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Given $X,Y \sim \mathcal{N}(0,1)$ and $Z=\sqrt{X^2 + Y^2}$, find the PDF of $Z$.

I know from digging around that this will follow a Rayleigh distribution since the sum of two squared normally distributed variables follow an exponential distribution and the root of an exponentially distributed variable follows something called a Rayleigh distribution. Wikipedia confirms this can be thought of as the root of the sum of two normally distributed variables with variance $\sigma$ $$ \begin{split} Z &\sim \mathcal{R}(1) \\ f_Z(z) &= ze^{-z^2/2} \end{split} $$

That said I really have no idea where to begin. My normal approach to is to think of things in terms of the CDF, i.e. $$ \mathbb{P}[Z < z] = \mathbb{P}\left[\sqrt{X^2+Y^2}<\sqrt{x^2+y^2}\right] $$ And then find an expression of the CDF that is easy to work with.

gt6989b
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J N
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2 Answers2

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Perhaps your approach is fruitful? $$ F_Z(z) = \mathbb{P}\left[\sqrt{X^2+Y^2} < z\right] = \iint_{D_z} f_X(x) f_Y(y) dxdy, $$ where $D_z$ is the region bounded by $\sqrt{x^2+y^2}<z$.

Since $X,Y \sim \mathcal{N}(0,1)$, we know $f_X(x) = f_Y(x)$ and can plug into the integral above and perhaps switch to polar coordinates...

gt6989b
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It is impossible to say anything about the distribution of $Z$ without extra hypothesis. I will assume that $X$ and $Y$ are independent so that the question does have an answer.

In this case we have $P(Z<z)=\iint_{(x,y): x^{2}+y^{2}<z^{2}} \frac 1 {2\pi} e^{-(x^{2}+y^{2})/2)}dxdy=\frac 1 {2 \pi} \int_0^{2\pi} \int_0^{z} e^{-r^{2}/2}rdr d\theta $. making the substitution $s=r^{2}$ we get $P(Z<z)=\frac 1 {2} \int_0^{z^{2}}e^{-s/2}ds= 1-e^{-z^{2}/2}$ for $z>0$.

Kavi Rama Murthy
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