CDF of product of three independent random variables

Question

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.

Answer 1

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