UNIFORM DISTRIBUTION

A random variable X is said to follow Uniform Distribution if

P_x(x)=1/(b-a), a<x<b

Mean of X=(b+a)/2 ,

Variance of X=(b-a)²/12

where,

X is the random variable.

x represents the values that X can take.

P_x(x) represents the Probability Density Function of X.

Examples:

1. Phase Angle of a random sinusoidal wave.

Applications:

In simulation it is used to generate other probability density functions.

ALGORITHM :

The algorithm used to generate Uniform random variables is Linear Congruential Method (LCM) which is also known as Residue Method.

LINEAR CONGRUENTIAL or RESIDUE METHOD

According to this method a sequence of integers X_i’shave a recursive relationship given by

X_i+1=(aX_i+c)mod m , i=1,2,3,4…

where,

X₀ is the initial value.

a is a constant multiplier.

c is called the increment.

m is the modulus.

The selection of the above mentioned parameters X₀ ,a, c, m drastically affect the statistical properties and the cycle length..

%Function to generate Uniform RV

function[c]=mrand(seed,c,a,m)

x=[];

%4 divides m so 4 divides a-1

%c and m are relatively prime

i=1;

x(i)=seed;

for i=2:m

x(i)=mod((a*x(i-1)+c),m);

end

x=x/m;

c=x;

%Example of usage--> k=mrand(5,11,29,2^11);

Execution and Graph:

k=mrand(11,17,21,1000);

hist(k,20);

GAUSSIAN DISTRIBUTION

The p.d.f of guassian distributed R.V is

Gaussian RV’s drom Central Limit Theorem:

1. For a single Gaussisan RV, the question is the number of Uniform RV’ s required. It has been shown that 12 Uniform RV’ s are convenient .

2. Later, the application of the relation,

Y=(Sum (X_i) – k/2)/sqrt(k/12) (k=12)

gives a Gaussian distributed RV.

3. The histogram of the obtained array of Gaussian Distributed RV’ s are then plotted.

Gaussian RV’ s from Box – Mueller Algorithm:

1. Generate two sets of Uniform RV’ s namely U1 and U2.

2. The corresponding Gaussian RV’ s are given by:

X₁=sqrt(- 2 ln U₁) cos (2* p* U2)

X₂=sqrt(- 2 ln U₁)sin(2* p* U₂)

% Program to generate normal distributions using Box-Muller algorithm and Central Limit
% theorem and compare the two distributions

numberofval=1000;
numberoftrials=10;

%Box - Muller Algorithm

for k=1:numberoftrials

u1 = rand(1,numberofval);
u2 = rand(1,numberofval);
for i=1:numberofval
X1(i) = sqrt((-2)*log(u1(i)))*cos(2*pi*u2(i));
X2(i) = sqrt((-2)*log(u1(i)))*sin(2*pi*u2(i));
end
varbox(k)=var(X1);
end
hist(X1);
title('PDF of Gaussian variate X1 genearted using Box-Muller algorithm');
figure(2);
hist(X2);
title('PDF of Gaussian variate X2 genearted using Box-Muller algorithm');

%Central Limit Theorem Method

iidnumber=12;
for k=1:numberoftrials
u=zeros(iidnumber,numberofval);
for i=1:iidnumber
u(i,:)=rand(1,numberofval);
end
norm = zeros(1,numberofval);
for j=1:numberofval
for i=1:iidnumber
norm (j) = norm(j) + u(i,j);
end
norm(j)=(norm(j) - iidnumber/2)/sqrt(iidnumber/12);
end
variid(k)=var(norm);
end
figure(3);
hist(norm);
title('PDF of Gaussian variate X generated using Central Limit Theorem');

% Comparison of the two normal distributions by comparing their variances with the
% theoretical variance of 1

errbox=varbox-1;
erriid=variid-1;
for i=1:numberoftrials
errideal(i)=0;
end
figure(4);
plot(errbox,'r');
hold on;
plot(erriid,'b');
plot(errideal,'g');
title('Comparison of the errors in variance of the two distributions');
legend('Box-Miller','Central Limit Theorem','Ideal');
xlabel('Trial number');
ylabel('Error in variance');

% Program to generate a Gaussian distribution from a uniform distribution

mean=5;
variance=3;
numberofval=1000;
u1=rand(1,numberofval);
expo = -2*log(u1); % Generation of exponential distribution with mean = 2
rayleigh = sqrt(expo); % Generation of Rayleigh distribution with sigma = 1
u2=rand(1,numberofval);
for i=1:numberofval
norm1(i) = rayleigh(i) * cos(2*pi*u2(i));
norm2(i) = rayleigh(i) * sin(2*pi*u2(i));
end
hist(norm1);
title('Normal distribution');
norm = norm1*sqrt(variance) + mean;
figure(2);
hist(norm);
title('Gaussian distribution with mean =5 and variance =3');

RAYLEIGH DISTRIBUTION

Generation of Rayleigh distribution from exponential :

This can be accomplished using the transformation

Y = Ö X

where X is an exponentially distributed random variable .

