1.0 KiB
1.0 KiB
Monte Carlo Integration
I = \int_a^b f(x) \, dx
Riemann Integration
Pick x_i = hi + a
I = \sum_{i=1}^{n} f(x_i) \frac{b-a}{n}
This isn't the only way to integrate.
Monte Carlo Integration
Sample X \sim U[a,b]
\hat{I} = \frac{b-a}{n} \sum_{i=1}^{n} f(x_i)
This is technically a random variable.
Expectation of the Estimator
E\{f(x)\} = \int_a^b f(x) \cdot U[a,b] \, dx = \int_a^b f(x) \frac{1}{b-a} \, dx = \frac{1}{b-a} \int_a^b f(x) \, dx
E\{\hat{I}\} = \frac{b-a}{n} \sum_{i=1}^{n} E\{f(x)\}
= \frac{b-a}{n} \sum_{i=1}^{n} \left( \frac{1}{b-a} \int_a^b f(x) \, dx \right)
= \frac{b-a}{n} \cdot \frac{n}{b-a} \int_a^b f(x) \, dx
E\{\hat{I}\} = \int_a^b f(x) \, dx
Key insight: The Monte Carlo estimator is unbiased — its expected value equals the true integral.
Multi-Dimensional Case
I = \iiint f(\vec{x}) \, d\vec{x}
\hat{I} = \frac{V}{n} \sum_{i=1}^{n} f(\vec{x}_i)
where V is the volume of the integration region.