Exponential and Chi-Square Random Variables

Recall Poisson R. V.

• In a fixed time interval of length T, if there are an average of arrivals, then “number of arrivals” has a Poisson distribution:

( )!

yep y

y

where y = 0, 1, 2, …

Similarly, given the average number of arrivals per unit time, say in * arrivals per minute…

Poisson R. V.

• …then in T minutes, we expect *T arrivals, and so “number of arrivals” in T minutes has a Poisson distribution:

**( )( )

!

y TT ep y

y

where y = 0, 1, 2, …

Consider the time between arrivals. That is, consider the “inter-arrival times”.

0.2 arrivals per minute

• If customers arrive at an average of 0.2 arrivals per minute, find the probability of 3 arrivals in a 10-minute period.

3 2(2)(3)

3!

ep

• Note *T = 2 arrivals and so

0 22(2)

(0)0!

ep e

• Find the probability of no arrivals in the 10-minute period.

Time till arrival?

• Consider W, the time until the first arrival.

Number of customers

Tt

• W is a continuous random variable. What can we say about its probability distribution?

Inter-arrival Distribution

• Note that F(w) = P(W < w) = 1 – P(W > w)

Number of customers

w t

• Time of first arrival W > w implies zero arrivals have occurred in the interval (0, w).

• Don’t we already know this probability?

Inter-arrival times

• If the average arrivals per unit time equals , the probability that zero arrivals have occurred in the interval (0, w) is given by the Poisson distribution

F(w) = P(W < w) = 1 – P(W > w)0( )

1 (0) 1 10!

www e

p e

Sometimes writtenwhere = 1/ is the average inter-arrival time(e.g., “minutes per arrival”).

/( ) 1 wF w e

Exponential Distribution

• A continuous random variable W whose distribution and density functions are given by

and

is said to have an exponential distribution with parameter (“average”)

/1 , 0( )

0, otherwise

we wF w

/1, 0

( )

0, otherwise

we wf w

Exponential Random Variables

• Typical exponential random variables may include:

• Time between arrivals (inter-arrival times)• Service time at a server (e.g., CPU, I/O device, or

a communication channel) in a queueing network.• Time to failure (“lifetime”) of a component.

0.2 arrivals per minute

0 5 10 15

0.1

0.2

lambda = 0.2

0 2 40

0.2

0.4

dgamma x 2.5( )

xcumulative distribution

0 5 10 15

0.5

1

0.2

0( ) (0.2 )wE W w e dw

As expected, since average time is 1/0.2 = 5 minutes/arrival.

( using integration-by-parts )

Distributions for W, time till first arrival:

5 minutes

Exponential mean, variance

• If W is an exponential random variable with parameter the expected value and variance for W are given by

2( ) and ( )E W V W

Also, note that

2 2( ) 2 , and in general,

( ) ( !)n n

E W

E W n

Problem 4.74

• Air samples in a city have CO concentrations that are exponentially distributed with mean 3.6 ppm.

• For a randomly selected sample, find the probability the concentration exceeds 9 ppm.

• If the city manages its traffic such that the mean CO concentration is reduced to 2.5 ppm, then what is the probability a sample exceeds 9 ppm?

Problem 4.82

• The lifetime of a component is exponentially distributed with an average = 100 hours/failure.

• Three of these components operate independently in a piece of equipment and the equipment fails if at least 2 of the components fail.

• Find the probability the equipment operates for at least 200 hours without failure.

Density Curves

0 5 10 15

0.5

1

dexp x 0.2( )

dexp x 0.5( )

dexp x 1( )

x

f(y) e -

0 2 40

0.2

0.4

dgamma x 2.5( )

x

Exponential distributions for some various rates .

( ) yf y e

1where = is average inter-arrival time.

Memoryless

• Note P(W > w) = 1 – P(W < w)

= 1 – (1 – e-w) = e-w

• Consider the conditional probability P(W > a + b | W > a ) = P(W > a + b)/P(W > a)

• We find that

( )

( | )

( )

a bb

a

P W a b W a

ee

eP W b

The only continuous memoryless random variable.

Gamma Distribution

• The exponential distribution is a special case of the more general gamma distribution:

where the gamma function is

1 /

, 0( ) ( )

0, otherwise

yy ey

f y

1

0( ) yy e dy

For the exponential, choose = 1 and note (1) = 1.

Gamma Density Curves

0 5 10 15

0.5

1

dexp x 0.2( )

dexp x 0.5( )

dexp x 1( )

x

0 2 40

0.5

1

dgamma x 1( )

dgamma x 2( )

dgamma x 3( )

x

the shape parameter,

0 5 10 15

0.5

1

Gamma function facts:(1) 1;

( ) ( 1) ( 1), 1;

( ) ( 1)!, .n n n Z

Exponential mean, variance

• If Y has a gamma distribution with parameters and the expected value and variance for Y are given by

2( ) and ( )E Y V Y

In the case of = 1, the values for the exponential distribution result.

Deriving the Mean

• By definition of the density function1 /

0

1 /

0

1 ( )( )

and so ( )

y

y

y ef y dy dy

y e dy

• Since this holds for any > 0, note that

1 /

0 ( 1) yy e dy

Deriving the Mean

• Now, consider the expected value1 /

0

/

0

( ) ( )( )

1

( )

y

y

y eE Y y f y dy y dy

y e dy

11

= ( 1)( )

( ) = , as claimed.

( )

Problem 4.88

• Find E(Y) and V(Y) by inspectiongiven that

2 24 , 0( )

0, otherwise

yy e yf y

Chi-Square Distribution• As another special case of the gamma distribution, consider letting = v/2 and = 2, for some positive integer

v.

This defines the Chi-square distribution. Note the mean and variance are given by / 2 1 / 2

/ 2, 0

( ) 2 ( / 2)

0, otherwise

v y

v

y ey

f y v

2( ) ( / 2)(2) , ( ) 2E Y v v V Y v

Statistical Testing

• For a sample of size n, with variance s2.

• To compare against a given value 02

• We find that the ratio (n – 1)s2/02 has the chi-square

distribution with v = n – 1 degrees of freedom.• Develop and test the “null hypothesis” based on the

chi-square probability distribution.

Get the details on hypothesis testing in MAT 432 in the Spring!

Practice Problems

• Work problems:4.69, 4.71, 4.73, 4.77, 4.78, 4.81, 4.83