Deriving PDF of Rayleigh random variable

In the post on Rayleigh channel model, we stated that a circularly symmetric random variable is of the form $Z = X + jY$ , where real and imaginary parts are zero mean independent and identically distributed (iid) Gaussian random variables. The magnitude $|Z|$ which has the probability density,

$p(z) = \frac{z}{\sigma^2}e^{\frac{-z^2}{2 \sigma^2}},\ \ \ z\ge 0$ is called a Rayleigh random variable. Further, the phase $\theta$ is uniformly distributed from $[0,\ 2\pi]$ . In this post we will try to derive the expression for probability density function (PDF) for $|Z|$ and $\theta$ .

The text provided in Section 5.4.5 of [ELECTRONIC-COMMUNICATION:PRADIP] is used as reference.

Joint probability

The probability density function of $x$ is

$p(x) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-x^2}{2\sigma^2}}$ .

Similarly probability density function of $y$ is

$p(y) = \frac{1}{\sqrt{2\pi\sigma^2}}e^{\frac{-y^2}{2\sigma^2}}$ .

As $X$ and $Y$ are independent random variables, the joint probability is the product of the individual probability, i.e,

$p(x,y) = \frac{1}{{2\pi\sigma^2}}e^{\frac{-(x^2+y^2)}{2\sigma^2}}$ .

The joint probability that the random variable $X$ lies between $x$ and $x+dx$ and the random variable $Y$ lies between $y$ and $y+dy$ is,

$P(x\le X+dx,y\le Y+dy)=\frac{1}{{2\pi\sigma^2}}e^{\frac{-(x^2+y^2)}{2\sigma^2}}dxdy$ .

Conversion to polar co-ordinate

Given that $(x,y)$ is in the Cartesian co-ordinate form, we can convert that into the polar co-ordinate $(z,\theta)$ where,
$Z=\sqrt{X^2+Y^2}$ and
$\Theta = \tan^{-1}\left(\frac{Y}{X}\right)$ .

Cartesian coordinate to Polar coordinate

Figure: Cartesian co-ordinate to polar co-ordinate

The area $dxdy$ is Cartesian co-ordinate form is equal to the area $zdzd\theta$ in the polar co-ordinate form.

$P(x\le X+dx,y\le Y+dy)=P(z\le Z+dz,\theta\le \Theta+d\theta)$ .

Simplifying,
$\begin{eqnarray}P(z\le Z+dz,\theta\le \Theta+d\theta)&=&\frac{1}{{2\pi\sigma^2}}e^{\frac{-(x^2+y^2)}{2\sigma^2}}zdzd\theta\\&=&{\frac{z}{{\sigma^2}}e^{\frac{-z^2}{2\sigma^2}}}dz{\frac{1}{2\pi}}d\theta\end{eqnarray}$ .

Summarizing the joint probability density function,

$p(z,\theta) = \frac{z}{{2\pi\sigma^2}}e^{\frac{-z^2}{2\sigma^2}}$ .

Since $z$ and $\theta$ are independent, the individual probability density functions are,
$p(z) = \frac{z}{{\sigma^2}}e^{\frac{-z^2}{2\sigma^2}},\ z\ge 0$ ,

$p(\theta) = \frac{1}{2\pi},\ -\pi \le \theta \le \pi$ .

Simulation Model

Simple Matlab/Octave simulation model is provided for plotting the probability density of $z$ and $\theta$ . The script performs the following:

(a) Generate two independent zero mean, unit variance Gaussian random variables

(b) Using the hist() function compute the simulated probability density for both $z$ and
$\theta$

(c) Using the knowledge of the equation (which we just derived), compute the theoretical probability
density function (PDF)

(d) Plot the simulated and theoretical probability density functions (PDF) and show that they are in good agreement.

Click here to download Matlab/Octave script for simulating the probability density function (PDF) of Rayleigh random variable

Figure: Simulated/theoretical PDF of Rayleigh random variable

PDF of uniformly distributed theta random variable

Figure: Simulated/theoretical PDF of uniformly distributed theta random variable

Reference

[ELECTRONIC-COMMUNICATION:PRADIP] Principles of Electronic Communications Analog – Digital, by Pradip Kumar Ghosh

Tagged as: pdf, Rayleigh

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{ 4 comments… read them below or add one }

1 mesange December 9, 2009 at 1:33 pm

Dear Krishna I hope you are doing good !!!
I have question relating to gaussian random variables, i need to plot the pdf of the following:

=summation(from i=0 to i = P) |H^i|²

where H are circularly symmetric complex Gaussian random variables …
I donno how to do it in matlab ,,, kindly help
thanks in advance,
mesange.

2 Krishna Sankar December 10, 2009 at 6:05 am: @mesange: Hope the following code snippet helps
>>h = 1/sqrt(2)*(randn(1,1000) + j*randn(1,1000)) % rayleigh random variable
>>hP = h.*conj(h); % finding the power
>>hist(hP) % plotting the histogram

3 PeterRay February 17, 2010 at 9:01 pm

Hello Krishna,
I am new to matlab. I need some help. Suppose we have ‘Nr x Nt’ MIMO channel matrix.And we assume iid rayleigh fading. So the magnitude of every entry will be rayleigh distributed. Now at the receiver side we have ‘Nr’ receive antennas and at every antenna we will have a set of values for every transmitted symbol. I wish to plot the pdf of the output values. (like the output SNR). How do I find the pdf then ? What kind of pdf will it be ? Further if I have to toss independent and non identical channels how to I do that ? we usually write “(1/sqrt(2))*(randn(Nr,Nt)+j*randn(Nr,Nt))” for a typical unit variance zero mean channel.Please correct me if my understanding is wrong. Any help would be appreciated.

4 Krishna Sankar April 4, 2010 at 3:49 am: @PeterRay: My replies
a) You can use the histogram (hist function in Matlab) to obtain the PDF.
b) One needs to the write the equations to figure out the nature of the PDF
c) Did you mean, you want to add correlation to the channel coefficients?

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Deriving PDF of Rayleigh random variable

Joint probability

Conversion to polar co-ordinate

Simulation Model

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