Difference between Rician distribution and Gaussian distribution

Question

could any one please tell me the difference between Rician and Gaussian Distribution and the advantages of using one over other please.With some mathematical proof would be truly appreciated Thank you.

Answer 1

This Wikipedia article begins as follows:

In probability theory, the Rice distribution or Rician distribution is the probability distribution of the magnitude of a circular bivariate normal random variable with potentially non-zero mean. It was named after Stephen O. Rice.

That should make clear that they're two different things. The normal (or "Guassian") distribution has as its support the whole real line, whereas the distribution of magnitudes of any random vector has to be supported within the half-line $[0,\infty)$. And if you look at these graphs of densities, you see that (as the above statement would lead you to think) they are asymmetric, unlike normal (i.e. "Gaussian") distributions, which are symmetric:

http://upload.wikimedia.org/wikipedia/commons/a/a0/Rice_distributiona_PDF.png

Answer 2

The Rice distribution is related to the Gaussian distribution as $R \sim \mathrm{Rice}(|\nu|,\sigma)$ if $R=\sqrt{X^2+Y^2}$ where $X \sim \mathcal{N}(\nu \cos(\theta), \sigma^2)$ and $Y \sim \mathcal{N}(\nu \sin(\theta), \sigma^2)$ are independent normal random variables. See Wikipedia.

As for an application of the Rice distribution, it is sometimes found in telecommunications, such as in satellite navigation (GPS, Galileo, GLONASS, BeiDou, ...). A GNSS receiver typically receives a signal as follows: $$x(t) = d(t) c(t-\tau) \cos(2\pi f_0t+\theta) + \eta(t),$$ where $d(t)$ is some data (value of $+1$ or $-1$), $c(t)$ is a known pseudorandom code that repeats continuously (value of $+1$ or $-1$), $\tau$ is the unknown transmission delay, $f_0$ is the carrier frequency (considered known here for our example, but in real life there is an uncertainty and its value should be searched), $\theta$ is the unknown carrier phase you don't know, and $\eta$ is a white Gaussian noise.

The goal of the GNSS receiver is to synchronise with $c(t)$ to measure the distance from the satellite (this is called the pseudorange), i.e. to determine when the code is starting, and to extract the content of $d(t)$ to get the satellite position, the time of the transmission, and other information required such that the receiver can determine its position.

The first thing you need to do is to synchronise with $c(t)$. For that you compute a correlation between the received signal and a local replica. If the local replica is aligned with the received signal, you get a high correlation peak, otherwise you get a very small value (the pseudorandom codes are designed for that).

The correlation consists in computing the following operation for different $\hat{\tau}$ until you get a high correlation peak (meaning that $\hat{\tau}$ is very close to $\tau$): $$y_{\mathrm{i},k}(\hat{\tau}) = \frac{1}{T} \int_{kT}^{(k+1)T} x(t) c(t-\hat{\tau}) \cos(2\pi f_0t)\mathrm{d}t.$$ Note that the data is constant for the duration of the integration, i.e. $d(t) = d_k$ for $kT \leq t < (k+1)T$, else it would not work (or at least you would have additional losses). Let's look at the case where the estimated delay is correct ($\hat{\tau} = \tau$): $$ \begin{align} y_{\mathrm{i},k}(\tau) &= \frac{1}{T} \int_{kT}^{(k+1)T} x(t) c(t-\tau) \cos(2\pi f_0t)\mathrm{d}t \\ &= d_k \frac{1}{T} \int_{kT}^{(k+1)T} c(t-\tau) c(t-\tau) \cos(2\pi f_0t+\theta) \cos(2\pi f_0t)\mathrm{d}t \\ &\quad + \frac{1}{T} \int_{kT}^{(k+1)T} \eta(t) c(t-\tau) \cos(2\pi f_0t) \mathrm{d}t \\ &= d_k \frac{1}{T} \int_{kT}^{(k+1)T} \cos(4\pi f_0t+\theta) + \cos(\theta)\mathrm{d}t + \eta_{\mathrm{i},k} \\ &\approx d_k \cos(\theta) + \eta_{\mathrm{i},k}, \end{align} $$ because $c(t-\tau) c(t-\tau) = 1$, $\frac{1}{T} \int_{kT}^{(k+1)T} \cos(4\pi f_0t+\theta) \mathrm{d}t \approx 0$ since the primitive of $\cos(4\pi f_0t+\theta)$ is $\frac{\sin(4\pi f_0t+\theta)}{4\pi f_0}$ and $f_0$ is a large value, and $\eta_{\mathrm{i},k}$ is a random variable with a distribution close to a Gaussian thanks to the integration, with a mean of $0$ and a variance of $\sigma^2$.
Therefore, we have $\eta_{\mathrm{i},k} \sim \mathcal{N}(0, \sigma^2)$, and consequently $y_{\mathrm{i},k}(\tau) \sim \mathcal{N}(d_k \cos(\theta), \sigma^2)$.

Here we have a problem because the result depends on $\cos(\theta)$, so if we are not lucky we may have $\theta \approx \pm \frac{\pi}{2}$, i.e. $\cos(\theta) \approx 0$, and thus not detect our signal. To avoid this, we need to get rid of the carrier phase, and for that we use the identity $\cos^2(a)+\sin^2(a)=1$.

Therefore, we will also compute $$ \begin{align} y_{\mathrm{q},k}(\tau) &= \frac{1}{T} \int_{kT}^{(k+1)T} x(t) c(t-\tau) \sin(2\pi f_0t)\mathrm{d}t \\ &= d_k \sin(\theta) + \eta_{\mathrm{q},k}, \end{align} $$ which gives us $y_{\mathrm{q},k}(\tau) \sim \mathcal{N}(d_k \sin(\theta), \sigma^2)$. And finally, we compute $$ z_k(\tau) = \sqrt{y_{\mathrm{i},k}^2(\tau) + y_{\mathrm{q},k}^2(\tau)}, $$ which gives us $z_k(\tau) \sim \mathrm{Rice}(1,\sigma)$ using the relation given at the beginning and $|d_k|^2 = 1$.

In summary, if the local sequence is aligned with the received one ($\hat{\tau} = \tau$), we have $z_k(\hat{\tau}) \sim \mathrm{Rice}(1,\sigma)$. If they are not aligned ($\hat{\tau} \neq \tau$), we have $z_k(\hat{\tau}) \sim \mathrm{Rice}(0,\sigma)$, which is also a Rayleigh distribution. The correlator output $z_k(\hat{\tau})$ is then compared to a threshold to determine if the signals are aligned or not. Thus, the distribution $\mathrm{Rice}(1,\sigma)$ is used to determine the probability of detection, and the distribution $\mathrm{Rice}(0,\sigma)$ is used to determine the probability of false alarm.

You can read more about all this in the Section 6.8.4 of the book A-GPS: Assisted GPS, GNSS, and SBAS by Frank van Diggelen, for example.