Knowing the distributions of $X$ and $Y$, find the distribution of $Z=XY^2$

Question

I know how to approach this kind of exercises, but I tried several times to calculate correctly the integrals, but I am doing some mistakes.

So, let $X$ and $Y$ two independent variables with distributions, respectively

$f(x) = 6x(1-x)$

And

$g(y)=2y$

And $x, y, \in [0,1] $

Calculate the distribution of $Z=XY^2$.

So, I have tried to do any kind of calculation, but I am doing something wrong.

Can anyone help me please?

Answer 1

Let $W=Y^2$. For $t\in(0,1)$ we have $$ \mathbb P(W\leqslant t) = \mathbb P(Y\leqslant \sqrt t) = \int_0^{\sqrt t}2y\ \mathsf dy = t, $$ so $W$ has density $f_W(t) = \mathsf 1_{(0,1)}(t)$. Now for the product $Z=XW$, we have \begin{align} \mathbb P(Z\leqslant z) &= \mathbb P(XW\leqslant z)\\ &= \mathbb P(W\leqslant z/X)\\ &= \int_0^\infty f_X(x)\int_0^{z/x}f_W(w)\ \mathsf dw\ \mathsf dx. \end{align} Differentiating with respect to $z$, the density of $Z$ is given by \begin{align} f_Z(z) &= \int_0^\infty f_X(x)f_W(z/x)\frac1x\ \mathsf dx\\ &= \int_z^1 6x(1-x)\frac1x\ \mathsf dx\\ &= 3(1-z)^2\cdot\mathsf 1_{(0,1)}(z). \end{align}

Answer 2

$W=Y^2$ and $P(W\le w)=w$ (as shown by Math1000) $F_Z(z|X=x)=P(Z\le z|X=x)=P(XW\le z|X=x)=P(W\le z/x|X=x)$.

For $x\gt z,\ F_Z(z|X=x)=z/x$. For $x\le z,\ F_Z(z|X=x)=1$.

Integrate over the density of $X$ to get $F_Z(z)=\int_0^z6(x-x^2)dx+z\int_z^16(1-x)dx=(3z^2-2z^3)+z(3z-6z^2+3z^3)=(z-1)^3+1$. Note that this is for $0\le z\le 1$ $F_Z(z)$ $=0$ for $z\lt 0$ and $=1$ for $z\gt 1$.

The density function for $Z$ is $3(z-1)^2$.