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I have a function $s(\omega)$ that is a sum of a function with random numbers $a_m$ and looks something like the following.

$$ s(\omega) = \sum_{m = 1}^{M} f(a_m, \omega) $$ where all the $a_m$ are draws from a probability distribution.

I have the expression of the expectation of the sum when $M \to \infty$ $$ \mathbb{E}\left[\sum_{m = 1}^{M} f(a_m, \omega)\right] = M \int_{-\infty}^{+\infty} f(x, \omega) p(x) dx $$, I changed the parametrization of $x= a_m$ to write the integral.

However, in practical simulations, M is not infinity and in my simulations I suspect that this approximation doesn't give me exact results. That is the reason why I want to model the error in this approximation so that I can use it as a distribution over $g$ to fit it with the average estimate of the sum. Is there an existing way to do this?

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The function $f$ looks like the following

$$ f(a_m, \omega) = \exp({-a_m \omega^2 }) $$

Where $a_m \sim \mathcal{N}(0, \sigma^2)$

Now if I want to know the expected value, I write it as,

$$ \mathbb{E}\left[\sum_{m = 1}^{M}f(a_m, \omega)\right]_{M\to\infty} = M \int_{-\infty}^{-\infty} \exp({-x \omega^2 })\frac{1}{2\pi\sigma^2} \exp({-\frac{x^2}{2\sigma^2}}) dx $$

I hope the notations are clear now a bit. I’m not very confident with statistical terms, so, I say $a_m$ are random draws from a normal distribution and the function is a sum of values of a function that’s computed at these random numbers. For example, it’s a superposition of responses from $M$ different components.

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  • $\begingroup$ Your notation is confusing. The second formula does not seem to have anything to do with expectation of a sum $s$. $\endgroup$
    – Tim
    Commented Mar 31, 2023 at 11:23
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    $\begingroup$ If the $a_m$ are not identically distributed then the equality $$\mathbb{E}\left[\sum_{m = 1}^{M} f(a_m, \omega)\right] = M \int_{-\infty}^{+\infty} f(x, \omega) p(x) dx $$ is not true. Do they follow the same distribution ? If so, what distribution is it ? $\endgroup$ Commented Mar 31, 2023 at 15:41
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    $\begingroup$ @StratosFair By simplifying the notation to use "$X$" in place of the random variable "$f(a_m,\omega)$" and without loss of generality (due to linearity of expectation) considering $m=1,$ the assertion is $E[X]=\int xp(x)\,\mathrm d x.$ In what sense is this "not true"? Have I perhaps oversimplified the question or misinterpreted it? $\endgroup$
    – whuber
    Commented Mar 31, 2023 at 19:15
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    $\begingroup$ That narrows it down, because $s$ clearly is a sum of iid Lognormal values. Unfortunately, when $M\gt 1$ there is no closed form for its distribution. You can still (easily) work out the moments of $s$ and compute measures of dispersion like the variance from those: see stats.stackexchange.com/a/89973/919. $\endgroup$
    – whuber
    Commented Apr 2, 2023 at 14:16
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    $\begingroup$ When we compute the variance, we always do it for finite $M$! It cannot be computed for infinite $M:$ you have to take a limit -- but obviously this limit is infinite in your case. $\endgroup$
    – whuber
    Commented Apr 3, 2023 at 1:16

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