Conventional wisdom says yes or mostly. But consider the following simple derivation:

Let $y = \prod_{i=1}^M x_i$ where $x_i\sim U(0,1)$. Then from independence, $E[y] = 2^{-M}$.

Now, if we let $z = \ln y = \sum_{i=0}^M \ln x_i$ we can use the central limit theorem to claim that $z\sim Norm(\mu=-M,\sigma^2 = M)$ where we get $\mu = M \int_0^1 \ln x dx = -M$ and $\sigma^2 = M\int_0^1 (\ln x + 1)^2 dx = M$ from the standard change-of-variable equations for $\ln x$ combined with the CLT. This leads to a Log-Normal distribution of y as: $$f_Y(y) = \frac{1}{y\sqrt{2\pi M}}e^{-\frac{(\ln y +M)^2}{2M}} $$ Great! We got a distribution for $y$. But if you try and calculate the expectation now, you get $e^{\mu + \frac{\sigma^2}{2}} = e^{-M/2} \neq 2^{-M}$. What's worse, if you look at the asymptotic behavior, $e^{-M/2} \gg 2^{-M}$ as $M\rightarrow \infty$.

This doesn't make sense to me. How can $z$ be considered "normal" in the limit of large numbers if it leads to a divergent systematic deviation in the calculation of the expectation of $y$? Is this an artifact of the finite support in $x_i$? If so, how to take it into account? More importantly, if $y$ is not really Log-Normal, to what degree are calculations of other properties valid? Ultimately, I need to be able to predict the most probable value (mode) of $y$. Can I quantify the extent to which $y$ is not quite Log Normal and present some kind of bounds on the accuracy of my calculations?


As you defined $z$, it will always be negative, and thus cannot be Normal. You can find its density as follows:

If $U$ is a Uniform $(0,1)$ r.v. then $P[-\log(U) \leq x] = P[U \geq e^{-x}] = 1 - e^{-x}$ for all $x \in [0,\infty)$, so $-\log U$ is an exponential r.v. with $\lambda=1$. That means that $$ -\sum_{i=1}^M \log(x_i) \sim \text{Gamma}(M,1), $$ i.e., the density of $-Z$ is $$ f_{-Z}(z) = \frac{z^{M-1} e^{-z}}{(M-1)!}, \ \ z > 0. $$ Then using a simple change of variable the density for $Z$ is $$ f_Z(z) = \frac{(-z)^{M-1} e^z}{(M-1)!}, \ \ z \leq 0. $$

I also don't see how you've used the CLT.

  • $\begingroup$ Indeed, that seems correct. But this gives me $f_Y(y) = \frac{(-\ln y)^{M-1}}{(M-1)!)}$ which gives a mode of 0 (because it diverges as $y\rightarrow 0$) which isn't right... $\endgroup$ – mmdanziger Apr 4 '14 at 8:02
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    $\begingroup$ Here is a check for $M=2$: \begin{align} P[U_1U_2 \geq x] &= \int_x^1 \int_{x/u_1}^1 du_2du_1 \\ &= \int_x^1(1-x/u_1)du_1 \\ &= 1 - x + x\ln x \end{align} so $$\frac{d}{dx} P[U_1U_2 \leq x] = \frac{d}{dx}(x - x\ln x) = -\ln x.$$ I think the result is correct as stated $\endgroup$ – user139388 Apr 6 '14 at 5:40
  • $\begingroup$ Yes, you're right. The mode is 0. $\endgroup$ – mmdanziger Apr 6 '14 at 9:09

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