integrationmonte carlo
I have this integral
$\int_{0}^{\infty}\int_{0}^{\infty} e^{-x-y-xy}dxdy$
I have to estimate the value of it using the Monte Carlo method, using as importance function the exponential PDF.
Does anyone have any ideas how I can do this? How can I apply a change of variables to go from 0 $\rightarrow$ $\infty$ to 0 $\rightarrow$ 1 (so I can also implement a small program to simulate the solution)?
Thank you.
When you see $10^{-29}$ on a machine with finite precision, think zero. What the first snippet calculates is a rounding error, not a variance. The random variable $U$ is never used, and the variable $x$ is never initialized.
Here is a function that will give you a vector of points uniformly distributed over the required triangular region:
function [rv]=RandTri(m)
% function to generate an output array of M x,y coordinates uniformly
% distributed over the upper triangle within the unit square
rv=rand(2,m);
rv1t=rv(1,:)+(rv(2,:)<rv(1,:)).*(-rv(1,:)+rv(2,:));
rv(2,:)=rv(2,:)+(rv(2,:)<rv(1,:)).*(-rv(2,:)+rv(1,:));
rv(1,:)=rv1t;
endfunction
And here is the plot generated by:
>> xy=RandTri(100);
>> plot(xy(1,:),xy(2,:),'o')
Now taking the average of your function over this sample provides an estimate of $$ I=\int_{y=0}^1\int_{x=0}^y x^2y\; p(x,y)\;dx\; dy=2\int_{y=0}^1\int_{x=0}^y x^2y\;dx\;dy $$
Best Answer
Found the solution.
So, I implemented both the "brute" method, and the method based on importance (or importance sampling).
Something like this: $\bar\theta = \frac{1}{n} \frac{1}{n} \sum_{i=1}^{n} e^{-X_i -Y_i -X_i Y_i} $ where $X_i$ and $Y_i$ are the random numbers. In my case, $n=100,000,000$ to get near the actual integral value.
So, using the MC method:
$\bar\theta = \frac{1}{n} \sum_{i=1}^{n} e^{-X_i Y_i}$.
The $X_i$ and $Y_i$ random variables are generated using the inverse method. I used the inverse of the exponential function, which is the logarithm function.
So in my program I generated numbers between 0 and 1, and passed them to the logarithm function, which gave me back numbers between 0 and $\infty$ (actually more like between 0 and 100, since $\infty$ is hard to reach on most computers).