# sum of independent Rayleigh random variables [closed]

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.

## closed as off-topic by Yiyuan Lee, Guy, Did, user63181, Claude LeiboviciMar 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}
• 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 $X+Y$. – Robert Israel Mar 26 '14 at 3:12
• $$\int_0^z \dfrac{t(z-t)}{b_1^2 b_2^2} \exp\left(-\dfrac{t^2}{2b_1^2} - \dfrac{(z-t)^2}{2b_2^2}\right)\; dt$$ – Robert Israel Mar 26 '14 at 5:22