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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](http://en.wikipedia.org/wiki/Rayleigh_distribution)?2012-05-10
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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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