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Doing some analysis of my problem, I have come up with the following equation (I tried to search this problem but no luck)

$S_N = \sum \limits_{i=1}^{N} \exp{\left(-\frac{X_i^2}{2\sigma^2}\right)}$

where $X_i \ \ \text{~} \ \ {N(0, \sigma^2)}$

It is obvious that $S_N \rightarrow \infty$ as $N \rightarrow \infty$

The solution I was thinking of is (it may be wrong)

$\frac{N}{S_N} \simeq \sqrt{2}$

which I got from Matlab by generating 1000000 numbers, and they were all same for different standard deviation, and different number of samples. Here is Matlab script I used, 

N = 1000000;
sd = 3; % arbitrary standard deviation
x = sd*randn(1, N);
S = sum(exp(-x.^2/(2*sd^2)));
N/S

How can I solve this problem? Can you recommend any approximation, or bound theorem to play this with? Thanks!!

ben Heo
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1 Answers1

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I tried it, and I got 1.7314 which looks more like $\sqrt 3$ to me.

Anyway, $$ E(S_N) = N E(\exp(-X^2/\sigma^2)) = N \frac1{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty \exp(-x^2/\sigma^2) \exp(-x^2/2\sigma^2) \, dx = N \frac1{\sqrt{2\pi}\sigma} \int_{-\infty}^\infty \exp(-3x^2/2\sigma^2) \, dx = \frac N{\sqrt 3} .$$

Oops, I tried it again. And then I got $\sqrt 2$. But your code is different than the formula - in the code you are doing $$ \sum_{n=1}^N \exp(-X_n^2/2\sigma^2) .$$ Anyway, similar calculation - you will get $N/\sqrt 2$ this time.

Did
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Stephen Montgomery-Smith
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