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Find the distribution of $\sqrt{X^2+Y^2}$ where $X$ and $Y$ are independent normally distributed $\mathcal{N}(0,1)$.

What is the best way to go about this? I tried finding the distribution of $X^2$ and $Y^2$ and then adding the two distributions, but this results in an undefined integral.

Or do we let $U=\sqrt{X^2+Y^2}$ and $V=X$ and then find the joint distribution of $U$ and $V$ and then find the marginal distribution of $U$?

I have the answer I am meant to reach; just can't get there.

guest10923
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2 Answers2

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$$\begin{eqnarray*} F_z(z) = \mathbb{P}[X^2+Y^2\leq R^2]&=&\frac{1}{2\pi}\iint_{x^2+y^2\leq R^2}e^{-\frac{x^2+y^2}{2}}\,dx\,dy \\ &=& \frac{1}{2\pi}\int_{0}^{2\pi}\int_{0}^{R}\rho e^{-\rho^2/2}\,d\rho\,d\theta = 1 - e^{-R^2/2} \end{eqnarray*}$$

hence the distribution of $Z=\sqrt{X^2+Y^2}$ has a probability density function supported on $\mathbb{R}^+$ and given by: $$f_Z(z)=z\, e^{-z^2/2}.$$

albusSimba
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Jack D'Aurizio
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Let $X=R\cos A$ and $Y=R\sin A$ where $R>0$ and $A\in (0,2\pi)$. You want to find the distribution of $R$.

So the Jacobian turns out to be $r$.

So the joint distribution of $R$ and $A$ is $$f_{R,A}(r,a)=\dfrac{r}{2\pi}e^{\dfrac{-r^2}{2}}$$.Note that $R$ and $A$ are independent.

Integrate out $A$ from $0$ to $2\pi$ to obtain the marginal of $R$ as $f_R(r)=re^\dfrac{-r^2}{2}$ where $r>0$

Landon Carter
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  • Thank you! I wouldn't have thought about using a polar transformation. – guest10923 Mar 30 '15 at 11:29
  • I remember when I faced this question for the first time, just a month ago. :) I tried this method and it worked !! :) Your method will also work but the steps will be messy. – Landon Carter Mar 30 '15 at 11:30
  • By the way, adding $X^2$ and $Y^2$ gives you the very famous chi-square distribution, whose root is something you will find difficult to find, probably. – Landon Carter Mar 30 '15 at 11:31