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I came across these two postings: $f(x,y)=12xy(1-y)$, $x,y \in (0,1)$. Find distribution of $Z=XY^{2}$., and How to deduce the CDF of $W=I^2R$ from the PDFs of $I$ and $R$ independent, and this article.

There are a lot of high level discussions relating to some topics that are out of my knowledge level.

It looks like the derivation seems very complicated. Is there a "easier way" (the way that a beginning level student can understand) to derive the pdf of $Z$ where $Z=XY^{2}$?

Pantelis Sopasakis
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Amos Ku
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    The straightforward approach to try would be: figure out which pairs of $X,Y$ values correspond to the event ${Z\le z}$, use this to write down an expression for $P(Z\le z)$ (which is the cdf of $Z$, as a function of $z$), and then differentiate that function to get the pdf. – Karl Apr 09 '20 at 06:25
  • That is what I started with as I found this method described in one of my book example, but I get lost soon afterwards because that example is for two uniform random variables. – Amos Ku Apr 09 '20 at 07:58

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Define $V=Y^2\sim \chi^2_1$ $$F_Z(z)=\mathbb P(Z\leq z)=\mathbb P(XY^2 \leq z) =\mathbb P(XV \leq z)=\int_0^{+\infty} P(XV \leq z |V=v) f_V(v) dv$$

$$=\int_0^{+\infty} P(X \leq \frac{z}{v} |Y=y) f_V(v) dv=\int_0^{+\infty} P(X \leq \frac{z}{v} ) f_V(v) dv$$

So by derivation

$$f_Z(z)=\int_0^{+\infty} \frac{1}{v} f_X(\frac{z}{v} ) f_V(v) dv$$

$$=\int_0^{+\infty} \frac{1}{v} f_X(\frac{z}{v} ) \frac{1}{\sqrt{2} \Gamma(\frac{1}{2})} v^{\frac{1}{2}-1} e^{-\frac{v}{2}} dv$$

$$=\int_0^{+\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}\frac{z^2}{v^2}} \frac{1}{\sqrt{2} \Gamma(\frac{1}{2})} v^{-\frac{3}{2}} e^{-\frac{v}{2}} dv$$

$$= \frac{1}{2\pi}\int_0^{+\infty} v^{-\frac{3}{2}} e^{-\frac{v}{2}-\frac{z^2}{2v^2}} dv$$

Now you need to solve the integral.

Masoud
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