Mean and Variance of the T & F DistributionsDate: 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 distribution. 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 http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/