Transformation of two i.i.d. uniform random variables

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G'day folks,

I'm trying to work through a problem in preparation for an exam and it's got me stumped. The question is:

Let $X_{1}$ and $X_{2}$ be i.i.d. $U(0,1)$ random variables. Let $Y_{1}=X_{1}X_{2}$ and $Y_{2}=X_{1}^{2}$.

Evaluate the joint pdf of $Y_{1}$ and $Y_{2}$.

My attempt:

We have $X_{1}=\sqrt{Y_{2}}$ and $X_{2}=\frac{Y_{1}}{\sqrt{Y_{2}}}$. The Jacobian of the inverse map is $J=-\frac{1}{2y_{2}}$.

Because $X_{1}$ and $X_{2}$ are independent, they have joint pdf $$f_{X_{1},X_{2}}(x_{1},x_{2})=f_{X_{1}}f_{X_{2}}$$

But this is just $1$, right?! For $0<x_{1}<1,\;0<x_{2}<1$.

That was my first point of confusion but oh well, maybe it'll come out in the wash. So I progress:

The part I'm having real trouble with is with the region that $Y_{1},Y_{2}$ are defined over.

The original region is the open region bounded by the $x_{1}$-axis, $x_{2}$-axis, the line $x_{1}=1$ and the line $x_{2}=1$, so a unit square.

In order to picture what was going on, I tried the following:

$(x_{1},x_{2})\mapsto(x_{1}x_{2},x_{1}^{2})$ so consider what happens to the boundaries described above. \begin{align*} (\phi,0)&\mapsto(0,0)\\ (0,\phi)&\mapsto(0,0)\\ (\phi,1)&\mapsto(\phi,\phi^{2})\\ (1,\phi)&\mapsto(\phi,1) \end{align*}

So all points along the axes are mapped to the origin, the boundary that was $x_{2}=1$ is now $x_{2}=x_{1}^{2}$ and the line $x_{1}=1$ is $x_{2}=1$.

I am quite confused! My intention was to use the following:$$f_{Y_{1},Y_{2}}(y_{1},y_{2})=|J|f_{X_{1},X_{2}}(x_{1}(y_{1},y_{2}),x_{2}(y_{1},y_{2}))$$

But I have no clue where to go from where I'm stuck. Any help with my understanding of the problem would be much appreciated!

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For the required region covered by $Y_1,Y_2$, note that $Y_1=X_2\sqrt{Y_1},\;$ so $0\lt Y_1\lt \sqrt{Y_2}\lt 1$. enter image description here

I think you have your first boundary mapping wrong, which should be $(\phi,0)\mapsto (0,\phi^2)$.

Your working otherwise looks fine and you have the pdf correct: $f_{Y_1,Y_2}(y_1,y_2) = \dfrac{1}{2y_2}.$