The question and solution is attached

Kindly eloborate the solution

- Feb 22nd 2011, 09:12 AMmoonnightingaleCan any body explain me this solution of Rayleigh distribution question
The question and solution is attached

Kindly eloborate the solution - Feb 24th 2011, 04:50 AMmoonnightingale
any comments on this plz

- Feb 24th 2011, 08:25 AMchaotic
Since you wanted at leats one comment :)

We are doing this now in class! Well, I can tell you what they are doing generally but I would have to look into it a little more. This is functions of random variables and the technique implored here is the CDF. If you are familiar with the definition of CDF then afterwards it's just integration. What you see being integrated is the pdf of the normal distribution with mean zero and variance sigma^2. Gaussian distribution is the same as normal distribution and because there are two variables x and y, it's a double integral and to solve it it looks like they converted the integral into polar coordinates.

Once I go over my notes I'll be able to tell you exactly what's happening. - Feb 24th 2011, 09:10 AMtheodds
You should ask a more specific question. What part of the solution are you having problems with? The third equality is a polar transformation, if that is the source of confusion. Or perhaps it is the multiple typos in the solution.

This is probably not of much interest, but I feel compelled to say it. $\displaystyle X^2 / \sigma^2$ and $\displaystyle Y^2 / \sigma^2$ are chisquare, so $\displaystyle X^2 + Y^2 \sim \sigma^2 \chi^2_2$ (by independence) which is exponential; square root an exponential to get a Rayleigh and we are done without using calculus :) - Feb 24th 2011, 10:41 AMmoonnightingale
- Feb 24th 2011, 11:50 AMCaptainBlack
If you think of the bivariate normal that you get from a pair of independent zero mean equal variance RV you van think of this as a distribution over the $\displaystyle $$x-y$ plane, which can be rewritten as in polars then your $\displaystyle z=r$. Then the density of $\displaystyle $$z$ comes out of the wash simply from the change of variables, in polars the symmetric bivariate normal distribution is constant as a function of $\displaystyle $$\theta$ and Rayleigh in $\displaystyle $$r$

CB