Let $X_1$ and $X_2$ have the following joint density:

$$f_{X_{1},X_{2}}(x_1,x_2) = \begin{cases} {\frac{4}{\pi}e^{-(x_1^2+x_2^2)}} & \text{$x_1 \gt 0$, $x_2 \gt 0$} \\{0} & \text{otherwise}\end{cases}$$

Letting $Y_1=X_1^2$ and $Y_2 =X_1^2 + X_2^2$, give the joint pdf of $Y_1$ and $Y_2$.

I have avoided using Jacobian Transformations in the past because it seemed complicated, but I think using it would be much easier than alternative methods in this case. I wanted to make sure I'm doing this correctly.

$$\begin{align*} J(x_1,x_2) &= \begin{vmatrix}\frac{\partial x_1^2}{\partial x_1}&&\frac{\partial x_1^2}{\partial x_2}\\\frac{\partial x_1^2+x_2^2}{\partial x_1}&&\frac{\partial x_1^2+x_2^2}{\partial x_2}\end{vmatrix}\\\\ &= \begin{vmatrix}2x_1&&0\\2x_1&&2x_2\end{vmatrix}\\\\ &= 4x_1x_2 \end{align*}$$

For $x_1$ and $x_2$ we have

$x_1=\sqrt{y_1}$ and $x_2=\sqrt{y_2-y_1}$

Thus I have,

$$\begin{align*} f_{Y_1,Y_2}(y_1,y_2) &= f_{X_1,X_2}\left(\sqrt{y_1},\sqrt{y_2-y_1}\right) \cdot \left |J(x_1,x_2)\right|^{-1}\\\\ &= \frac{4}{\pi}\cdot e^{-(\sqrt{y_1}^2+\sqrt{y_2-y_1}^2)}\cdot\frac{1}{4x_1x_2}\\\\ &= \frac{1}{\pi\sqrt{y_1}\sqrt{y_2-y_1}}\cdot{e^{-y_2}} \\\\ \end{align*}$$

So the pdf I obtained is

$$f_{Y_{1},Y_{2}}(y_1,y_2) = \begin{cases} \frac{e^{-y_2}}{\pi\sqrt{y_1}\sqrt{y_2-y_1}} & \text{$y_1 \gt 0$, $y_2 \gt 0$} \\{0} & \text{otherwise}\end{cases}$$

Is this correct?


If $Y_1 = X_1^2$ and $Y_2=X_1^2+X_2^2$ then $Y_2 \gt Y_1$ so

$$f_{Y_{1},Y_{2}}(y_1,y_2) = \begin{cases} \frac{e^{-y_2}}{\pi\sqrt{y_1}\sqrt{y_2-y_1}} & \text{$y_1 \gt 0$, $y_2 \gt y_1$} \\{0} & \text{otherwise}\end{cases}$$

Checking, I get that $$\int_0^\infty \int_x^\infty \frac{e^{-y_2}}{\pi\sqrt{y_1}\sqrt{y_2-y_1}} = 1$$

  • $\begingroup$ There should not be an $x$ in the bounds, and the order for integration must be indicated. SO: $$\int_0^\infty\int_{y_1}^\infty \dfrac{\mathrm e^{-y_2}}{\pi\,\sqrt{y_1}\,\sqrt{y_2-y_1}}\,\mathrm d y_2\,\mathrm d y_1 = 1$$ $\endgroup$ Oct 27 at 4:20

1 Answer 1


It is correct!

Tip: Let $Y_1 = g_1(X_1, X_2) = X_1^2$ and $Y_2 = g_2(X_1, X_2) = X_1^2 + X_2^2$, so your Jacobian is $$J_g(x_1,x_2) = \begin{vmatrix} \dfrac{\partial g_1}{\partial x_1} & \dfrac{\partial g_1}{\partial x_2} \\ \dfrac{\partial g_2}{\partial x_1} & \dfrac{\partial g_2}{\partial x_2} \end{vmatrix}$$

But, if you build yout Jacobian with inverses ($X_1 = h_1(Y_1, Y_2) = \sqrt{Y_1}$ and $X_2 = h_2(Y_1,Y_2) = \sqrt{Y_2 - Y_1}$) you get:

$$J_h(y_1,y_2) = \begin{vmatrix} \dfrac{\partial h_1}{\partial y_1} & \dfrac{\partial h_1}{\partial y_2} \\ \dfrac{\partial h_2}{\partial y_1} & \dfrac{\partial h_2}{\partial y_2} \end{vmatrix}$$

and your pdf is: $$f_{Y_1,Y_2}(y_1,y_2) = f_{X_1,X_2}(h_1,h_2) \lvert J_h(y_1,y_2) \rvert$$

without inverting the Jacobian.

You can try this to check your answer.


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