# PDF for square of Rician random variable? [closed]

Let 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 $0<\delta<1.$

Note: when $v^2=0$, $X$ has Rayleigh distribution.

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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 $\mathbb{E}[Y]=1$

$$\mathbb{E}[Y^{\delta}]=\Gamma[1+\delta].$$

• This question does not seem to concern the Mathematica software package, so it may be more appropriate for Cross Validated. – MarcoB Mar 1 '16 at 2:59

dist = RiceDistribution[v, Sqrt[a/2]];

PDF[dist, x] where

DistributionParameterAssumptions[dist] // Simplify

(*  v >= 0 && Sqrt[a] > 0  *)


The distribution for Y = X^2 is

dist2 = TransformedDistribution[X^2, X \[Distributed] dist];


with PDF

PDF[dist2, y] The expectation (moment) is

m[d_] = Expectation[y^d, y \[Distributed] dist2] Alternatively,

m[d] == Moment[dist2, d]

(*  True  *)


For the case with v = 0

(m[d] /. v -> 0) // FunctionExpand

(*  a^d Gamma[1 + d]  *)