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
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.
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
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 and .
Then how do you apply the Law of LN in this case?
and then divide that by n so we have"
Is this the right way, or do we say the limiting sequence converges in distribution to 0?