Exponential and Chi-Square Random Variables

Recall Poisson R. V. In a fixed time interval of length T, if there are an average of l arrivals, then “number of arrivals” has a Poisson distribution: where y = 0, 1, 2, … Similarly, given the average number of arrivals per unit time, say in l* arrivals per minute…

Poisson R. V. …then in T minutes, we expect l*T arrivals, and so “number of arrivals” in T minutes has a Poisson distribution: 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. Note l*T = 2 arrivals and so 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 t T 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 l, 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) Sometimes written where b = 1/ l is the average inter-arrival time (e.g., “minutes per arrival”).

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”) b .

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 Distributions for W, time till first arrival: ( using integration-by-parts ) As expected, since average time is 1/0.2 = 5 minutes/arrival.

Exponential mean, variance If W is an exponential random variable with parameter b, the expected value and variance for W are given by Also, note that

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 b = 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 Exponential distributions for some various rates l.

Memoryless Note P(W > w) = 1 – P(W < w) = 1 – (1 – e-lw) = e-lw Consider the conditional probability P(W > a + b | W > a ) = P(W > a + b)/P(W > a) We find that 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 For the exponential, choose a = 1 and note G(1) = 1.

Gamma Density Curves Gamma function facts:

Exponential mean, variance If Y has a gamma distribution with parameters a and b, the expected value and variance for Y are given by In the case of a = 1, the values for the exponential distribution result.

Deriving the Mean By definition of the density function Since this holds for any a > 0, note that

Deriving the Mean Now, consider the expected value

Problem 4.88 Find E(Y) and V(Y) by inspection given that

Chi-Square Distribution As another special case of the gamma distribution, consider letting a = v/2 and b = 2, for some positive integer v. This defines the Chi-square distribution. Note the mean and variance are given by

Get the details on hypothesis testing in MAT 432 in the Spring! Statistical Testing For a sample of size n, with variance s2. To compare against a given value s02 We find that the ratio (n – 1)s2/s02 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