Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $k$.

share|improve this question
    
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

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.