# $X \sim \operatorname{Rice}(\nu,\sigma)$, what is the distirbution of $X^2$?

Let $$X = |\nu e^{j\theta}+W|$$, where $$W \sim \mathcal{CN}(0,2\sigma^2)$$, then $$X\sim \operatorname{Rice}(\nu,\sigma)$$. What is the distribution of $$X^2$$?

Note that $$X$$ also can be written in terms of real and imaginary parts: $$X = \sqrt{\Re(X)^2 + \Im(X)^2}$$, where $$\Re(X) \sim \mathcal N(\nu\cos\theta,\sigma^2)$$ and $$\Im(X) \sim \mathcal N(\nu\sin\theta,\sigma^2)$$:

$$\begin{equation} \tag{1} f_{X}(x) = I_0\bigg(\dfrac{x·\nu}{\sigma^2}\bigg)\dfrac{x}{\sigma^2} e^{-\dfrac{x^2+\nu^2}{2 \sigma}}. \end{equation}$$

I know that if $$\sigma = 1$$ the distribution is $$X^2 \sim \chi_{2}^{,2} (\nu^2)$$, i.e. non-central chi squared distribution with noncentrality parameter $$\nu^2$$ and 2 degrees of freedom... But what happens for an arbitrary factor $$\sigma$$?

I tried to follow the theory and compute the PDF of $$X^2$$ as following:

Let $$Y=X^2$$ $$\begin{equation} \tag{2} f_{Y}(y) =\frac{f_X(\sqrt{y})+f_X(-\sqrt{y})}{2\sqrt{y}} \end{equation}$$

However I ended up with the result $$f_Y(y)=0$$ ($$I_0$$ function is symmetric), which I assume is not correct.

So I found out that if $X\sim Rice(\nu,\sigma)$, than $Y = \big(\frac{X}{\sigma}\big)^2 \sim \chi_2^2\Big(\big(\frac{\nu}{\sigma}\big)^2\Big)$ with $f_Y(y)$.
So $X^2 = \sigma^2 Y$, i.e. $f_{X^2}(x) = \frac{1}{\sigma^2}f_Y\big(\frac{x}{\sigma^2}\big)$. However, still do not know if it can be expressed in terms of a distribution.