# Thread: Limiting distribution involving chi square distribution

1. ## Limiting distribution involving chi square distribution

Let Zn be a chi square distribution with n degrees of freedom. Let Wn = Zn/n^2. Find the limiting distribution of Wn.

I know the MGF of a chi square distribution is $(1 - 2t)^{\frac{-n}{2}}$

$E(e^{tW_n}) = E(e^{t \{Z_n}{n^2})= (1-\frac{2t}{n^2})^{\frac{-n}{2}}$

I'm trying to use the taylor expansion (1+b/n)^c = e^bc as n approaces infinity, but can't since the denominator is n^2, and not n, and I'm stuck here. I'm assuming I made a mistake somewhere, but can't figure out where.

Any help would be appreciated.

2. Originally Posted by statmajor
Let Zn be a chi square distribution with n degrees of freedom. Let Wn = Zn/n^2. Find the limiting distribution of Wn.

I know the MGF of a chi square distribution is $(1 - 2t)^{\frac{-n}{2}}$

$E(e^{tW_n}) = E(e^{t \{Z_n}{n^2})= (1-\frac{2t}{n^2})^{\frac{-n}{2}}$

I'm trying to use the taylor expansion (1+b/n)^c = e^bc as n approaces infinity, but can't since the denominator is n^2, and not n, and I'm stuck here. I'm assuming I made a mistake somewhere, but can't figure out where.

Any help would be appreciated.
See here: - Wolfram|Alpha

So now you need to think about a distribution that has a moment generating function equal to 1 .... (see here Dirac delta function - Wikipedia, the free encyclopedia)

3. na, use the limit of exp(x).

By that I meant $\biggl(1+{a\over x}\biggr)^x\to e^a$ as x goes towards infinity.

However, there's nothing to this problem

${Z_n\over n}\to 1$ by the STONG law of large numbers

Since $Z_n$ can be represented as a sum of i.i.d. chi-squares with 1 df.

So ${Z_n\over n^2}\to 0$

4. ## Re: Limiting distribution involving chi square distribution

But the exponential formula cannot be used as it is since we have X^2, not x. Or it doesn't matter?

For the problem, also how do we compute the final answer? The E(Wi)=1/i, where i goes to infinite, so lim E(Wi)=0.

How do you prove the statement about the Strong Law of Numbers?

Then if we want the limiting distribution, can we do the following:
I know from the distribution that $\frac{\bar{Z_n}}{n} =1$

$P(lim_n=inf |\bar{Z_n} -1|< \epsilon)$
$P(lim_n=inf |(\frac{\sum_{i=1}^{n}Z_i}{n} -1 |)< \epsilon)$

Since we have finite Mean and finite variance, we can just apply the SLLN, right?

So, by doing a similar process for the original sequence we can get that the Mean $M=\frac{1}{i}$ and $\sigma=\frac{2}{i^3}$.
Then how do you apply the Law of LN in this case?
$\frac{Z_1+Z_1+...}{n}=\frac{nZ_1}{n}$
and then divide that by n so we have"
$P(lim_n=inf |\frac{Z_1}{n}-1|< \epsilon$

Is this the right way, or do we say the limiting sequence converges in distribution to 0?

Thank you