Suppose we want to find the area of a region \(X\), where \(X \subset Y = [a,b] \times [c,d]\). Let \(|X|\) and \(|Y|\) denotes the areas of \(X\) and \(Y\), respectively. Let \(Y_1\) and \(Y_2\) be independent and uniformly distributed random variables on the intervals \([a,b]\) and \([c,d]\), respectively. Then, \[ \text{Pr}( (Y_1,Y_2) \in X) = \frac{|X|}{|Y|} \] Isolating \(|X|\), we have: \[ \begin{align*} |X| & = |Y| \cdot \text{Pr}( (Y_1,Y_2) \in X) \\ & = (b-a) (d-c) \cdot \text{Pr}( (Y_1,Y_2) \in X) \\ & = (b-a) (d-c) \cdot \text{E}( \text{I}\{(Y_1,Y_2) \in X\}). \\ & = \text{E}( (b-a) (d-c) \cdot \text{I}\{(Y_1,Y_2) \in X\}). \\ \end{align*} \] Now that \(|X|\) is written as an expectation, we can use Monte Carlo integration to estimate it.

Specifically, the “hit-and-miss” method is a follows: Uniformly sample B points in \(Y\). The Monte Carlo estimate of \(|X|\) is the proportion of points in \(X\) times \(|Y|\) (i.e., \((b-a) \cdot (d-c)\)).

Example 1: Estimate \(\pi\)

Suppose one wishes to find the value of \(\pi\) using the hit-and-miss method. Consider the unit cirle, that is, a circle having radius 1 centered at the origin. Theoretically, the area of the circle is \(\pi\). Let’s us the hit-and-miss algorithm to estimate \(\pi\). The square \(Y = [-1,1] \times [-1,1]\) has area \(2 \times 2 = 4\) and the unit circle is a subset of it.

B <- 1000000
x <- 4 * ( runif(B)^2 + runif(B)^2 <= 1 )
pi.est <- mean(x)
## [1] 3.142788 <- pi.est + c(-1,1)*qnorm(0.975)*sd(x)/sqrt(B)
## [1] 3.139571 3.146005
pi   # Actual value
## [1] 3.141593

Example 2: Area of my Texas home


Could this be applied to \(\mathcal{R}^3\) to get volumes? What problems might there be as we increase the number of dimensions?