t - distribution

**gtaplayr** · September 17th, 2009, 11:41 PM

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

**mr fantastic** · September 18th, 2009, 06:46 AM

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

$E(X) = \int_{-\infty}^{+ \infty} x f(x) \, dx$ where f(x) is the pdf of the t-distribution with r degrees of freedom.

$Var(X) = E(X^2) - [E(X)]^2$ and $E(X^2) = \int_{-\infty}^{+ \infty} x^2 f(x) \, dx$ where f(x) is the pdf of the t-distribution with r degrees of freedom.

**matheagle** · September 19th, 2009, 01:47 PM

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

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

**awkward** · September 21st, 2009, 05:07 PM

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 $var(X)$ with a minimum of effort.

1. Show that E(X) = 0, so $var(X) = E(X^2)$ .

2. Show that $E(1 + X^2 / r) = 1 + E(X^2) / r$ .

3. Write down the integral for $E(1 + X^2 / r)$ . Make the substitution $t = \sqrt{\frac{r}{r-2}}\; x$ . Notice that the resulting integral is of the form $c \int_{-\infty}^{\infty} f(t) \,dt$ , 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 $\int f(t) \, dt = 1$ for any pdf f(t) to evaluate the integral in 3.

5. Substitute into the equation in 2. and solve for $E(X^2)$ .

**gtaplayr** · September 22nd, 2009, 11:40 PM

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

**mr fantastic** · September 23rd, 2009, 02:08 AM

Originally Posted by gtaplayr

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

**awkward** · September 23rd, 2009, 02:43 PM

Originally Posted by gtaplayr

So where does $E(Z)$ , $E(Z^2)$ , $E(1/\sqrt{U})$ , $E(1/U)$ where $Z\sim,N(0,1)$ and $U\sim\chi^2_{r}$ , 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 $\Gamma(x+1) = x \Gamma(x)$ to get the answer in its final form.

**Lomofun** · September 23rd, 2009, 05:08 PM

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 $E(Z)$ , $E(Z^2)$ , $E(1/\sqrt{U})$ , $E(1/U)$ where $Z\sim,N(0,1)$ and $U\sim\chi^2_{r}$

**noob mathematician** · September 23rd, 2009, 06:23 PM

Originally Posted by awkward

Here is a recipe which I think will allow you to find $var(X)$ with a minimum of effort.

1. Show that E(X) = 0, so $var(X) = E(X^2)$ .

2. Show that $E(1 + X^2 / r) = 1 + E(X^2) / r$ .

3. Write down the integral for $E(1 + X^2 / r)$ . Make the substitution $t = \sqrt{\frac{r}{r-2}}\; x$ . Notice that the resulting integral is of the form $c \int_{-\infty}^{\infty} f(t) \,dt$ , 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 $\int f(t) \, dt = 1$ for any pdf f(t) to evaluate the integral in 3.

5. Substitute into the equation in 2. and solve for $E(X^2)$ .

Hi, does this work for F-distribution too?

**mr fantastic** · September 23rd, 2009, 11:01 PM

Originally Posted by noob mathematician

Hi, does this work for F-distribution too?

What are you trying to do? Find the mean and variance?

**noob mathematician** · September 24th, 2009, 03:37 AM

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?

**mr fantastic** · September 24th, 2009, 03:48 AM

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 $Y = X^2$ ~ $F(\nu_1 = 1, \nu_2 = r)$ and you've been shown how to calculate $E(X^2)$ .

**gtaplayr** · September 24th, 2009, 08:19 PM

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:
$\int \exp^{tz} \frac{1}{\sigma\sqrt2\pi} \exp^{-1/2(z-\mu/\sigma)} dz$
$M(t)=\exp{\mu.t+1/2\sigma^2t^2}$
To show that $E(Z)=M'(0)=0$ and $E(Z^2)=M''(0)=1$

I used the Chi moment generating function:
$M(t)=(1-2t)^{-r/2}$
To show that $E(U)=M'(0)=r$
$E(1/U)=1/E(U)=1/r$

This is where i am stuck
I am unsure how to calculate $E(1/\sqrt{U})$

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