This activity introduces the Gamma function and distribution and shows how it is the parent of the exponential distribution which measures the time to the first occurrence and the time between occurrences in a Poisson process. Learning about exponential random variables helps complete our study of continuous random variables. Because of its ability to measure arrival times, the exponential distribution is very important when we try to simulate a system using a computer in order to predict its behavior and to forecast the consequences of important business decisions. Simulation is one of the most important applications of probability and statistics in business and industry.
oo
__ {
| (a) = / xa-1 e-x dx
}
0
__ __ __
By algebraic manipulation we see that | (a+1) = a| (a) = a(a-1)| (a-1) ...
__ __
and | (1) = 1. Thus, | (n) = (n-1)!.
This shows that the gamma function is a continuous extension to the factorial function.
Another interesting property of the gamma function is its value when a = 1/2. It can be shown that
__
| (1/2) = \/pi.
For our purposes, the most
important
aspects of the gamma
function are the probability
distributions it generates.
{ 1
| ---__ xa-1 e-x/B x > 0
f(x) = | Ba | (a)
|
{ 0 elsewhere
Its moment generating function can be computed as follows:
oo
{ 1
MX(t) = / etx ---__ xa-1 e-x/B
} Ba | (a)
0
oo
1 {
= ---__ / xa-1 e-x(1-tB)/B dx
Ba | (a)}
0
If we let y = x(1 - tB) then xa-1 becomes (y/(1 - tB))a-1 and dx becomes dy/(1 - tB). Substituting we get
oo
1 { 1
MX(t) = --------- / ---__ ya-1 e-y/B dy
(1 - tB)a } Ba | (a)
0
= (1 - tB)-a
The last integral is 1 since it is the integral of the density function for a gamma random variable over its full range of values.
Using the moment generating function we can verify that
the expected value and variance
of the gamma
distribution are aB and aB2 respectively.
{ 1
| --- e-x/B x > 0
f(x) = | B
{ 0 elsewhere
and E[X] = B and Var(X) = B2. Note that the moment
generating function for an exponential distribution is 1/(1 - tB).
A Saint Mary's student is studying in her room and hoping that the phone will ring. She gets on the average 4 calls an hour. She would like to go down to the vending machines (a 10 minute trip) and asks you for the probability that she will miss a phone call if she leaves her room right after she hangs up on the previous call.
Let X be the time between phone calls. Since the occurrence of phone calls is a poisson process, X has an exponential distribution with B = 1/4 (i.e. the average time between calls is 15 minutes since she gets on the average 4 calls per hour. She doesn't talk very long when she gets a call). She wants to know the probability that the next phone call will come in the next 10 minutes = 1/6 hour. Thus,
1/6 1/6
{ 1 { 1/6
P(X <= 1/6) = / --- e-x/B dx = / 4 e-4x dx = -e-4x | = 1-e-2/3
} B } 0
0 0
These will be provided in class.
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