
t distribution
It was given that a random variable T is said to have a tn distribution when:
$\displaystyle T=\frac{Z}{\sqrt{U/n}}$
but while proving for the density function (pdf) of a tn distribution, why is is equal to the joint distribution $\displaystyle \int_{0}^{\infty}f_{T,U}(t,u).du$?

You first find the joint denisty of Z and U, where you needed to say that
$\displaystyle U\sim \chi^2_n$ and $\displaystyle Z\sim N(0,1)$ (and they are indep.)
Then pick a dummy varable, say W=Z or W=U.
(IF you let W=U, then you can do this, but you better check your bounds of integration.)
Use calc3 to find the density of T and W.
THEN integrate out the W and you have the density of U.

Well initially I was trying to solve it by 1st letting:
$\displaystyle Y=\sqrt{U/n}$
$\displaystyle f_Y(y)=f_u(ny^2)(2ny)$
then follow by:
$\displaystyle T=\frac{Z}{Y}$
$\displaystyle f_T(t)=\int_{\infty}^{\infty}yf_Y(y)f_z(yt).dy$
but it turned out to be very complicated

this is in many books
I bet you can find it online too.
I did a quick search yesterday, but didn't find it.
I have it some where in my office and I know I can do it,
but I don't have the time.
It's in Hogg and Craig.