Let X have a t-distribution with r degrees of freedom.
How would I go about showing:
E(X) = 0 , r > 1
and
Var(X) = r/(r-2) , r > 2
$\displaystyle E(X) = \int_{-\infty}^{+ \infty} x f(x) \, dx$ where f(x) is the pdf of the t-distribution with r degrees of freedom.
$\displaystyle Var(X) = E(X^2) - [E(X)]^2$ and $\displaystyle E(X^2) = \int_{-\infty}^{+ \infty} x^2 f(x) \, dx$ where f(x) is the pdf of the t-distribution with r degrees of freedom.
Here is a recipe which I think will allow you to find $\displaystyle var(X)$ with a minimum of effort.
1. Show that E(X) = 0, so $\displaystyle var(X) = E(X^2)$.
2. Show that $\displaystyle E(1 + X^2 / r) = 1 + E(X^2) / r$.
3. Write down the integral for $\displaystyle E(1 + X^2 / r)$. Make the substitution $\displaystyle t = \sqrt{\frac{r}{r-2}}\; x$. Notice that the resulting integral is of the form $\displaystyle c \int_{-\infty}^{\infty} f(t) \,dt$, where c is a constant (depending on r, but independent of t), and f is the pdf of a t distribution with r-2 degrees of freedom.
4. Use the fact that $\displaystyle \int f(t) \, dt = 1$ for any pdf f(t) to evaluate the integral in 3.
5. Substitute into the equation in 2. and solve for $\displaystyle E(X^2)$.
Like Mr. Fantastic, I find this question confusing. You don't need to know any of those things, provided you know the formula for the pdf of a t distribution. If you don't know the pdf, then that's a different problem.
One thing I forgot to mention in my original post: you will need the identity $\displaystyle \Gamma(x+1) = x \Gamma(x)$ to get the answer in its final form.
If you correctly answer this question, with full working, 1% will be added to your final mark.
Let X have a t-distribution with r degrees of freedom. Show that E(X) = 0, r > 1 and V ar(X) = r/(r − 2), r > 2 by first finding $\displaystyle E(Z)$, $\displaystyle E(Z^2)$, $\displaystyle E(1/\sqrt{U})$, $\displaystyle E(1/U)$ where $\displaystyle Z\sim,N(0,1)$ and $\displaystyle U\sim\chi^2_{r}$
Ok this is what iv done so far
Iv used the normal moment generating function:
$\displaystyle \int \exp^{tz} \frac{1}{\sigma\sqrt2\pi} \exp^{-1/2(z-\mu/\sigma)} dz$
$\displaystyle M(t)=\exp{\mu.t+1/2\sigma^2t^2}$
To show that $\displaystyle E(Z)=M'(0)=0$ and $\displaystyle E(Z^2)=M''(0)=1$
I used the Chi moment generating function:
$\displaystyle M(t)=(1-2t)^{-r/2}$
To show that $\displaystyle E(U)=M'(0)=r$
$\displaystyle E(1/U)=1/E(U)=1/r$
This is where i am stuck
I am unsure how to calculate $\displaystyle E(1/\sqrt{U})$