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Let $Z=X+Y$; where $X\sim \mathscr N(0,\sigma^2_1)$ i.e. a Gaussian random variable and $Y$ follows the Rayleigh distribution: $$ f_Y(y) = \frac{y}{\sigma^2_2}\exp\left(-\frac{y^2}{2\sigma^2_2}\right) \mathbf{1}_{y \geqslant 0} $$ What will be the distribution of $Z$?

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Would you like to know if the distribution of $x+y$ belongs to a known class of distributions? Also: are $x,y$ independent? – Ilya May 9 '12 at 13:19

The probability density function for $Z$, $f_Z(z)$ is obtained as a convolution of $f_X$ and $f_Y$, assuming $X$ and $Y$ are independent: $$ f_Z(z) = \int_{0}^\infty f_Y(y) f_X(z-y) \mathrm{d} y = \frac{1}{\sqrt{2 \pi} \sigma_1 \sigma_2^2} \int_0^\infty y \exp\left( -\frac{y^2}{2 \sigma_2^2}\right) \exp\left( - \frac{(z-y)^2}{2 \sigma_1^2} \right) \mathrm{d} y $$ Evaluation of this integral is straightforward, but tedious, and is done via integration by parts, with the following result: $$ f_Z(z) = \frac{\sigma_2 z}{\left(\sigma_1^2+\sigma_2^2\right)^{3/2}} \mathrm{e}^{-\frac{z^2}{2 \left(\sigma_1^2+\sigma_2^2\right)}} \Phi\left( \frac{\sigma_2}{\sigma_1} \frac{ z}{ \sqrt{\sigma_1^2+\sigma_2^2}}\right) + \frac{\sigma_1}{\sqrt{2 \pi} \left(\sigma_1^2+\sigma_2^2\right)} \mathrm{e}^{-\frac{z^2}{2 \sigma_1^2}} $$ where $\Phi(x)$ is the cumulative distribution function of the standard normal random variable.

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The correct answer is

$$\frac{\exp\left(-\frac{|x|}{\sigma_w \sigma_h} \right)}{2 \sigma_w \sigma_h}$$

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this is the correct answer – Ghanim Sep 30 '15 at 17:47
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Please use LaTeX for math. And don't comment on your own answers to say they are right. – vonbrand Sep 30 '15 at 18:13

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