Central Limit Theorem in matlab - matlab

Question

I am trying to prove CLT in matlab by comparing histogram for sum of three RV and normal distribution.
Here is my code:
clc;clear;
len = 50000;
%y0 : Exponential Distribution
lambda = 3;
y0=-log(rand(1,len))./lambda;
%y1 : Rayleigh Distribution
mu = 0;
sig = 2;
var1 = mu + sig*randn(1,len);
var2 = mu + sig*randn(1,len);
t1 = var1 .^ 2;
t2 = var2 .^ 2;
y1 = sqrt(t1+t2);
% %y2: Normal Distribution
y2 = randn(1,len);
%y3 : What result excpected to be:
mean0 = (sum(y0)+ sum(y1)+ sum(y2)) / (len * 3);%how do I calculate this?
var0 = 1;%how do I calculate this?
y3 = mean0 + var0*randn(1,len);
delta = 0.1;
x3 = min(y3):delta:max(y3);
figure('Name','Normal Distribution');
hist(y3,x3);
%Central Limit Theorem:
%what result is:
res = y0+y1+y2;
xn = min(res):delta:max(res);
figure('Name','Final Result');
hist(res,xn);
I have two main problems.
How can I calculate mean and variance for y3 (what result should be)
Is my code correct?

Answer 1

Since y0, y1 and y2 are row vectors, you have to do:
mean0 = mean([y0 y1 y2]);
variance0 = var([y0 y1 y2]);
When you create [y0 y1 y2] you are creating a big vector with all your previous samples in a single vector (As if they were samples form one single distribution).
Now just plug it into the functions you want (mean and variance) as showed above.
About the statistical part: I think you are getting some things wrong.
The Central Limit Theorem applies for the sum of variables distributed according to a same distribution. It can be indeed be any distribution D, but all variables must have that same distribution D. You are trying to sum different distributions.
The theorem says:
I've coded an example for variables distributed according to an exponential distribution.
Run it and you observe that when you increase N, the resulting distribution tends to the expected normal distribution. For N=1 you have your exponential distribution (very different from a normal distribution), but for N=100 you already have a distribution that is very close to the expected normal distribution (you can see how the mean and variance are basically the same now).
CLT for Exponentials with N=1
CLT for Exponentials with N=3
CLT for Exponentials with N=10
CLT for Exponentials with N=100
The expected normal distribution (convergence distibution of CLT)
clc;clear;
len = 50000;
lambda = 3;
%yA : Exponential Distribution A
yA=-log(rand(1,len))./lambda;
%yB : Exponential Distribution B
yB=-log(rand(1,len))./lambda;
%yC : Exponential Distribution C
yC=-log(rand(1,len))./lambda;
%yD : Exponential Distribution D
yD=-log(rand(1,len))./lambda;
%yE : Exponential Distribution E
yE=-log(rand(1,len))./lambda;
%yF : Exponential Distribution F
yF=-log(rand(1,len))./lambda;
%yG : Exponential Distribution G
yG=-log(rand(1,len))./lambda;
%yH : Exponential Distribution H
yH=-log(rand(1,len))./lambda;
%yI : Exponential Distribution I
yI=-log(rand(1,len))./lambda;
%yJ : Exponential Distribution J
yJ=-log(rand(1,len))./lambda;
%y1 : What result you expect it to be (centred Gaussian with same variation as exponential):
mean0 = 0;
var0 = var(yA);
y1 = mean0 + sqrt(var0)*randn(1,len);
delta = 0.01;
x1 = min(y1):delta:max(y1);
figure('Name','Normal Distribution (Expected)');
hist(y1,x1);
%Central Limit Theorem:
%what result is:
res1 = (((yA)/1) - mean(yA))*sqrt(1);
res2 = (((yA+yB)/2) - mean(yA))*sqrt(2);
res3 = (((yA+yB+yC)/3) - mean(yA))*sqrt(3);
res4 = (((yA+yB+yC+yD)/4) - mean(yA))*sqrt(4);
res5 = (((yA+yB+yC+yD+yE)/5) - mean(yA))*sqrt(5);
res10 = (((yA+yB+yC+yD+yE+yF+yG+yH+yI+yJ)/10) - mean(yA))*sqrt(10);
delta = 0.01;
xn = min(res1):delta:max(res1);
figure('Name','Final Result for N=1');
hi st(res1,xn);
xn = min(res2):delta:max(res2);
figure('Name','Final Result for N=2');
hist(res2,xn);
xn = min(res3):delta:max(res3);
figure('Name','Final Result for N=3');
hist(res3,xn);
xn = min(res4):delta:max(res4);
figure('Name','Final Result for N=4');
hist(res4,xn);
xn = min(res5):delta:max(res5);
figure('Name','Final Result for N=5');
hist(res5,xn);
xn = min(res10):delta:max(res10);
figure('Name','Final Result for N=10');
hist(res10,xn);
%for N = 100
y100=-log(rand(100,len))./lambda;
res100 = ((sum(y100)/100) - mean(yA))*sqrt(100);
xn = min(res100):delta:max(res100);
figure('Name','Final Result for N=100');
hist(res100,xn);