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} $ If we convolve $P(x)$ and $P(y)$ then we will be able to get the $P(z)$. What will be the distribution of $Z^2$ as $Z$ is not a known distribution?
Remark: this question is a follow-up to this earlier question.