# Expected covered area as integral of the probability of being covered?

If I have a disk $$\mathcal{D}$$, where each point is "covered" randomly with probability given by $$P(x \text{ is covered}) = p(x)$$ so that each point of the disk has a position-dependent probability of being covered, I understand I can write that the expected value of the total covered area

$$\mathbb{E}(\text{total covered area}) = \int_{\mathcal{D}}P(x \text{ is covered}) \mathrm{d}x$$

but I am not sure how this relates to the definition of expected value. It seems relatively clear but I'm not sure how to prove it. Does it follow straightforwardly?

## 2 Answers

Let $$\mathcal{P}_n=\big\{\mathcal{B}_i,x_i\big\}_{i=1}^n$$ be a sequence of tagged partition of $$\mathcal{D}$$ such that $$\max_{1\leq i \leq n}\mu(\mathcal{B}_i)\longrightarrow 0$$ as $$n\longrightarrow \infty$$. Naturally, we'll $$\text{cover all } \mathcal{B}_i \iff x_i \text{ gets covered}$$ Define $$\mathbb{X}_i=1$$ if $$\mathcal{B}_i$$ gets covered and $$\mathbb{X}_i=0$$ otherwise. Then $$\sum_{i=1}^{n}\mu(\mathcal{B}_i)\mathbb{X}_i$$ is the total covered area whose expected value is $$\mathbb{E}\bigg(\sum_{i=1}^n\mu(\mathcal{B}_i)\mathbb{X}_i\bigg)=\sum_{i=1}^n\mu(\mathcal{B_i})\mathbb{E}(\mathbb{X}_i)=\sum_{i=1}^np(x_i)\mu(\mathcal{B}_i)\longrightarrow \int_{\mathcal{D}}p(x)\mathrm{d}x$$ as $$n \longrightarrow \infty$$

• Ok I see the $\mu$ is becomes the $\mathrm{d}x$, and the $p$ is the probability of being covered.
– apg
Oct 25, 2021 at 20:54

This is in fact known as Robbins' theorem.

See Section 2 of Garwood, The variance of the overlap of geometrical figures with reference to a bombing problem, Biometrika 34, 1-17 (1947).

or

Theorem 8.24 of I. Molchanov Theory of Random Sets (Springer, 2006).