monte carlo integration on a R^5 hypercube in MATLAB - matlab

Question

I need to write MATLAB code that will integrate over a R^5 hypercube using Monte Carlo. I have a basic algorithm that works when I have a generic function. But the function I need to integrate is:
∫dA
A is an element of R^5.
If I had ∫f(x)dA then I think my algorithm would work.
Here is the algorithm:
% Writen by Jerome W Lindsey III
clear;
n = 10000;
% Make a matrix of the same dimension
% as the problem. Each row is a dimension
A = rand(5,n);
% Vector to contain the solution
B = zeros(1,n);
for k = 1:n
% insert the integrand here
% I don't know how to enter a function {f(1,n), f(2,n), … f(5n)} that
% will give me the proper solution
% I threw in a function that will spit out 5!
% because that is the correct solution.
B(k) = 1 / (2 * 3 * 4 * 5);
end
mean(B)

Answer 1

In any case, I think I understand what the intent here is, although it does seem like somewhat of a contrived exercise. Consider the problem of trying to find the area of a circle via MC, as discussed here. Here samples are being drawn from a unit square, and the function takes on the value 1 inside the circle and 0 outside. To find the volume of a cube in R^5, we could sample from something else that contains the cube and use an analogous procedure to compute the desired volume. Hopefully this is enough of a hint to make the rest of the implementation straightforward.

Answer 2

I'm guessing here a bit since the numbers you give as "correct" answer don't match to how you state the exercise (volume of unit hypercube is 1).
Given the result should be 1/120 - could it be that you are supposed to integrate the standard simplex in the hypercube?
The your function would be clear. f(x) = 1 if sum(x) < 1; 0 otherwise

Answer 3

%Question 2, problem set 1
% Writen by Jerome W Lindsey III
clear;
n = 10000;
% Make a matrix of the same dimension
% as the problem. Each row is a dimension
A = rand(5,n);
% Vector to contain the solution
B = zeros(1,n);
for k = 1:n
% insert the integrand here
% this bit of code works as the integrand
if sum(A(:,k)) < 1
B(k) = 1;
end
end
clear k;
clear A;
% Begin error estimation calculations
std_mc = std(B);
clear n;
clear B;
% using the error I calculate a new random
% vector of corect length
N_new = round(std_mc ^ 2 * 3.291 ^ 2 * 1000000);
A_new = rand(5, N_new);
B_new = zeros(1,N_new);
clear std_mc;
for k = 1:N_new
if sum(A_new(:,k)) < 1
B_new(k) = 1;
end
end
clear k;
clear A_new;
% collect descriptive statisitics
M_new = mean(B_new);
std_new = std(B_new);
MC_new_error_999 = std_new * 3.921 / sqrt(N_new);
clear N_new;
clear B_new;
clear std_new;
% Display Results
disp('Integral in question #2 is');
disp(M_new);
disp(' ');
disp('Monte Carlo Error');
disp(MC_new_error_999);