# Prove that the Cartesian coordinates of a Rayleigh distribution are independent

I'm stuck on a question with a weird twist on the usual Rayleigh distribution.

Instead of assuming that the components of the distribution are independent, we are given alternative conditions and asked to show independence. The setup is as follows:

• Let $R \geq 0$ be a random variable with density $f_R(r) = re^-\frac{r^2}{2}$
• Let $\Theta$ be uniformly distirbuted on $[-\pi,\pi]$ so that $f_\Theta(\theta) = \frac{1}{2\pi}$
• Show that the variables $X = R\cos(\theta)$ and $Y = R\sin(\theta)$ are independent and have density $f_X(t) = f_Y(t) = \frac{e^{-\frac{t^2}{2}}}{\sqrt{2\pi}}$

My current approach is to show that $f_{XY}(x,y) = f_X(x)f_y(y)$. I have used the multivariate change of density formula to obtain the correct distribution for $f_{XY}(x,y)$, which is $\frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}$. However I cannot figure out how to obtain expressions for $f_X(x)$ and $f_Y(x)$.

Please note The question is actually designed to show that $\int_{-\infty}^\infty e^{-\frac{t^2}{2}} = \sqrt{2\pi}$, so we cannot assume that this is true. This effectively means that we cannot integrate $f_{XY}$ over the range of Y to get $f_X$.

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## 1 Answer

All you need is that the joint density that you've calculated correctly (except for a typo in the exponent) factorizes. That's the definition of independence. In the present case you can argue from symmetry that the factor $\frac1{2\pi}$ has to be split between $x$ and $y$ so the marginal densities must be $\frac1{\sqrt{2\pi}}\mathrm e^{-x^2/2}$ and $\frac1{\sqrt{2\pi}}\mathrm e^{-y^2/2}$, but even if you couldn't, you could still argue that the joint density factorizes even without knowing how to split the normalization constant between the two factors.

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Thanks for this! I am wondering: does this mean that a pair of random variables $X$ and $Y$ are independent if I can express their joint PDF as the product of two functions that only depend on $X$ and $Y$? – Elements Oct 19 '12 at 17:30
@Elements: Yes, that's what it means. If you have $p(x,y)=f(x)g(y)$, then integrating shows that the marginal distributions are $p(x)=f(x)G$ and $p(y)=g(y)F$, where $F$ and $G$ are the integrals of $f$ and $g$, respectively; so $p(x,y)=p(x)p(y)/(GF)$, but $GF=1$ by normalization, so $p(x,y)=p(x)p(y)$. – joriki Oct 19 '12 at 19:48
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