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