Trying to find empirical cdf

%Sn = X1 + X2 ... + Xn

%Xi's are independent random variables 

%uniform on the interval [-a,a]

a = 1;

n1 = 4;
n2 = 20;
n3 = 50;

x1n1 = -a + (a - (-a)) * (rand(n1,1));
x1n2 = -a + (a - (-a)) * (rand(n2,1));
x1n3 = -a + (a - (-a)) * (rand(n3,1));

sort(x1n1);
sort(x1n2);
sort(x1n3);

y1n1 = (1/2)*(1 + erf( (x1n1-n1*mean(x1n1)) / (sqrt(n1)*var(x1n1)) ));
y1n2 = (1/2)*(1 + erf( (x1n2-n2*mean(x1n2)) / (sqrt(n2)*var(x1n2)) ));
y1n3 = (1/2)*(1 + erf( (x1n3-n3*mean(x1n3)) / (sqrt(n3)*var(x1n3)) ));

hold on
stairs(x1n1,y1n1)
stairs(x1n2,y1n2)
stairs(x1n3,y1n3)
grid on
title('a=1')
xlabel('x')
ylabel('CDF')
hold off

Attempting to find the empircal cdf of a function with a different amount of random variables. But the function looks nothing like the typical empirical cdf graph. I even used the mean() and var() function yet I can't seem to find the error.

[Star Strider]

If you have the Statistics and Machine Learning Toolbox, consider using the ecdf function.

[Torsten]

@Kylenino Espinas

You are aware that you don't sum the x1ni anywhere, aren't you ?

Answer 1

[Gyan Vaibhav]

Hi Kylenino,

So the problem here is with your computation which I think is wrong probably due to the formula.

If you have the Statistics and Machine Learning Toolbox, consider using the “ecdf” function as stated earlier by Star Strider. It can be simply done by replacing the corresponding lines in your code.

[y1n1, x1n1] = ecdf(x1n1);
[y2n2, x1n2] = ecdf(x1n2);
[y3n3, x1n3] = ecdf(x1n3);

Here is the documentation link for the “ecdf” function:

https://in.mathworks.com/help/stats/ecdf.html

Alternatively, you can code it as follows:

% Calculate empirical CDF
y1n1 = zeros(size(x1n1));
y1n2 = zeros(size(x1n2));
y1n3 = zeros(size(x1n3));

for i = 1:length(x1n1)
    y1n1(i) = sum(x1n1 <= x1n1(i)) / n1;
end

for i = 1:length(x1n2)
    y1n2(i) = sum(x1n2 <= x1n2(i)) / n2;
end

for i = 1:length(x1n3)
    y1n3(i) = sum(x1n3 <= x1n3(i)) / n3;
end

Hope this helps.