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Mean and Variance of the T & F Distributions

Date: 02/16/2007 at 01:30:33
From: Bobby
Subject: Mean and Variance of the T and F distributions

In my statistics class we identified the mean of the t-distribution
(with degree of freedom v) as 0 and its variance as v/(v-2).  We also
have that the mean of the f-distribution (with degree of freedom v &
w) is w/(w-2) and its variance is (2w^2(v+w-2))/(v(w-2)^2(w-4)).

I've tried to prove these results, but have been unable to find a way
to arrive at them.  Since both the t and f distributions are 
combinations of independent random distributions, I tried using both
transformations and looking for E(Y) and V(Y) directly.

Date: 02/16/2007 at 06:11:53
From: Doctor George
Subject: Re: Mean and Variance of the T and F distributions

Hi Bobby,

Thanks for writing to Doctor Math.

This is not a simple problem, and the complete details are hard to 
follow in a text format.  I'll outline the method for you.

Your textbook may explain the calculation of the mean and variance
for the Gamma distribution.  The technique is similar for the T and F
distributions, so that is a good place for you to start.  If you
understand it that will help you with the other distributions.

The T distribution is a little easier than the F distribution since it
has just one parameter for degrees of freedom, but it is common to
derive the F distribution first because of how the two are related.
You may want to work on the T distribution first.

F distribution
First, the mean is the integral of x*f(x) over the entire domain of 
the density function.  The variance is Var(X) = E[(X-mu)^2].  It can
be shown that

  Var(X) = E(X^2) - E(X)^2.

This is generally a more useful expression than the definition.  The 
term E(X^2) is the integral of (x^2)*f(x) over the entire domain.

The trick to finding the mean of F(p,q) is to define values p' and q',
and rearrange the terms of the integrand so that

  x * f(x|p,q) = C * f(x|p',q')

where C is a constant.  Since f(x|p',q') is itself a density function,
its integral over the entire domain is 1, so the mean of F(p,q) is just C.

Finding E(X^2) to get Var(X) requires another application of the 
same trick, but it gets even messier.

T distribution
Rather than work with the definition of the student T distribution it
would be better to work with its density function.

  E(T) = 0 by symmetry.

Now we need to compute E(T^2) to find the variance.  This is easy if
you already know the mean of the F distribution because T^2 has an F

Without making use of the F distribution we need to compute the 
integral for E(T^2) to find the variance.  There are two steps.

1. Apply the integration by parts technique using u = t and dv = the
remaining terms.  The u*v part of the result will be zero by symmetry.
The Integral(v*du) part will look much like the density function.

2. Carefully select a value n' as an offset from n in order to
transform the Integral(v*du) into the density function times a 
constant factor.  The new integral now equals one, and the factor 
becomes E(T^2).

Does that make sense?  Write again if you need more help.

- Doctor George, The Math Forum 
Associated Topics:
College Statistics

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