[Math] Deriving properties of the t distribution

probabilityprobability distributionsprobability theorystatistics

Let X~$t_p$

  1. Derive the Mean and Variance of X (EX and Var(X).
  2. Show that $\chi^2$~$F_{1,p}$.
  3. Show that X converges to N(0,1)

I'm not so sure how to do part 1 since it involves Gamma distribution and I don't really know how to deal with the integral portion of the pdf.

For part 2 I tried plugging in 1,p into the pdf but I'm getting something that is completely different from what's needed.

For part 3 I used the Almost sure convergence theorem but I'm getting stuck at showing how the probability=1.

I'd really need some help. thanks.

Best Answer

A random variable $X\sim T_n$ has the following density function:

$\qquad f_{t_n}=\frac{1}{\sqrt{n}}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{n}{2})}(1+\frac{x^2}{n})^{-\frac{n+1}{2}}$

1) Using the definition of $\mathrm{E}[X]$, we get

$\qquad \mathrm{E}[X]=\frac{1}{\sqrt{n}}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{n}{2})}\int_{-\infty}^{+\infty}x(1+\frac{x^2}{n})^{-\frac{n+1}{2}}dx$

First, notice that $\mathrm{E}[X]$ is undefined if $0<n\le1$. So let's take it from $n>1$.

Now, if we look at our integrand, we see that the function $f(x)=x(1+\frac{x^2}{n})^{-\frac{n+1}{2}}$ is an odd function. So the integral equals $0$. The means we found the value of the expected value of are r.v $X$, that is:

$\qquad \mathrm{E}[X]=0$

The variance is a little trickier. Using what we've showed about $\mathrm{E}[X]$, we get:

$\qquad \mathrm{Var}[X]=\frac{1}{\sqrt{n}}\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{n}{2})}\int_{-\infty}^{+\infty}x^2(1+\frac{x^2}{n})^{-\frac{n+1}{2}}dx$

If we do the change of variable $y=(1+\frac{x^2}{n})^{-1}$ we get:

$\qquad n\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{n}{2})}\int_0^1y^{\frac{n}{2}-2}(1-y)^{\frac{1}{2}}$

Now if we remember a property of the gamma function, which states that:

$\qquad \int_0^1x^{p-1}(1-x)^{q-1}=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$

In our case $p=\frac{n}{2}-1$ and $q=\frac{3}{2}$. And simplifying all that gamma stuff, we get:

$\qquad \mathrm{Var}[X]=\frac{n}{n-2}$

2) Here, you probably meant $X^2\sim F_{1,p}$. That follows almost inmediatly from the definition of both distributions.

$\qquad T_n=\frac{Z}{\sqrt{\frac{1}{p}\sum_{i=1}^p Y_i^2}}$ $\qquad (1)$

where $Z\sim N(0,1)$ and $Y_i\sim N(0,1)$ for al $i=1,...,n$. Just squared that expression and you'll get the distribution of $F_{1,p}$.

3) The result you want to prove makes use of the Strong Law of Large Numbers. Review it and notive that if

$\qquad Y=\sum_{i=1}^pY_i^2$,

then $\frac{Y}{p}\to1$ as $p\to\infty$. Substitute in $(1)$ and you get the result.

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