# Approximate distribution of product of N normal i.i.d.? Special case μ>10σ, σ>0

Given $N\geq30$ i.i.d. $X_n\approx\mathcal{N}(\mu_X,\sigma_X^2)$, and $|\mu_X|\geq10\sigma_X$, $\sigma > 0$, looking for:

1. accurate closed form distribution approximation of $Y_N=\prod\limits_{1}^{N}{X_n}$
2. asymptotic (exponential?) approximation of same product

This is a special case $|\mu_X|\geq10\sigma_X$, $\sigma > 0$ of a more general question.

• For (2), you could work on the log scale and apply CLT. I believe whuber mentioned lognormal approximation on an earlier question in this regard. – Glen_b -Reinstate Monica Apr 6 '15 at 2:11
• @Glen_b: The limiting distribution has to be the log-normal distribution since $Y_N=\exp\sum_i\log X_i$ and the CLT applies to $\sum_i\log X_i$. – Xi'an Apr 6 '15 at 6:47
• @Xi'an Yes, that was exactly the point I was making in my first sentence. The second sentence was pointing out whuber had already suggested that was the case elsewhere. – Glen_b -Reinstate Monica Apr 6 '15 at 8:14
• Because you still include this case in your previous question (explicitly), this post clearly is a duplicate. – whuber Apr 6 '15 at 16:17
• Andrei, the duplicate question is an exact duplicate. Unless I am misreading the text, I see no room for discussion about that. As of this moment the other question still asks "Special case of the question, for $|\mu_X|\geq10\sigma_X.$" Moreover, because it still poses that question, somebody has posted an answer to it at the duplicate. If your intention is that your questions should not be duplicates then you must remove the material that is common. – whuber Apr 6 '15 at 19:16

Note that after re-scaling form Y to F to G, we can reformulate the question:

$Z=\mathcal{N}(0,1)$, $X=\mu_X+\sigma_XZ$, $Y_N=\mu_X^N\prod{(1+\frac{\sigma_X}{\mu_X}Z)}$,

$Y_0=\mu_X^N$, $\alpha=\frac{\sigma_X}{\mu_X}$,

$F =\frac{Y_N}{Y_0}=\prod{(1+\alpha Z)}$ - factor function,

$G=log(F)/(N\alpha)$ - growth function,

where $\alpha N$ is scaling parameter, apparent from $\lim(\alpha \rightarrow 0)$.

The question then becomes to find approximation of $\mu_G$ and $\sigma_G$, for $G$:

$$G_N=\frac{1}{\alpha N}\sum\limits_{1}^{N} log(1+\alpha Z_n)$$

Here is an intuitive outline for possible solution - an attempt of qualitative approximation to the Monte-Carlo results.

1.1 assume $\alpha$ is small and truncated taylor series can be used: $log(1+x) \approx x - \frac{1}{2}x^2$,

then $G$ becomes: $G \approx \frac{1}{\alpha N}(\sum \alpha Z - \frac{1}{2}\sum \alpha^2 Z^2)$

or $G \approx G_A +G_B$, where $G_A=\frac{1}{N}\sum Z$, $G_B=-\frac{1}{2}\frac{\alpha}{N}\sum Z^2$

1.2 assume berry-esseen can be applied to $G_A$, then $G_A$ becomes:

$G_A \approx 0 + \frac{1}{\sqrt N}Z_A$, with $Z_A=\mathcal{N}(0, 1)$

1.3 assume $\chi^2$ can be approximated as $\mathcal{N}(N, 2N)$, then $G_B$ becomes:

$G_B \approx -\frac{1}{2}\alpha - \frac{1}{\sqrt{2N}}\alpha Z_B$, with $Z_B=\mathcal{N}(0, 1)$

1.4 assume $G_A$ and $G_B$ are correlated with $\rho_N=\mathrm{Cov}(G_A,G_B)$ then we finally have approximate resulting $G$: $$G_N \approx -\frac{1}{2}\alpha +\frac{1}{\sqrt{N}}(1+\rho_N\alpha + \frac{1}{2}\alpha^2)^{\frac{1}{2}}Z$$

This solution is already in use (granted - with some accuracy concerns).

• Please do not post questions as answers. Check also stats.stackexchange.com/help and stats.stackexchange.com/tour pages to learn about etiquette of this page e.g. not posting duplicate questions, not posting questions as answers or comments etc. – Tim Apr 7 '15 at 19:12
• I already have 3 variations of this question and do not want to multiply them any more. – Andrei Pozolotin Apr 7 '15 at 19:28
• That is exactly what I mean: you have 3 versions of the same question and also post answer to one of them that asks additional questions on the same topic. – Tim Apr 7 '15 at 19:49
• my intent is to invite someone else to provide "similar-in-approach", but "much-better-worked-out" answer to the same topic, w/o loss of focus – Andrei Pozolotin Apr 7 '15 at 20:03
• So this qualifies as a question rather then answer and should be posted as so. – Tim Apr 7 '15 at 21:29