Let the variable $Z$ equal $Z = XY$ where $X$ and $X$ are two i.i.d. continuous random variables which distributions are given by $f_X()$ and $f_Y$. The distribution of $Z$ is given by:

$$f_Z(z) = \int_{-\infty}^\infty f_X(x)f_Y(z/x)\frac{1}{|x|}dx$$

, when $f_X = f_Y = f$ the equation becomes

$$f_Z(z) = \int_{-\infty}^\infty f(x)f(z/x)\frac{1}{|x|}dx$$

If $Z$ is the product of $4$ i.i.d random variables $Z=\prod_{i=1}^4 X_i$, then the distribution of $Z$ is

$$f_Z(z) = \int_{-\infty}^\infty\left(\int_{-\infty}^\infty\left( \int_{-\infty}^\infty f(x_1)f(x_2/x_1)\frac{1}{|x_1|}dx_1\right) f(x_3/x_2)\frac{1}{|x_2|}dx_2\right)f(z/x_3)\frac{1}{|x_3|}dx_3$$

  • Is it correct? If yes, Is there a better way to write this down?

  • What if $Z$ is the product of $n$ i.i.d random variables $Z=\prod_{i=1}^n X_i$? How could we write down the formula?


$$f_Z(z)=\int_{\mathbb R^{n-1}}f(x_1)f(x_2)\cdots f(x_{n-1})f\left(\frac{z}{x_1x_2\cdots x_{n-1}}\right)\frac{\mathrm dx_1\mathrm dx_2\cdots\mathrm dx_{n-1}}{|x_1x_2\cdots x_{n-1}|}$$

  • $\begingroup$ Thanks for your answer. Could you please develop a bit on how you found out this formula. Thanks a lot! $\endgroup$ – Remi.b Mar 29 '14 at 13:27
  • $\begingroup$ Yes: change of variables + Jacobian. $\endgroup$ – Did Mar 29 '14 at 13:35

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