( \ \ ).x k y kk e x dx e y dy by letting y x

1 1 2

0

0 0

( ) | ( 1) ( 1) ( 1)y k y k y kk e y dy e y e k y dy k k

0

( ) ( 1)! (1) // // (1) 1

\ , \ \ ( ) ( 1)!

yk k and e dy

That is k k

1/ .

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• Properties of Gamma random variables If are independent gamma random variables

with respective parameters then

is a gamma random variable with parameters

• The gamma random variable with parameters is equivalent to the exponential random variable with parameter

• If are independent exponential random variables, each having rate , then

is a gamma random variable with parameters

, 1, ,iX i n ( , ),ik

1

n

ii

X

1

( , ).n

ii

k (1, )

.

, 1, ,iX i n

1

n

ii

X

( , ).n

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4.3.2 Examples of the Gamma distribution

• Example 32 (Steel Girder Fractures) :

• Y= the number of fractures within 1m of the girder

1

1

, ,

( , ) ( ?)

k

k

i

X X with

X G k why

i

iid E( )

then

X

( )

( 1) ( ) ( ?)

Then

Y Poisson

and

P X P Y k why

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Figure 4.15 Distance to fifth fracture has agamma distribution with parameters k = 5 and = 4.3

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5

1

( ), \ 4.3,

, \ \

\ . . \ \ 5 \ \ 4.3.

5[ ] 1.16[ ]

4.3{ 1} (1) 0.4296 \ ( \ \ )

i

ii

X E

Then X X is

Gamma r v with k and

kE X m

P X F from numerical computation

\ ( ). \ , \ { 5} 0.4296Let Y P Then P Y

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4.4 Weibull distribution4.4.1 Definition of the Weibull distribution

• Useful for modeling failure and waiting times (see Examples 33 & 34)

• The pdf:

• Mean and variance:

1 ( )( ) ( ) 0, 0aa xf x a x e for a

2

2

1 1( ) (1 )

1 2 1( ) (1 ) (1 )

E Xa

V Xa a

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Curves of the Weibull distribution

1 ( )( ) ( )aa xf x a x e

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The c.d.f. of Weibull distribution:

The pth quantile of Weibull distribution:Find x such that

( )( ) 1axF x e

( ) .F x p( )

( )

1/

1

1

( ) ln(1 )

( ln(1 ))

a

a

x

x

a

a

e p

e p

x p

px

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4.5 The Beta distribution

• Useful for modeling proportions and personal probability (See Examples 35 & 36.)

• Pdf:

• Mean and variance:

1 ( 1)( )( ) (1 )

( ) ( )a ba b

f x x xa b

( )

1( )

1

aE X

a ba b

V Xa b a b a b

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Figure 4.21 Probability density functions of the beta distribution

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Figure 4.22 Probability density functions of the beta distribution

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• Exponential and Weibull random variables have as their set of possible values.

• In engineering applications of probability theory, it is occasionally helpful to have available family of distributions whose set of possible values is finite interval.

• One of such family is the beta family of distributions.

(0, )

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• Example: the beta distribution and rainstorms. Data gathered by the U.S. Weather Service in Alberquerque,

concern the fraction of the total rainfall falling during the first 5 minutes of storms occurring during both summer and nonsummer seasons. The data for 14 nonsummer storms can be described reasonably well by a standard beta distribution with a=2.0 and b=8.8.

• Let X be the fraction of the storm’s rainfall falling during the first 5 minutes. Then, the probability that more than 20% of the storm’s rainfall during the first 5 minutes is determined by 1.0

1.0 7.8

0.2

(2.0 8.8){ 0.2} (1 ) (86.24)(0.0045) 0.39.

(2.0) (8.8)P X u u du