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Joint PDF of $(X,Y)$ is $f(x,y)=12xy(1-y)$ for $0 < x < 1, 0 < y < 1$ and $0$ elsewhere. Let $Z=XY^2$ and $W=Y$ be a joint transformation of $(X,Y)$.

Find the Jacobian, and hence the joint PDF.

Find the PDF of $Z=XY^2$ from the joint PDF of $(Z,W)$.

When I solve this I keep getting a divergent integral for the marginal pdf of $Z$, am I doing something wrong?

My Working:

The support of $Z$ and $W$ is $0< z <1$ and $0< w <1$

$v_2(z,w)=Y=W, Z=XY^2$ , $v_1(z,w)=X=Z/W^2$

$$J= \begin{pmatrix} dX/dZ & dX/dW\\ dY/dZ & dY/dW\\ \end{pmatrix} = \begin{pmatrix} 1/W^2 & -2Z/W^3\\ 0 & 1\\ \end{pmatrix}=1/W^2 $$

$f(z,w)=|J|.f(v_1(z,w),v_2(z,w))=(1/w^2)[(12z/w^2)(w)(1-w)]=(12z/w^3)(1-w)$

$f_1(z)=\int_0^1f(z,w)\,dw$, which is a divergent integral

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    $\begingroup$ It would also help if you show your work, whatever you have done so far. $\endgroup$
    – StubbornAtom
    May 24, 2018 at 4:49
  • $\begingroup$ If you do not expose your attempts, we cannot distinguish a call for help from an attempt at having one's homework solved for free. $\endgroup$
    – Xi'an
    May 24, 2018 at 7:19
  • $\begingroup$ I added my working out :) $\endgroup$
    – Ricardo723
    May 24, 2018 at 10:36
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    $\begingroup$ Hint: for starters, check whether your claimed joint density for $X$ and $Y$ is a valid joint density. After fixing the error, re-do your work and check if the corrected joint density that you find for $Z$ and $W$ is a valid joint density. For example, where do make a note of the fact that if $Y = W= 0.5$, (say), then $Z$ cannot possibly exceed $0.25$, and so your expression for $f_{Z,W}$ cannot be valid for all $z, w \in (0,1)$? This restriction will affect your marginal pdf calculation. $\endgroup$
    – Dilip Sarwate
    May 24, 2018 at 14:38

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