# Monte Carlo Integration

#### econmajor

1. The problem statement, all variables and given/known data
$$f(x)=NormalPDF(x,2,1)+NormalPDF(x,2,(1/2)^2)$$.

where NormalPDF(a,b) is the PDF for a normal distribution with mean a and variance b.

Use Monte Carlo Integratoion to find: $$\int_{-10}^{10}f(x)dx$$
2. Relevant equations
The solution to this integration is 2.
I use the method described in this video:

3. The attempt at a solution
What I have done is as follows:
- draw n (=5000) random numbers uniformly distributed from -10 to 10. in R: runif(n,-10,10)
- evaluate the function f for each of the n randomly distributed numbers so I end up with n different values of f
- find the mean of those values and that is my integral.

I end up with 0.1 instead of 2. What do I do wrong? When I experiment with $$\int_{0}^{1}\exp(-x^2/2)$$ and use the same method I get the correct result

When I multiply by 20 then I get the correct answer. I assume it has something to do with my integration Interval. But I can't see why it gets me the correct answer when mulitiplying by 20

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#### andrewkirk

Homework Helper
Gold Member
By generating a large number of random numbers $x$ in the integration interval [-10,10] and taking the average of $f(x)$ you have estimated the mean of $f(x)$ in the interval. The mean of a function $f$ over an interval $[a,b]$ is defined as
$$\frac{\int_a^b f(x)dx}{b-a}$$
So, to estimate the integral, you need to multiply your estimate of the mean (0.1) by $(b-a)$ which in this case is $(10--10)=20$.

"Monte Carlo Integration"

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