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I've been reading a text on probability theory and it thoroughly looked at the distribution of the sum of i.i.d. random variables but on the last page, they made a comment about the distribution of the product of i.id. random variables that confused me:

They said that if you had i.i.d. random variables $Y_1,Y_2,...$ that are uniformly distributed on (0,1), then the distribution of $Y_1Y_2\cdots Y_n$ is approximately $e^{X}$, where X is normal with mean $n\mu$ and standard deviation $\sigma\sqrt{n}$ (for some $\sigma>0$ and real $\mu$. Can someone please explain why this is?

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    $\begingroup$ Write the product as the exponential of a sum of i.i.d. random variables and apply the central limit theorem to these. $\endgroup$ – Did Nov 14 '11 at 21:17
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    $\begingroup$ In other words, apply the central limit theorem to the logarithm of the product. $\endgroup$ – Michael Hardy Nov 14 '11 at 21:37
  • $\begingroup$ Let $W_i=\log(Y_i)$. In principle you even almost don't need to compute the mean and variance of $W_i$, since you are not asked to identify $\mu$ and $\sigma$. (But you do need to know the mean and variance exist.) However, the calculation of the mean and variance of $\log(Y_i)$ is not hard by integration by parts. $\endgroup$ – André Nicolas Nov 15 '11 at 0:16
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Let $Y_i\sim U(0,1)$ be IID. Now let: $$ \mathcal{Y}=\prod_{i=1}^n Y_i $$ Next, define: $$ \mathcal{Y} = \exp(\ln(\mathcal{Y})) = \exp\left(\sum_{i=1}^n \ln(Y_i)\right)=\exp(\mathfrak{X}) $$ where we let $X_i=\ln(Y_i)$ and defined $$ \mathfrak{X}=\sum_{i=1}^n \ln(Y_i) $$ Next, we can assume $X_i$ has mean $\mu=\mathbb{E}[X_i]$ and variance $\sigma^2=\mathbb{V}[X_i]$. Now we use the central limit theorem, which says: $$ \lim_{n\rightarrow\infty}\sum_{i=1}^n X_i \,\xrightarrow[\,]{\;d\;}\, \mathcal{N}(n\mu,n\sigma^2) $$ which means that $$ \mathfrak{X}\,\xrightarrow[n\rightarrow\infty]{\;d\;}\,\mathcal{N}(n\mu,n\sigma^2)$$ So, for large enough $n$, we get $\mathcal{Y}=e^\mathfrak{X}$, where $\mathfrak{X}$ is approximately normally distributed with mean $\mathbb{E}[\mathfrak{X}]=n\mu$ and standard deviation $\mathbb{V}[\mathfrak{X}]=\sigma\sqrt{n}$


See also this question, for more on the exact distribution of $X_i$.

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