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Assuming $x_1,x_2,\ldots, x_n$ are $n$ independent variables from standard Gaussian distribution $N(0,1)$. Then we construct a new variable by $y=\Pi_{i=1}^n x_i$. Can anyone show the probability density function of $y$?

olivia
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  • See if anything here is helpful to you. – hexaflexagonal Jul 20 '15 at 03:15
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    go to the logarithm, then convoluate the probability densities. $p(x_i = x) = a_i e^{- b_i (x-\mu_i)^2}$, $p(\ln x_i = t) = e^{\displaystyle- b_i (e^t-\mu_i)^2}$, $\ln y = \sum \ln x_i$, $\ln p(\ln y = t) = \ast \ln p(\ln y)\ast \ldots p(\ln x_n)$ – reuns Jul 20 '15 at 03:16
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    The problem with the logarithm approach is that the $x_i$ can be negative. – Ian Jul 20 '15 at 03:22
  • Well, to solve the problem of negative numbers in the $\log$, you could take the absolute values and take the PDF for that. Then recognize by symmetry teh probability of the product being negative is 1/2. – Michael Jul 20 '15 at 04:24

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