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 $Z$?
What is the distribution of sum of a Gaussian and a Rayleigh distributed independent r.v.?
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0Would you like to know if the distribution of $x+y$ belongs to a known class of distributions? Also: are $x,y$ independent? – 2012-05-09
2 Answers
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.
The correct answer is
$\frac{\exp\left(-\frac{|x|}{\sigma_w \sigma_h} \right)}{2 \sigma_w \sigma_h}$
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2Please use LaTeX for math. And don't comment on your own answers to say they are right. – 2015-09-30