How to integrate a distribution function for a specific interval?

I computed two distribution function:

[distribution_1,x]=ecdf(X); Distribution_2= f(distribution_1);

Now I want to integrate Distribution_2 over the interval [a,b] and tried to use q = integral(fun,xmin,xmax) but I have trouble creating the function handle "fun" in my specific case (since I cannot create the connection to x?)

Thankful for any advice!

[Robert]

What does the function f do?

[Nina]

function f distorts distribution_1

[Robert]

Is Distribution_2 also a CDF? Just to be clear, are you trying to integrate the CDF or integrate the corresponding PDF (which you could do by differencing two points on the CDF).

[Nina]

Distribution_2 is also a CDF :-/

[Nina]

Concerning the distortion function f: Distribution_2=1-(1-distribution_1.^(1/(1+g))).^(1+h) eg. with g=0.1 and h=0.2

Answer 1

[Robert]

To make your empirical CDF into a function that integral can use, you should interpolate distribution_2. Because of the way the CDF is defined, I recommend you interpolate based on the previous value. The first value of x is a duplicate of the second, but for use with interp1(..., 'previous') it should be a (finite) number less than any you plan to use. -realmax should do the trick.

x(1) = -realmax;
f_dist2 = @(new_x) interp1(x, distribution_2, new_x, 'previous', 'extrap');
integral(f_dist2, -1, 1)

[Robert]

Here is the code I used for testing:

X = randn(100, 1); % sample data
[distribution_1, x] = ecdf(X);
g=0.1;
h=0.2;
distribution_2 = 1-(1-distribution_1.^(1/(1+g))).^(1+h);

cla
stairs(x, distribution_1)
hold all
stairs(x, distribution_2)
hold off

x(1) = -realmax;
f_dist2 = @(new_x) interp1(x, distribution_2, new_x, 'previous', 'extrap');
f_dist2(0)
f_dist2([-10, 10])
integral(f_dist2, -1, 1)
integral(f_dist2, -10, 10)

[Nina]

Thank you very much! This was exactly what I needed =)

[Nina]

Another problem came up for the code when using integral(f_dist2,0,inf). Do I need to take into account anything extraordinary in this infinity case? Would be very pleased to receive another answer :)

Below the warning I get when running the code: Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is 1.7e-04. The integral may not exist, or it may be difficult to approximate numerically to the requested accuracy.

Answer 2

[David Goodmanson]

Hi Nina,

If this is not a misinterpretation of the problem, then it seems to work. I am assuming that the function ecdf is already on the Matlab path.

g = .1; h = .2;
fun = @(x) 1-(1-ecdf(x).^(1/(1+g))).^(1+h)

integral(@(x) fun(x), 2,3)

function a = ecdf(x)    % for demo purposes
  a = sin(x);
end

I'm curious about the idea of integrating a cdf, could you perhaps comment on that?

[Nina]

Hi David, thanks for your answer, your code would work if X=x, but this is not the case here: Interpolation can solve this problem as described by Robert below :)