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Given, $X$, $Y$, $Z$ are three Rayleigh distributed independent random variables. I need to calculate the product CDF of the variables, i.e. $F_\lambda(x)=\text{?}$ where, $\lambda = XYZ$.

I am trying to solve the problem as follows.

\begin{align} F_\lambda(p) & = \Pr(\lambda < p) \\[8pt] & = \Pr(XYZ < p) \\[8pt] & = \Pr(X < \frac{p}{YZ}) \\[8pt] & = 1 - \int_0^\infty \int_0^\infty e^{-p/(YZ\delta_1)} \frac 1 {\delta_1} f_{1/Y}(a) f_{1/Z}(b) \, da\,db \end{align}

I didn't go further as I think the way I am solving might not be correct.

Could you please help? Thank you.

MK71
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    Sometimes knowing the answer will help one figure out how to get there. (But sometimes not.) Using Mathematica if the three independent Rayleigh distributions all have the same parameter $\sigma$, then the cdf of $XYZ$ is $\frac{8 \sigma ^6-p^2 G_{0,3}^{3,0}\left(\frac{p^2}{8 \sigma ^6}| \begin{array}{c} -1,0,0 \ \end{array} \right)}{8 \sigma ^6}$ where $G$ is the Meijer G function. – JimB Jul 01 '21 at 03:48
  • You might want to look at https://math.stackexchange.com/questions/55684/pdf-of-product-of-variables. – JimB Jul 01 '21 at 04:15
  • Hi! The solution should help a lot. Thank you. – MK71 Jul 01 '21 at 04:20
  • Could you please share a little detail about how to get to this solution? – MK71 Jul 01 '21 at 04:28
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    Here is the Mathematica code: dist = TransformedDistribution[x y z, {x \[Distributed] RayleighDistribution[s], y \[Distributed] RayleighDistribution[s], z \[Distributed] RayleighDistribution[s]}]; CDF[dist, p]. – JimB Jul 01 '21 at 04:33
  • Thank you so much! – MK71 Jul 01 '21 at 04:37

1 Answers1

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I believe the density function for $U=XYZ$ can be expressed as $$f_U(u)=\int_0^\infty\int_0^\infty f_X(\frac{u}{yz})\frac{f_Y(y)}{y} \frac{f_Z(z)}{z} dy dz$$ in terms of the density functions for $X,Y,Z$.

While cdf $F_U(u)=\int_0^\infty\int_0^\infty F_X(\frac{u}{yz}) f_Y(y) f_Z(z) dy dz$.

herb steinberg
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  • Thank you! Is this equation for the PDF of the product of X, Y, and Z? How about the CDF? – MK71 Jul 01 '21 at 04:24
  • I edited result to include cdf. – herb steinberg Jul 01 '21 at 17:35
  • Thank you so much! – MK71 Jul 02 '21 at 00:02
  • Could you share a little detail about how did you get to this cdf equation? I will also be grateful if you can share any insight on how to find any closed form solution to this equation. – MK71 Jul 02 '21 at 00:07
  • $P(U\le u)=\int\int P(X\le\frac{u}{yz}|(Y=y)\cap(Z=z))f_Y(y)f_Z(z)dydz$ assuming independence of $X,Y,Z$. I have made no attempt at getting a closed form solution for these Rayleigh distributions. – herb steinberg Jul 02 '21 at 00:16
  • Much appreciated! – MK71 Jul 02 '21 at 00:45
  • I will be grateful if you advise about one more thing. How the PDF equation updates when $U = \frac{{XY}}{Z}$? Is it like $\int_0^\infty {\int_0^\infty {{f_X}\left( {\frac{{uz}}{y}} \right)\frac{{{f_Y}\left( y \right)}}{y}} \frac{{{f_Z}\left( z \right)}}{z}} dydz$ – MK71 Jul 02 '21 at 03:13
  • It is clearer when you use cdf. $\int\int F_X(\frac{uz}{y})f_Y(y)f_Z(z)dydz$ So pdf is $\int\int f_X(\frac{uz}{y})\frac{z}{y}f_Y(y)f_Z(z)dydz$ – herb steinberg Jul 02 '21 at 03:52
  • Thank you so much! – MK71 Jul 02 '21 at 04:40
  • Hello Sir! May I have your advice on the following question? https://math.stackexchange.com/questions/4217119/pdf-related-to-three-random-variables – MK71 Aug 05 '21 at 23:33