I am trying to use Monte Carlo Integration, which is nicely described in the answer here (Confusion about Monte Carlo integration).
I am using Monte Carlo Integration to evaluate $\int_0^1x^2\,dx$.
I set $w(x) = f(x)/g(x) = x^2/2x = x/2$
Then, I solved for $(1/n)\sum_i^nw(x_n) = (1/n)\sum_i^nx/2$
However, I do not know if this is a good solution. I would like to know of other ideas, and why they are better.
The reason I do not think this is a good solution is if I actually solve for:
$\int_0^1x^2\,dx$ = $x^3/3 = 1/3-0 = 1/3$
But when I look at my answer, the mean will be 1/4 (not 1/3). Had I used a different equation, such as $(1/n)\sum_i^nx/1.5$, then I will get a mean of 1/3, which is very close to what the true integration is. Thank you!
Best Answer
The algorithm for the Monte-Carlo approach is the following.
We can then compare the error from the exact value versus the result from the Monte Carlo approach. Depending on how we coded up the Monte-Carlo Method, we can also estimate the error.
Example 1: Using the method above with $n = 1000$, we get the plot:
This produces $I = 0.3323369730904328$.
We can also use other improved Monte Carlo Methods.
Example 2: Using a Quasi-Monte-Carlo method with $n = 1000$, we get the plot:
This produces $I = 0.3332527188916689$.
Example 3: Using an Adaptive-Monte-Carlo method with $n = 1000$, we get the plot:
This produces $I = 0.333727841861132$.
Example 4: Using an Adaptive-Quasi-Monte-Carlo method with $n = 1000$, we get the plot:
This produces $I = 0.3333327629439999$.
In all cases we can compare that to the actual result of:
$$\displaystyle \int_0^1 x^2~ dx = \dfrac{1}{3}$$