Representing a Rayleigh distribution by Gamma distribution

gamma distributionprobability distributions

Rayleigh distribution is formulated as $$P(x\mid\sigma)=\frac{x}{\sigma^2}\exp\left(-\frac{x^2}{\sigma^2}\right) \,,\tag{1}$$ where $\sigma^2$ is the variance. $Z$ is a complex variable specified as $Z=a+jb$.

A paper said that Gamma distribution $\Gamma(k,\theta)$ can represents the Rayleigh distribution when $k=\theta=1$.

I know that $\Gamma(k,\theta)=\frac{x^{k-1}e^{-\frac{x}{\theta}}}{\theta^k\Gamma(k)}$ and $\Gamma(1,1)=e^{-x}$ which is unequal to the above Rayleigh distribution $(1)$.

I don't understand why the Gamma distribution can represent the Rayleigh distribution and if the conclusion is correct how to prove that?

Thanks a lot! : )

Best Answer

The statement is wrong! (better, the statement is correct but something was left implied)

  1. Some notes:
  • The density you call Rayleigh is not a nice density. The Rayleigh's density is the following, for $x>0$

$$f_X(x;\sigma^2)=\frac{x}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}$$

  • The variance is $V[X]=(2-\frac{\pi}{2})\sigma^2$
  1. to Represent a Rayleigh with a Gamma you have to do a variable transform:

If X is a $Gamma(n,\theta)$ and you transform it with

$Y=\sqrt{X}$ then your statement is correct.

$f(y;1;1) \sim Ray(\frac{1}{2})$

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