Let $$f: \mathbb{R}^n \rightarrow \mathbb{R}$$ be a measurable function and let $$p$$ be a probability density on $$\mathbb{R}^n$$. We consider the problem of numerically evaluating

$\begin{equation} I = \mathbb{E}_p[f] = \int_{\mathbb{R}^n} f(x) p(x) dx. \end{equation}$

Using i.i.d. random variables $$X_i \sim p$$, Monte-Carlo integration gives the following estimation of $$\mathbb{E}_{p}[f]$$ :

$\begin{equation} Z_N = \frac{1}{N} \sum_{i=1}^N f(X_i). \end{equation}$

Let $$\sigma^2 = \mathbb{E}_p[f^2] - \mathbb{E}_p[f]^2$$. $$Z_N$$ is a random variable and one has:

\begin{equation} \left\{ \begin{aligned} &\mathbb{E}[Z_N] = \mathbb{E}_p[f] \\ &Var(Z_N) = \sigma^2 / N \end{aligned} \right. \end{equation}

The quality of a Monte Carlo estimation is determined by its variance. Therefore, the number of required samples $$N$$ for a good estimation, increases with $$\sigma^2$$.

Deterministic integral evaluation methods are based on the properties of $$f$$ and their costs are strongly dependent on the dimension. However the Monte-Carlo integration can be reformulated in a dimension independent way. Let $$\pi$$ be the pushforward measure on $$\mathbb{R}$$ of $$p$$ by $$f$$. Recall the definition of the pushforward measure, for any measurable subset $$A$$:

$\begin{equation} \pi(A) = \int_{\mathbb{R}^n} 1_{A}(f(x)) p(x) dx \end{equation}$

Let $$Y_i = f(X_i)$$. Then $$Y_i \sim \pi$$ and are i.i.d. . The approximation of $$I$$ can be rewrite as

$\begin{equation*} Z_N = \frac{1}{N} \sum_{i=1}^N Y_i \end{equation*}$

which is the Monte-Carlo method applied to $$\int_{\mathbb{R}} x\pi(x)dx$$ with $$p = \pi$$, $$n = 1$$ and $$f : x \rightarrow x$$.

Hence Monte-Carlo integration convergence speed depends only on the pushfowrad measure $$\pi$$, making it suited for high dimensional problems.

### Example: volume estimation

This is well illustrated by the following example: let $$B$$ be a measurable subset of $$A=[0,1]^n$$ with volume $$V$$, let $$f = 1_{B}$$ and let

$I = \int_{[0,1]^n} f(x)dx$

Here, $$p = \lambda$$ the Lebesgue measure on $$[0,1]^n$$, therefore the variables $$X_i$$ are independent and uniformly distributed in $$[0,1]^n$$. Therefore the variables $$Y_i=f(X_i)$$ are independent Bernoulli random variables of parameter $$V/Vol(A)$$. Thus the Monte Carlo method is exactly the same for any $$n$$ and depends only on $$V$$.