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Suppose you have

$$ Y = X^2+Y^2 $$

where $X$ and $Y$ are both Gaussian with zero mean and variance $\sigma^2/2$ (you can think of $y$ as the square norm of $Z = X + jY$). The pdf should be Erlang, but I'm not sure about the values of $\lambda$ and $k$.

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In the section of related distributions, $Y^{2}$ is Rayleigh distributed. It seems not difficult to find the distribution for $Y$ since it's always positive. Hope this could offer you a new perspective. – newbie Aug 18 '11 at 19:11
y = x^2 + y^2 forces x and y to be bounded, which is not the case for Gaussians. – zyx Aug 18 '11 at 19:28
You've used the same letter, $Y$, for two different things. – Michael Hardy Aug 18 '11 at 20:53
up vote 1 down vote accepted

The easiest way to solve this would be compute characteristic function $\mathbb{E}(\mathrm{e}^{i t Y}) = \mathbb{E}(\mathrm{e}^{i t (X_1^2 + X_2^2)}) = \mathbb{E}(\mathrm{e}^{i t X_2^2}) \mathbb{E}(\mathrm{e}^{i t X_1^2})$.

For normal variate:

$$ \mathbb{E}(\mathrm{e}^{i t x^2}) = \int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\sqrt{\pi} \sigma} \mathrm{e}^{i t x^2} \mathrm{e}^{- \frac{x^2}{\sigma^2}} = \frac{1}{\sigma} \left( \frac{1}{\sigma^2} - i t\right)^{-\frac{1}{2}} = \frac{1}{\sqrt{1- i t \sigma^2}} $$

Then you would compare this to the characteristic function of $\Gamma$ distribution to determite that it is the characteristic function of of $\Gamma(\alpha = \frac{1}{2}, \beta = \sigma^2)$.

Added: I should add, after seeing Michael's response that combining we get $\mathbb{E}(\mathrm{e}^{i t Y}) = (1-i t \sigma^2)^{-1}$, and this is gamma distribution with $\alpha=1$, also known as exponential distribution.

Alternatively you should use the fact that $X_1 =^d \frac{\sigma}{\sqrt{2}} Z_1$, where $Z_1$ is standard normal variate. Then $Y = X_1^2+X_2^2 = \frac{\sigma^2}{2} (Z_1^2+Z_2^2)$. Notice that $Z_1^2 + Z_2^2$ follows $\chi^2_2$, i.e chi-squared with two degrees of freedom, hence $Y$ is a rescaled $\chi^2_2$ variate.

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I see that Sasha has written a very careful answer with a conspicuous omission: What you get is an exponential distribution, with expected value $\sigma^2$ (since the variance you started with is $\sigma^2/2$.

The chi-square distribution with two degrees of freedom is a memoryless exponential distribution.

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