This is a quick question. If $X\sim\operatorname{Normal}$, is $X^2$ rayleigh distributed? I ask this question is because from wiki, it says $X^2$ is called the Chi-Squared Distribution with a degree freedom of $1$.
How does that reduce to the rayleigh distribution?
Update:
According to wiki:
A second example of the distribution arises in the case of random complex numbers whose real and imaginary components are i.i.d. (independently and identically distributed) Gaussian with equal variance and zero mean. In that case, the absolute value of the complex number is Rayleigh-distributed.
No, the Rayleigh distribution involves multiplying by $x$ in the definition for the density at $x$, so $X^2$ is not Rayleigh distributed and in general Chi-squared and Rayleigh distributions are always different. See http://en.wikipedia.org/wiki/Rayleigh_distribution#Related_distributions for various non-linear relationships that do hold between Rayleigh and other distributions, including Chi-squared.