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The joint distribution of $X$ and $Y$ is given by $f(x,y)=12xy(1-y)$ if $x,y \in (0,1)$. Find the distribution of $Z=XY^{2}$.

I tried to find the joint distribution of $Z$ and $X$:

$$g(z,y)=f \left(\frac{z}{y^{2}},y\right) \left| \frac{\partial x}{\partial z} \right|=12*\frac{z}{y^2}*y(1-y)*\frac{1}{y^2}=12z\left( \frac{1}{y^3} - \frac{1}{y^2}\right)$$

From there, I tried to find the marginal distribution of $Z$:

$$ \int_0^1 12z\left( \frac{1}{y^3} - \frac{1}{y^2}\right) \,dy$$

However, this integral does not converge. I'm not exactly sure what I'm missing. Any advice is appreciated!

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1 Answer 1

up vote 4 down vote accepted

In your second equality, you replace $f(\frac z{y^2},y)$ with $12\frac z{y^2}y(1-y)$, but this is incorrect as $\frac z{y^2}$ might be larger than 1, in which case $f(\frac z{y^2},y) = 0$. Taking this into account, your integral turns into $$ \int_{\sqrt z}^1 12z\bigg( \frac1{y^3}-\frac1{y^2} \bigg) \,dy $$ which now converges.

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