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What is the probability density function of the following product of uniformly distributed random variables: $Y=X_1\dotsb X_n$, where $X_n \thicksim U[1,2]$ (Uniform distribution); the $X_n$ are independent.

OBS: It is not a duplicate. I found the answer only for the $U[0,1]$ and here, there is no analytical equation for the pdf.

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  • $\begingroup$ Why are you interested in this probability distribution, just out of interest? $\endgroup$ – Kenny Wong May 11 '17 at 2:31
  • $\begingroup$ are they independent ? $\endgroup$ – user379195 May 11 '17 at 4:50
  • $\begingroup$ @AnUser I haven't got an explicit formula to give you off hand. However, somebody posted a very nice method for computing these sorts of things yesterday on math.SE: math.stackexchange.com/questions/2274582/… $\endgroup$ – Kenny Wong May 11 '17 at 8:35
  • $\begingroup$ @SoumikGhosh yes, they are independent. I updated this information in the Edit. $\endgroup$ – AnUser May 11 '17 at 14:17
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Let's assume independence.

If $n$ is large enough then the product can be considered as lognormal by the central limit theorem since the logarithm of the product is a sum of iid. random variables with finite mean and finite variance.


Let $X$ be of uniform over $[1,2]$ then the pdf of $\ln(X)$ is $e^x$ over $[0,\ln(2)]$, $0$ otherwise. The mean is $\ln(4)-1$ and the standard deviation is $\sqrt{\ln^2(2)-\ln^2(4)+3}$.


The mean of $\ln(\prod_{i=1}^nX_i)$ is $\mu=n(\ln(4)-1)$ and its variance $\sigma^2=n(\ln^2(2)-\ln^2(4)+3)$ and the density is

$$f_{\prod_{i=1}^n X_i}(x)\approx\frac{1}{x\sigma\sqrt{2\pi}}e^{-\frac{(\ln(x)-\mu)^2}{2\sigma^2}}.$$


Finally, the random variable $$\frac{\ln(\prod_{i=1}^nX_i)-n(\ln(4)-1)}{\sqrt n\sqrt{\ln^2(2)-\ln^2(4)+3}}$$

is very close of standard normal if $n$ is large.

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  • $\begingroup$ To illustrate, could you plot your limit pdf when say $n=5$ and $n=25$ and something larger? $\endgroup$ – wolfies May 11 '17 at 14:38
  • $\begingroup$ Nice transform ! $\endgroup$ – user2879934 May 11 '17 at 14:53
  • $\begingroup$ @wolfies: I will do a simulation. $\endgroup$ – zoli May 11 '17 at 15:07

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