Generation of Rayleigh distribution from normal distributions :

The formula used for this transformation is

Y = Ö ( X₁² + X₂² )

where X₁ and X₂ are normal distributions .

% Program to generate Rayleigh ditribution using two methods
numberofval=1000;
%First Method - rayleigh distribution from uniform distribution

u = rand(1,numberofval);
for i=1:numberofval
expo(i) = -2*log(u(i)); % Generation of exponential distribution
end
for j=1:numberofval
ray1(j) = sqrt(expo(j)); %Generation of Rayleigh distribution from exponential
end
hist(ray1);
title('PDF of Rayleigh distribution generated from uniform distribution');

% Second Method - rayleigh distribution from normal distributions
iidnumber=12;
for i=1:iidnumber
u(i,:)=rand(1,numberofval);
end
norm1 = zeros(1,numberofval);
for j=1:numberofval
for i=1:iidnumber
norm1 (j) = norm1(j) + u(i,j);
end
norm1(j)=(norm1(j) - iidnumber/2)/sqrt(iidnumber/12);
end

for i=1:iidnumber
v(i,:)=rand(1,numberofval);
end
norm2 = zeros(1,numberofval);
for j=1:numberofval
for i=1:iidnumber
norm2 (j) = norm2(j) + v(i,j);
end
norm2(j)=(norm2(j) - iidnumber/2)/sqrt(iidnumber/12);
end

for i=1:numberofval
ray2(i) = sqrt( norm1(i)^2 + norm2(i)^2 );
end
figure(2);
hist(ray2);
title('PDF of Rayleigh distribution generated from normal distributions');

The figure below shows the effect of Rayleigh fading channel in the place

of AWGN channel.

EXPONENTIAL DISTRIBUTION

The exponential random variable with arrival rate λ > 0 is considered here.

This is a simple technique and has all the advantages of the inverse transform method. It is also reasonably fast.

f(x) = λexp(-λx) ; F(x) = 1- exp(-λx)

U = 1 – exp(-λz) ; z = (-1/λ)ln(1-U)

If U is uniform in (0,1) then 1-U is also uniform in (0,1). Hence z = (-1/λ)lnU .

Algorithm uses the Transform Domain Approach.

Let 1/l be the mean of the Random Process
We use the transform domain approach
After performing the calculations, we get,

z=-ln(U)/ l

where U is the ensemble of Uniform Random Numbers.

Program:

%Funtion to generate Exponentially distributed RV

function[x]=mexp2(m,lamb);

m1=rand(1,m);

for i=1:length(m1)

if m1(i)==0

m1(i)=0.001;

end

x(i)=-1/lamb*log(m1(i));

end

hist(x,25);

Execution and Graph:

x= mexp2(1000,3);

POISSON DISTRIBUTION

Algorithm:

1) Set A=1, k=0;

2) Generate U(0,1) i.e. uniformly dist RV

3) Set A=U*A

4) If A<exp(-l), deliver X=k and go to step 1

else, set k=k+1 and return to step 3

PROGRAM :

function poisson(N,mean)

n=1;

while(n<=N)

a=exp(-mean);

b=1;

i=0;

ui1=rand(1);

b=b*ui1;

while(b>a)

i=i+1;

ui1=rand(1);

b=b*ui1;

end

po(n)=i;

n=n+1;

end

m=1;

max(po)

for t=0:1:max(po)

p(m)=((mean^t)/fact(t))*exp(-mean);

m=m+1;

end

c=[0:1:max(po)];

[d,q]=hist(po,c);

e=d/N;

h=[0:max(po)/[length(c)-1]:max(po)];

bar(h,e,'r');

title('Histogram of Poisson distributed random variable');

hold on;

t=0:1:m-2;

plot(t,p(t+1));

for w=1:length(q)

pi(w)=((mean^q(w))/fact(q(w)))*exp(-mean);

er(w)=e(w)-pi(w);

end

xlabel('-------> x');

ylabel('p(x)');

Here mean = 9

GAMMA DISTRIBUTION

The gamma distribution has two parameters namely α and β. The algorithm for generating gamma distributed random variables proposed by Cheng in 1977 is given below.

