Find the P.D.F for these Variables(Jacobian transformation)

probabilityprobability distributions

$Q)$ Let the continuous variables $X$ and $Y$ which are following the uniform distribution on $D=\{(x,y) \vert 0 < x < y < 1 \}$

Find the joint probability density function for variables $Y$ and $Z$ (Here the $Z = {X \over Y}$)

I tried this by Jacobi transformation $(X,Y) \to (Y,Z)$

Since $f(X,Y) = 2$ on $D$, Then the answer would be $2\vert J\vert$.

Then, only we have to do just find the $\vert J \vert $(Here the $J$ is a determinant of the Jacobian matrix) Here the $J$ = $
\begin{vmatrix}
x_y & x_z \\
y_y& y_z \\
\end{vmatrix}
$

So since the $Z = {X \over Y}$, $J$ = $
\begin{vmatrix}
Z & Y \\
1& -{X \over Z^2} \\
\end{vmatrix}
$ =$ 2Y$
(Here the $Y={X \over Z}$, $X = ZY$)

Therefore my answer is $f(Y,Z)= 4Y$.

But the answer sheet claimed $J$ = $
\begin{vmatrix}
x_y & x_z \\
y_y& y_z \\
\end{vmatrix} =
\begin{vmatrix}
Z & Y \\
1& 0 \\
\end{vmatrix} = -Y
$ so the answer was $f(Y,Z)= 2Y$.

What the point did I have a mistake? I have a doubt about the its claim since the variable $Y$ itself depends on the $Z$. So do I have to differentiate $Y= {X \over Z}$, not $Y$?

Any help would be appreciated. thanks.

Best Answer

So since the $Z = {X \over Y}$, $J = \begin{Vmatrix}Z & Y \\ 1& -{X \over Z^2}\end{Vmatrix} = 2Y$ (Here the $Y={X \over Z}$, $X = ZY$)

What the point did I have a mistake?

You must express $X$ and $Y$ in terms of $Y$ and $Z$.

$X=YZ$ and $\underline{Y=Y}$ so

$$J=\begin{Vmatrix}x_y&x_z\\[2ex] y_x&y_z \end{Vmatrix}=\begin{Vmatrix}\tfrac{\partial yz}{\partial y}&\tfrac{\partial yz}{\partial z}\\ \tfrac{\partial y}{\partial y}&\tfrac{\partial y}{\partial z}\end{Vmatrix}=\begin{Vmatrix}z&y\\[2ex]1&0 \end{Vmatrix}=\lvert -y\rvert$$

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