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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} $ 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.

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    @draks The OP did not mention it but this question is a direct followup of [this one](http://math.stackexchange.com/q/143056/6179).2012-05-10

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If the distribution of $Z$ has density $f$, then the distribution of $T=Z^2$ has density $g$ where, for every $t\gt0$,

$ \color{green}{g(t)=\frac1{2\sqrt{t}}\left(f(\sqrt{t})+f(-\sqrt{t})\right)}. $

This is a simple consequence of the definition of $f$ and of the change of variables formula applied to the expression of $\mathrm E(u(Z^2))$ as an integral, for every bounded measurable function $u$.

Since @Sasha gave you the density $f$, you are done.