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If we have a number $X\sim f(x|\lambda)$ where $$f(x|\lambda) = \lambda e^{\lambda x}, x \geq 0$$ then $\sqrt{X} \sim r(y|\sigma)$ where $$r(y|\sigma) = \frac{y}{\sigma^{2}} e^{-y^{2} /(2 \sigma^{2})}, y \geq 0$$ which is the Rayleigh distribution. Why is this the case? It is easy to prove this mathematically with the transformation technique, as shown here, but I'd like to have a more intuitive understanding on why this arises.

For example the emergence of the Rayleigh distribution can be readily explained by considering the absolute value of two normally, uncorrelated vector components.

Dilip Sarwate
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Q.P.
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    Doesn't your last remark answer your question? – whuber May 04 '21 at 18:43
  • I feel the two must be related, and perhaps my question is somewhat circular. I suppose then I am asking: $\sqrt{x} = \sqrt{a^{2} + b^{2}}$, which implies that $a^{2} + b^{2}$ must belong to an exponential distribution. – Q.P. May 04 '21 at 20:33
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    The sum of squares of two iid zero-mean Normal variates has an exponential distribution. – whuber May 04 '21 at 22:11
  • As far as I am concerned it is all about the densities. $e^{-x^2/2}$ and similar functions are difficult to integrate, while $e^{-x}$ is easy and $xe^{-x^2/2}$ is not much harder. So the question becomes what something with a density proportional to $xe^{-x^2/2}$ might represent. There are at least two answers: (a) take the square root of a suitable exponential distribution using a change of variable, or (b) take a 2D polar version of a standard normal with uniformly distributed angle, in effect multiplying the normal density by $2\pi x$ and then standardising. – Henry May 05 '21 at 00:24

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