1. ## t - distribution

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

2. Originally Posted by gtaplayr
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.

3. The mean is easy because the density is symmetric about zero.
So xf(x) is an odd function, hence

$\displaystyle \int_{-\infty}^{\infty}xf(x)dx=0$

4. Originally Posted by gtaplayr
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
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 $\displaystyle E(X^2)$.

5. So where does $\displaystyle E(Z)$, $\displaystyle E(Z^2)$, $\displaystyle E(1/\sqrt{U})$, $\displaystyle E(1/U)$ where $\displaystyle Z\sim,N(0,1)$ and $\displaystyle U\sim\chi^2_{r}$, come into this?

6. Originally Posted by gtaplayr
So where does $\displaystyle E(Z)$, $\displaystyle E(Z^2)$, $\displaystyle E(1/\sqrt{U})$, $\displaystyle E(1/U)$ where $\displaystyle Z\sim,N(0,1)$ and $\displaystyle U\sim\chi^2_{r}$, come into this?
Why? Unless you have in mind a specific way of solving the original question that involves those things, the question has been solved.

7. Originally Posted by gtaplayr
So where does $\displaystyle E(Z)$, $\displaystyle E(Z^2)$, $\displaystyle E(1/\sqrt{U})$, $\displaystyle E(1/U)$ where $\displaystyle Z\sim,N(0,1)$ and $\displaystyle U\sim\chi^2_{r}$, come into this?
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.

8. Originally Posted by mr fantastic
Why? Unless you have in mind a specific way of solving the original question that involves those things, the question has been solved.
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 $\displaystyle U\sim\chi^2_{r}$

9. Originally Posted by awkward
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 $\displaystyle E(X^2)$.
Hi, does this work for F-distribution too?

10. Originally Posted by noob mathematician
Hi, does this work for F-distribution too?
What are you trying to do? Find the mean and variance?

11. Originally Posted by mr fantastic
What are you trying to do? Find the mean and variance?
The expected value. Since they have the same result right?

12. Originally Posted by noob mathematician
Hi, does this work for F-distribution too?
Originally Posted by mr fantastic
What are you trying to do? Find the mean and variance?
Originally Posted by noob mathematician
The expected value. Since they have the same result right?
Well, I suppose it works to a point since $\displaystyle Y = X^2$ ~ $\displaystyle F(\nu_1 = 1, \nu_2 = r)$ and you've been shown how to calculate $\displaystyle E(X^2)$.

13. Originally Posted by mr fantastic
Why? Unless you have in mind a specific way of solving the original question that involves those things, the question has been solved.
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 $\displaystyle E(1/\sqrt{U})$