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.

Really appreciate your help, thanks.

  • $\begingroup$ Perhaps you could provide the functional form/parameterisation of the Rice distribution that you are using. There are almost always multiple versions around ... $\endgroup$
    – wolfies
    May 8, 2015 at 16:17
  • $\begingroup$ Better? @wolfies $\endgroup$ May 8, 2015 at 16:27

1 Answer 1


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.


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