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Our joint pdf is $f(x,y) = \frac{1}{\sqrt{2π}} e^\frac{x^2+y^2}{2}$

Now we let $ U = X^2 + Y^2 $ and $ V = Y$, we can then get our Jacobian as $ J = \frac{1}{\sqrt{u-v^2}} $

Since this transformation isnt one to one on this range, we need to divide into to ranges $S_{1} = (-∞,0] $ and $ S_{2} = (0, ∞) $

Thus I get $f(u,v) = \frac{1}{\sqrt{u-v^2}}(\frac{1}{2π}e^\frac{-u}{2}) $

This seems wrong, for some reason. Please tell me if this is incorrect thus far.

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  • $\begingroup$ Use polar coordinates if you want to do this by change of variables. $\endgroup$ – StubbornAtom Feb 2 at 20:03
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By definition, it's Chi squared with two degree of freedoms. The one you've found is the joint PDF of $U$ and $V$. You need to marginalize it to obtain $f_U(u)$: $$f_U(u)=\int_{-\sqrt u}^\sqrt u \frac{1}{\sqrt{u-v^2}}\frac{e^{-u/2}}{2\pi}dv=\frac{e^{-u/2}}{2\pi}\overbrace{\int_{-\sqrt{u}}^{\sqrt u}\frac{1}{u-v^2}dv}^\pi=\frac{1}{2}e^{-u/2}$$ where $u\geq 0$. This is at the same time an exponential RV with $\lambda=1/2$, and Chi-Squared RV with $k=2$.

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  • 1
    $\begingroup$ thank you, this showed me my range was wrong for my marginal. appreciate it $\endgroup$ – Kevin G Feb 2 at 20:19
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If I'm not mistaken, this should have chi-square distribution with 2 degrees of freedom. If X and Y are standard normal, then this should follow from the definition of the chi square distribution. See here.

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