I was wondering if there is a closed-form expression for the sum of n iid Rayleigh RVs. Here is what I tried: An exponential density function $$f_x(x)=\frac{1}{2 \sigma^2}exp(-\frac{x^2}{2\sigma^2})$$

can be transformed from a Rayleigh density $$f_z(z)=\frac{z}{\sigma^2}exp(-\frac{z^2}{2\sigma^2})$$ using $f_x(x)=\frac{1}{(2 \sqrt{x})}f_z(\sqrt{x})$ because $x=z^2$.

From this, the density of the sum of n iid values of x (exponentially distributed) can be determined analytically using convolution. There is a thread on this already.

$$f_t(t)=\left(\frac{1}{2\sigma^2}\right)^n \frac{t^{n-1}}{(n-1)!}exp(-\frac{t^2}{2\sigma^2})$$

I thought I could then transform this density to arrive at the density of the sum of Rayleigh RVs using $y=\sqrt{t}$, because Rayleigh and exponential RVs are related this way. The resulting density is

$$f_y(y)=2 y f_t(y^2)$$

Using Mathcad, if I compare the shape of the pdf $f_y(y)$ to the pdf obtained using n-fold convolution of $f_t(t)$, how I would normally do it, the results are not the same. I suppose this could be a numerical methods problem.
However, I have never seen a closed-form solution to this problem, so I expect I could have a problem with my reasoning here. Any help on this is appreciated. Thanks!

  • $\begingroup$ See Wikipedia on 'Rayleigh distribution' for PDF. $\endgroup$ – BruceET Jan 18 '18 at 18:57
  • $\begingroup$ I viewed the Wikipedia page regarding the Rayleigh PDF. While this seems to thoroughly describe the properties of Rayleigh RVs, I can't see that it provides a closed-form expression for the sum of n such variables. Further, it is not evident that it provides the information required to derive the expression I am seeking. Some clarification would be appreciated. Thanks. $\endgroup$ – John M. Jan 18 '18 at 22:54
  • $\begingroup$ I did not intend to imply that Wikipedia worked your problem. Based on what I read in you Questions, it seemed to me that the article might be helpful. If not, sorry about that. $\endgroup$ – BruceET Jan 19 '18 at 1:20

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