monte carlo在MATLAB中集成在R ^ 5超立方体上

时间:2011-09-25 08:02:51

标签: matlab integration multidimensional-array montecarlo

我需要编写将使用蒙特卡罗在R ^ 5超立方体上集成的MATLAB代码。我有一个基本算法,当我有一个通用函数时,它可以工作。但我需要整合的功能是:

∫dA

A是R ^ 5的元素。

如果我有∫f(x)dA,那么我认为我的算法会起作用。

以下是算法:

% 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) 

3 个答案:

答案 0 :(得分:2)

无论如何,我想我明白这里的目的是什么,虽然它看起来有点像人为的练习。考虑通过MC尝试查找圆形区域的问题,如here所述。这里的样本是从单位正方形中抽取的,函数取圆圈内的值1和外面的0。要在R ^ 5中查找立方体的体积,我们可以从包含多维数据集的其他内容中进行采样,并使用类似的过程来计算所需的体积。希望这足以让其余的实现变得简单明了。

答案 1 :(得分:1)

我猜这里有点,因为你给出的“正确”答案的数字与你说的练习方式不符(单位超立方体的数量是1)。

鉴于结果应该是1/120 - 是否应该将standard simplex集成到超立方体中?

你的功能很清楚。如果sum(x)

答案 2 :(得分:0)

%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);