t distribution

• Aug 21st 2009, 10:52 PM
noob mathematician
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 $\displaystyle \int_{0}^{\infty}f_{T,U}(t,u).du$?
• Aug 21st 2009, 10:59 PM
matheagle
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.
• Aug 21st 2009, 11:36 PM
noob mathematician
Well initially I was trying to solve it by 1st letting:
$\displaystyle Y=\sqrt{U/n}$
$\displaystyle f_Y(y)=f_u(ny^2)(2ny)$

$\displaystyle T=\frac{Z}{Y}$
$\displaystyle f_T(t)=\int_{-\infty}^{\infty}|y|f_Y(y)f_z(yt).dy$