[Math] the distribution of sum of a Gaussian and a Rayleigh distributed independent r.v.

Question

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 $Z$?

Answer 1

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.