1. Exponential and Chi-Square Random Variables

2. 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…

3. 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”.

4. 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.

5. 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?

6. 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?

7. 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”).

8. 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 .

9. 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.

10. 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.

11. 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

12. 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?

13. 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.

14. Density Curves
Exponential distributions for some various rates l.

15. 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.

16. 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.

17. Gamma Density Curves
Gamma function facts:

18. 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.

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

20. Deriving the Mean
Now, consider the expected value

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

22. 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

23. 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!

24. Practice Problems
Work problems: 4.69, 4.71, 4.73, 4.77, 4.78, 4.81, 4.83