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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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Welcome to MathStackExchange. Your question is hard to read. Do you mean $f(x;\sigma) = \frac{x}{\sigma^2} e^{-x^2/2\sigma^2}, \quad x \geq 0, $ when you're talking about Rayleigh distribution? –  draks ... May 10 '12 at 12:28
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@draks The OP did not mention it but this question is a direct followup of this one. –  Did May 10 '12 at 14:21

1 Answer 1

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.

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