Limiting Distribution

**akolman** · March 21st 2009, 04:21 PM

Hello,

I'm stuck with this problem.

Let $Z_n$ be $\chi^2(n)$ and let $W_n=\frac{Z_n}{n^2}$ . Find the limiting distribution of $W_n$ .

I tried getting the mgf of $W_n$ and then take the limit as $n \to \infty$ to the mgf of the limiting distribution
This is what I got, but I don't know if it's right.
$M_{W_n}(t)=\left( 1- \frac{2t}{n^2} \right)^{-\frac{n}{2}}$ for $t < \frac{n^2}{2}$ .

I don't know how to arrange the mgf so I can get an e as I take the limit....

Thanks in advance.

**mr fantastic** · March 21st 2009, 07:21 PM

Originally Posted by akolman

Hello,

I'm stuck with this problem.

Let $Z_n$ be $\chi^2(n)$ and let $W_n=\frac{Z_n}{n^2}$ . Find the limiting distribution of $W_n$ .

I tried getting the mgf of $W_n$ and then take the limit as $n \to \infty$ to the mgf of the limiting distribution
This is what I got, but I don't know if it's right.
$M_{W_n}(t)=\left( 1- \frac{2t}{n^2} \right)^{-\frac{n}{2}}$ for $t < \frac{n^2}{2}$ .

I don't know how to arrange the mgf so I can get an e as I take the limit....

Thanks in advance.

1. Note that $\left( 1 - \frac{2t}{n^2}\right)^{-n/2} = \frac{1}{\left( 1 - \frac{2t}{n^2}\right)^{n/2} }$ .

2. Note that $\left( 1 - \frac{2t}{n^2}\right)^{n/2} = \left( 1 - \frac{\sqrt{2t}}{n}\right)^{n/2} \left( 1 + \frac{\sqrt{2t}}{n}\right)^{n/2}$ $= \left( 1 + \left[- \frac{\sqrt{2t}}{n}\right]\right)^{n/2} \left( 1 + \frac{\sqrt{2t}}{n}\right)^{n/2}$

3. Substitute $m = \frac{n}{2}$ :

$= \left( 1 + \left[- \frac{\sqrt{\frac{t}{2}}}{m}\right]\right)^{m} \left( 1 + \frac{\sqrt{\frac{t}{2}}}{m}\right)^{m}$ .

4. Now take the limit and find that the limiting value of the mgf is 1.

Therefore the limiting distribution function is the Dirac Delta function (Dirac delta function - Wikipedia, the free encyclopedia).

**matheagle** · March 21st 2009, 07:43 PM

You can express $Z_n=\sum_{k=1}^n Y_k$ where each $Y_k$ is a $\chi^2$ with one degree of freedom.

Thus ${Z_n\over n}={\sum_{k=1}^n Y_k\over n}\to E(Y_1)=1$ by the Strong Law of Large Numbers.

(By the way this shows how a $t_n\to N(0,1)$ as $n\to\infty$ .)

Hence ${Z_n\over n^{1+\delta} }={\sum_{k=1}^n Y_k\over n^{1+\delta}}\to 0$ for all $\delta>0$ .

MGF of one means that the rv is 0 with probability 1, same answer.

$E(e^{Xt})=1$ means $P(X=0)=1$ .

**matheagle** · March 21st 2009, 09:39 PM

I wouldn't use conjugates

$\biggl(1-{2t\over n^2}\biggr)^{n/2}= \Biggl[\biggl(1-{2t\over n^2}\biggr)^{n^2/(2t)}\Biggr]^{t/n}$

The term inside the ( ) is approaching 1/e and as the exponent t/n goes to zero,
as n goes to infinity, you have the MGF going towards 1.

**Moo** · March 22nd 2009, 12:17 AM

Hello,

MGF of one means that the rv is 0 with probability 1, same answer.

I don't understand how you get to the MGF=1

Can you explain please ?

And by the way, I may be a little picky, but one has to say that in the Law of large numbers, it's an almost sure convergence... (our lecturer insisted so much on it lol)

Since it converges to 0 almost surely, then it converges to 0 in probability, and hence it converges to 0 in distribution...

This solves the problem

and looking at akolman's previous threads, I'm pretty sure he can use this link between the convergences

(and I must say, akolman, that I love the questions you ask here, because it helps me study my lessons xD)

**matheagle** · March 22nd 2009, 06:02 PM

I (almost) certainly know what almost sure convergence is.
LOL

**Moo** · March 23rd 2009, 02:42 PM

Originally Posted by matheagle

I (almost) certainly know what almost sure convergence is.
LOL

I don't see what's funny in it.

**matheagle** · March 23rd 2009, 03:52 PM

almost certainly = almost sure

**mr fantastic** · March 23rd 2009, 06:01 PM

The question is answered and moo is puzzled

. A perfect state of affairs at which to close the thread.

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