# Using the Central Limit Theorem to prove a limit

• Mar 23rd 2009, 06:16 AM
phillips101
Using the Central Limit Theorem to prove a limit
Use the central limit theorem to prove that, as M tends to infinite:

exp(-2M) * SUM(exp(2Mn)/n!) --> 1/2 . The sum goes from n=0 to n=M.

I don't know how to format it correctly, sorry.

I sort of know what to do with this question. I need to find the right variables so the left side of the CLT boils down to what's above, and the right would be the normal distribution, presumably from 0 to infinite to get a half. A hint or a correction if I'm wrong will be brilliant to start me off, thanks :)

James
• Mar 23rd 2009, 10:44 AM
Laurent
Quote:

Originally Posted by phillips101
Use the central limit theorem to prove that, as M tends to infinite:

exp(-2M) * SUM(exp(2Mn)/n!) --> 1/2 . The sum goes from n=0 to n=M.

I don't know how to format it correctly, sorry.

I sort of know what to do with this question. I need to find the right variables so the left side of the CLT boils down to what's above, and the right would be the normal distribution, presumably from 0 to infinite to get a half. A hint or a correction if I'm wrong will be brilliant to start me off, thanks :)

James

There's a mistake in your formula (the left-hand side is greater than 1, and it diverges), I think it should be:

$\displaystyle e^{-M}\sum_{n=0}^M \frac{M^k}{k!}\to_{M\to\infty} \frac{1}{2}$

Anyway, your intuition is correct ; consider applying the CLT to Poisson random variables of parameter 1 (remember the sum of n independent such variables is Poisson distributed with parameter n).
• Mar 23rd 2009, 11:09 AM
phillips101
Quote:

Originally Posted by Laurent
There's a mistake in your formula (the left-hand side is greater than 1, and it diverges), I think it should be:

$\displaystyle e^{-M}\sum_{n=0}^M \frac{M^k}{k!}\to_{M\to\infty} \frac{1}{2}$

Anyway, your intuition is correct ; consider applying the CLT to Poisson random variables of parameter 1 (remember the sum of n independent such variables is Poisson distributed with parameter n).

That's what I thought! I really didn't think it would converge, and none of the standard distributions gave the right answer.

Thanks for the confirmation. I'll inquire about the error.