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How do I find the probability density distribution (pdf) of the sum of independent Rayleigh random variables (whose probability density functions are known)? where is the reference? Could anybody please possibly paste the original convolution integral (definition) which expresses the pdf of the sum of independent Rayleigh random variables with different scale parameters? Thanks in advance.

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closed as off-topic by Yiyuan Lee, Guy, Did, user63181, Claude Leibovici Mar 21 '14 at 9:27

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I don't think there's a name for it. For example, according to Maple the sum of two independent Rayleigh random variables with the same scale parameter $b$ has PDF $$ f(z) = \dfrac{z}{2b^2} e^{-z^2/(2b^2)} + \dfrac{\sqrt{\pi} z^2}{4b^3} \text{erf}\left(\frac{z}{2b}\right) e^{-z^2/(4b^2)} - \dfrac{\sqrt{\pi}}{2b} \text{erf}\left(\frac{z}{2b}\right) e^{-z^2/(4b^2)}$$

EDIT: if the scale parameters are $b_1$ and $b_2$, Maple gets $$\eqalign{\sqrt {\pi/2}{{\rm e}^{-1/2\,{\frac {{z}^{2}}{{b_{{1}}}^{2}+{b_{{2}}} ^{2}}}}} {{\rm erf}\left(1/2\,{\frac {\sqrt {2}b_{{2}}z}{b_{{1}}\sqrt {{b_{{1}}}^{2}+{b_{{2}}}^{2}}}}\right)} \left( {z}^{2}-{b_{{1}}}^{2}-{b_{{2}}}^{2} \right) b_{{2}}b_{{1}} \left( {b_{{1}}}^{2}+{b_{{2}}}^{2} \right) ^{-5/2}\cr+ \sqrt {\pi/2}{{\rm e}^{-1/2\,{\frac {{z}^{2}}{{b_{{1}}}^{2}+{b_{{2}}}^{2}} }}} {{\rm erf}\left(1/2\,{\frac {z\sqrt {2}b_{{1}}}{b_{{2}}\sqrt {{b_{{1}}}^{2}+{b_{{2}}}^{2}}}}\right)} \left( {z}^{2}-{b_{{1}}}^{2}-{b_{{2}}}^{2} \right) b_{{2}}b_{{1}} \left( {b_{{1}}}^{2}+{b_{{2}}}^{2} \right) ^{-5/2}\cr+z{b_{{ 1}}}^{2}{{\rm e}^{-1/2\,{\frac {{z}^{2}}{{b_{{1}}}^{2}}}}} \left( {b_{ {1}}}^{2}+{b_{{2}}}^{2} \right) ^{-2}+{{\rm e}^{-1/2\,{\frac {{z}^{2}} {{b_{{2}}}^{2}}}}}z{b_{{2}}}^{2} \left( {b_{{1}}}^{2}+{b_{{2}}}^{2} \right) ^{-2}\cr} $$

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  • $\begingroup$ It appears right. if b1^2+b2^2=1, the pdf could be simplified. It seems that it is not easy to determine the integral (between 0 and infinity). Could you please possibly paste the original convolution integral which leads to such result. $\endgroup$ – Jinlu Mar 26 '14 at 1:36
  • $\begingroup$ Actually I didn't do an integral, just (in Maple's Statistics package) defined $X$ and $Y$ as random variables with the appropriate distributions and asked for the PDF of $X+Y$. $\endgroup$ – Robert Israel Mar 26 '14 at 3:12
  • $\begingroup$ Desperately, the convolution integral which leads to the Robert's formula via Maple is sought. Anybody can provide it? Thank you so much in advance. $\endgroup$ – Jinlu Mar 26 '14 at 4:23
  • $\begingroup$ $$\int_0^z \dfrac{t(z-t)}{b_1^2 b_2^2} \exp\left(-\dfrac{t^2}{2b_1^2} - \dfrac{(z-t)^2}{2b_2^2}\right)\; dt$$ $\endgroup$ – Robert Israel Mar 26 '14 at 5:22
  • $\begingroup$ Wonderful. Many thanks for your post, Prof. Robert. Would you please possibly expalin a litte more about the upper limit being the argument (z) not being the infinity. Usually, the upper limit of the convolution integral for the pdf of the sum of independent random varaibles is the variable "infinity", am I right? Why here it is the argument (z) for the Rayleigh random variable? Is it beacuse the fact that the difference (z-t) is neagive is not allowed for the Rayleigh distribution? I am not so sure about this. Looking forward to hearing from you. Thanks alot for your help. $\endgroup$ – Jinlu Mar 26 '14 at 8:52

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