# A Local CLT with large variance

For n an even integer, $0 \leq i \leq$ ${n} \choose{j}$, $1 \leq j \leq n$ let $X_{i,j}$ be a random variable taking values $\frac{n}{2}-j,0,j - \frac{n}{2}$ with equal probability. Let $S_{n}$ be the sum of these $2^{n}$ random variables.

My question is, what is the 'correct' local limit theorem for this sum as n goes to infinity? That is, what is a local limit theorem that is in some sense sharp?

To those who have not heard of the term: A local limit theorem is one that describes probabilities of the sum being a specific number rather than being in a region of size roughly proportional to the square root of the variance.

To those who might think this is a little specialized: You're right, of course. I wanted to post a question with a concrete answer, and this is the simplest one that seems to be 'on the edge' of where local CLTs hold. The versions you see in textbooks fail for this (those that I'm familiar with fail in a few places), but 'just barely'. Also, a local CLT does at least hold here. I'm interested in other borderline cases as well.

To those who might think this is trivial: It is true that a CLT for this sum follows from standard textbook theorems (e.g. the Lindeberg CLT). It is even true that the martingale local CLT can be used to get rates here - unfortunately, they seem to be wrong (that is, quite far from sharp).

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Have you tried doing any sort of simulation? What does it look like the answer "should" be? – Michael Lugo Dec 12 '09 at 1:26
Thanks Michael. I haven't tried simulating. My naive thought was that we would need at least $2^{n}$ or so draws from the n'th distribution, n from say 1 to 20, to get any idea what the rate is... and that is already about a trillion random variables. Is there a more reasonable way of doing this (or a way to get that on a computer)? I know almost nothing of programming. – user2282 Dec 12 '09 at 14:44
You probably forgot to mention that $X_{i,j}$ are independent. – zhoraster Oct 17 '10 at 9:33