ALGORITHM :

1. Generate U1 and U2 as IID U(0,1).

2. Let V = a ln[ U1/(1-U1)], Y = αexp(V), Z = U1^2 * U2, and W = b + qV- Y.

3. If W+d-θZ >= 0, return X = Y. Otherwise proceed to step 4.

4. If W >= lnZ, return X = Y. Otherwise, go back to step 1.

Where a = 1/sqrt(2α – 1) , b = α – ln4 , q = α + 1/a ,

θ = 4.5 and d = 1+lnθ

Here X is the gamma distributed random variate.

PROGRAM :

function gamma(N,alpha,beta)

a=1/sqrt(2*alpha-1);%INITIALIZE THE CONSTANTS

b=alpha-log(4);

q=alpha+(1/a);

theta=4.5;

d=1+log(theta);

n=1;

while(n<=N)

u1=rand(1);%STEP 1 OF THE ALGORITHM

u2=rand(1);

v=a*log(u1/(1-u1));%STEP 2 OF THE ALGORITHM

y=alpha*exp(v);

z=(u1^2 )*u2;

w=b+q*v-y;

if(w+d-theta*z >=0)%STEP 3

ga(n)=beta*y;

a=1/sqrt(2*alpha-1);

b=alpha-log(4);

q=alpha+(1/a);

theta=4.5;

d=1+log(theta);

n=n+1;

else

if(w>=log(z))%STEP 4

ga(n)=beta*y;

n=n+1;

a=1/sqrt(2*alpha-1);

b=alpha-log(4);

q=alpha+(1/a);

theta=4.5;

d=1+log(theta);

end

[l,h]=hist(ga,15*beta);

e=l/N;

s=[min(ga):(max(ga)-min(ga))/(15*beta-1):max(ga)];

bar(s,e,'r');%PLOT THE HISTOGRAM OF THE GENERATED VARIATES

hold on;

g=1;

for k=1:alpha-1

g=g*k;

end

m=1;

for j=0:0.1:max(ga)

p(m)=(1/(g*(beta^alpha)))*(j^(alpha-1))*exp(-j/beta);%THE ACTUAL DISTRIBUTION

m=m+1;

end

j=0:0.1:max(ga);

c=1:m-1;

plot(j,p(c));%PLOT THE ORIGINAL CURVE

title('Histogram of Gamma distributed random variable');

for w=1:length(h)

pi(w)=(1/(g*(beta^alpha)))*(h(w)^(alpha-1))*exp(-h(w)/beta);

er(w)=e(w)-pi(w); %CALCULATE THE ERROR BETWEEN THE TWO

end

xlabel('--------------> x');

ylabel('p(x)');

LAPLACIAN DISTRIBUTION

LAPLACIAN DISTRIBUTED RV USING TRANSFORM METHOD:

The transform method is used to generate Laplacian distributed random variates. An example of a Laplacian distributed random variable is human voice.

Method :

f(x) = (1/2√2 )*exp(- x /√2)

F(x) = 1 – exp(x/√2)/2 for x > 0

= exp(x/√2)/2 for x < 0

U = exp(Z√2)/2

Hence Z = √2ln(2U) for x < 0

U = 1 – exp(Z√2)/2

Hence Z = - √2ln(2U) for x > 0

ALGORITHM :

1. Generate U(0,1).

2. Calculate Z1 = √2ln(2U) and Z2 = - √2ln(2U)

PROGRAM:

close all;

clear all;

y=0;

for i=1:1000

b1=0;

b2=0;

u=rand(1);

b1=-(sqrt(2))*log(2*u);

b2=(sqrt(2))*log(2*u);

y=[y b1 b2];

end

%Plotting of histogram

c=length(-5:1.1:5);

s=hist(y,c);

bar(-5:1.1:5,s/2000);

m=1;

for z=-5:0.05:5

if z<0

pz(m)=1/(2*sqrt(2))*(exp(z/(sqrt(2)))); %Laplacian distribution

else

pz(m)=1/(2*sqrt(2))*(exp(-z/(sqrt(2)))); %Laplacian distribution

end

m=m+1;

end

%Plotting of Laplacian distribution using pdf

hold on;

z=-5:0.05:5;

a=1:m-1;

plot(z,pz(a),'r');

grid on;

title('LAPLACIAN DISTRIBUTION');

xlabel('--------->X , Y');

ylabel('--------->p(X) , p(Y)');

legend('p(X)','p(Y)');