# Limiting distribution involving chi square distribution

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• October 20th 2009, 03:58 PM
statmajor
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.
• October 20th 2009, 04:15 PM
mr fantastic
Quote:

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)
• October 20th 2009, 04:22 PM
matheagle
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 ${Z_n\over n^2}\to 0$