How exactly Rayleigh, exponential and Chi-square distributions are related? I have read on Wikipedia and understood that

*

*If $R$ is Rayleigh distributed with standard deviation = 1 then only $R^2$ has a chi-squared distribution.


*Also from various sources, I had seen that if $R$ is Rayleigh distributed with standard deviation other than 1 then $R^2$ has exponential distribution.
My query is that is my understanding given in above two points correct?
Any help in this regard will be highly appreciated.
 A: Start observing that the definition of a Reileigh $R(\sigma)$ is the following
$$R=\sqrt{X^2+Y^2}$$
where $X,Y$ are iid gaussian $N(0;\sigma^2)$

Proof:
$$f_{XY}(x,y)=\frac{1}{\sigma^22\pi}e^{-(x^2+y^2)/(2\sigma^2)}$$
passing in  polars you get
$$f_{\Theta P}(\theta,\rho)=\frac{1}{2\pi}\mathbb{1}_{(0;2\pi)}(\theta)\times \frac{\rho}{\sigma^2} e^{-\rho^2/(2\sigma^2)}\mathbb{1}_{(0;\infty)}(\rho)$$
where $\Theta\sim U(0;2\pi)$ is the distribution of the angle and $P\sim\text{Rayleigh}(\sigma)$ is the distribution of the radius $R=\sqrt{X^2+Y^2}$

thus it is evident (and anyway easy to prove) that if $\sigma=1$, $R^2$ is the sum of two independent standard gaussian, that is
$$R^2(1)\sim \chi_{(2)}^2$$
It is not difficult to prove that
$$R^2(\sigma)\sim\text{Exp}\left( \frac{1}{2\sigma^2} \right)$$

Proof:
being $Y=X^2$ a monotonic transformation when $x\ge 0$ we have
$x=\sqrt{y}$; $|x'|=1/(2\sqrt{y})$ and thus
$$f_Y(y)=\frac{\sqrt{y}}{\sigma^2}\frac{1}{2\sqrt{y}}e^{-y/(2\sigma^2)}=\frac{1}{2\sigma^2}e^{-y/(2\sigma^2)}$$

Using MGF properties, immediately follows that
$$\Sigma_i R_i^2\sim\text{Gamma}\left(n; \frac{1}{2\sigma^2} \right)$$
You can find other related distribution on wiki here
