We know that the square of a Rayleigh random variable has exponential distribution, i.e.,
Let the random variable $X$ have Rayleigh distribution with PDF
$$f_X(x)=\frac{2x}{\alpha}e^{-x^2/{\alpha}}.$$
Then the random variable $Y=X^2$ has the PDF given by $$f_Y(y)=\frac{1}{\alpha}e^{-y/{\alpha}}.$$
For an exponentially distributed r.v. $Y$ with mean $\mathbb{E}[Y]=1$
$$\mathbb{E}[Y^{\delta}]=\Gamma[1+\delta].$$
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Now, if the random variable $X$ has Rician distribution (unit power in direct and scattered paths), whose PDF is given by
$$f_X(x)=\frac{2x}{\alpha}\text{exp}\left(\frac{-(x^2+v^2)}{\alpha}\right)I_0\left(\frac{2xv}{\alpha}\right)$$
with $\frac{v^2}{\alpha}=1$ and $I_0(z)$ is the modified Bessel function of the first kind with order zero.
what is the PDF of $Y=X^2$?
And what is $\mathbb{E}[Y^{\delta}]$ when $\delta<1.$
Note: when $v^2=0$, $X$ has Rayleigh distribution.
Best Answer
Using a Dirac-delta method $$ f_Y(y)=\int_0^\infty dx\ \frac{2x}{\alpha}\text{exp}\left(\frac{-(x^2+v^2)}{\alpha}\right)I_0\left(\frac{2xv}{\alpha}\right)\delta(y-x^2) $$ $$ \int_0^\infty dx\ \frac{2x}{\alpha}\text{exp}\left(\frac{-(x^2+v^2)}{\alpha}\right)I_0\left(\frac{2xv}{\alpha}\right)\frac{\delta(x-\sqrt{y})}{2\sqrt{y}}=\frac{1}{\alpha}\exp\left(\frac{-(y+\nu^2)}{\alpha}\right)I_0\left(\frac{2\nu\sqrt{y}}{\alpha}\right)\ , $$ which for $\nu=0$ correctly reproduces the exponential PDF. Note that $\int_0^\infty f_Y(y)dy=1$, so the PDF is correctly normalized.
$$ \mathbb{E}[Y^\delta]=\int_0^\infty dy\ y^\delta \frac{1}{\nu^2}\exp\left(\frac{-(y+\nu^2)}{\nu^2}\right)I_0\left(\frac{2\sqrt{y}}{\nu}\right) =\nu ^{2 \delta } \Gamma (\delta +1) L_\delta(-1)\ , $$ as computed by Mathematica (for $\nu^2/\alpha=1$), where $L_\delta(x)$ is a Laguerre function [I checked a few numerical values and it seems it works just fine].