Suppose that $T$ is a t-distribution with $n$ df. How can we find $E(|T|)$ through the relation between $T$ and the $F$ distribution?

Question

Suppose that $T$ is a t-distribution with $n$ df. How can we find $E(|T|)$ through the relation between $T$ and the $F$ distribution?

The usual way to do this is by brute force integration, which is tedious. However, I saw in a book that this can be done with its relation with the $F$ distribution in that:

If $$ X \sim t_n $$

then

$$ X^2 \sim F(1, n) $$

Is there a way to do it by this representation?

Answer 1

You could get $E(T)$ using the expression $X^2\sim F(1,n)$, if you known following formula of moments of F-distributed RV: $$E[F(a,b)]^k=\Bigl(\frac{b}{a}\Bigr)^k\frac{\Gamma(a/2+k)\Gamma(b/2-k)}{\Gamma(a/2)\Gamma(b/2)},\qquad k<\frac{b}2.\tag{1}$$ Using (1) it is easy to deduce that $$E[|t_n|]=E\{[F(1,n)]^{1/2}\}=\frac{\sqrt{n}\Gamma((n-1)/2)}{\sqrt{\pi}\Gamma(n/2)}. $$

Answer 2

Comment (continued):

Here is the numerical integration in R statistical software for 5 degrees of freedom. Matlab, Mathematica, etc. may have methods even easier to use.

 integrand = function(x){x*2*dt(x,5)}
 integrate(integrand, 0, Inf)  # Note use of 'Inf'; R 'decides' where to truncate
 ## 0.9490167 with absolute error < 5.8e-05

Reality check: Simulate a million observations from $\mathsf{T}(\nu=5),$ take their absolute values, and average.

 x = abs(rt(10^6, 5));  mean(x)
 ## 0.9489664

However, numerical integration does not give a general expression for any number $\nu$ of degrees of freedom.

By contrast according to Wikipedia, the distribution $\mathsf{F}(\nu_1=1, \nu_2=5)$ has mean $\nu_2/(\nu_2 - 2) =$ $5/(5-2) =$ $5/3 \approx 1.6667,$ with square root $1.290994,$ which is not even a poor approximation of $0.9490.$