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Let $X$, $Y$ be i.i.d. random variables with distribuition $\mathcal{N}(0,1/2)$ and $Z = X^2 + Y^2$. I'd like to prove based on $X$ and $Y$ pdf's that $Z$ has exponential distribuition.

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    $\begingroup$ Hint: convert the integral to polar coordinates, where $Z$ becomes the square of the radius. $\endgroup$ – whuber Feb 11 '17 at 23:27
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First use the joint pdf of $X$ and $Y$ and switch to polar coordinates, then $$ \mathbb{P}(Z\leq z)=\mathbb{P}(X^2+Y^2\leq z)=\frac{1}{\pi}\int_{x^2+y^2\leq z}e^{-x^2+y^2}\;dxdy=\frac{1}{\pi}\int_{0}^{2\pi}\int_0^{\sqrt{z}}e^{-r^2}r\;drd\theta$$ $$=2\int_0^{\sqrt{z}}re^{-r^2}\;dr $$ Now if we set $u=r^2$ then we get $$ \mathbb{P}(Z\leq z)=\int_0^ze^{-u}\;du$$

so $Z$ is exponentially distributed with rate parameter $\lambda = 1$.

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  • $\begingroup$ How do we know the joint CDF of $X^2$ and $Y^2$? $\endgroup$ – gwg May 16 at 21:55
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Z has a chi square distribution with the number of degrees of freedom to make it the special case of the exponential. X and Y are required to be independent.

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  • $\begingroup$ Just edited to add that they are i.i.d. $\endgroup$ – Felipe Augusto de Figueiredo Feb 11 '17 at 23:04
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    $\begingroup$ This post merely restates the result. The question asks to show how this is the case using the distribution functions. $\endgroup$ – whuber Feb 11 '17 at 23:28
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    $\begingroup$ @whuber But Michael Chernick's answer is what you once described to me re another question as a statistician's answer (in contrast to the answer that I had given to the question which you characterized as a mathematician's (or probabilist's) answer. So why is Michael Chernick's answer put down as a mere restatement of the result to be proved? To my mind, it is the perfect statistician's answer. $\endgroup$ – Dilip Sarwate Feb 12 '17 at 0:46
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    $\begingroup$ @Dilip I suppose that in some cases the distinctions between "why," "what," and "how" might be subtle. In this particular instance, the context of the question--with particular reference to its explicit request to use pdfs--makes it clear to me that this post, although it's partially correct, is not responsive. $\endgroup$ – whuber Feb 12 '17 at 15:56

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