Prove that $\frac{1}{n}D_{n}\to \frac{\pi}{4}$ a.s.

Let $$X^{n}:=(X_{1}^{n},X_{2}^{n})$$ and $$(X^{n})_{n}$$ be IID random variables where $$X^{n}$$~$$\mathcal{U}(K)$$ on a probability space $$(\mathbb R^{2}, \mathcal{B}(\mathbb R^{2}), P)$$ where $$\forall A \in \mathcal{B}(\mathbb R^{2}):P(A)=\frac{1}{4}\lambda^{2}(A\cap K)$$, while $$K$$ is the unit circle with radius $$1$$.

Show that $$\frac{1}{n}|\{i\in\{1,...,n\}|X_{i}\in K\}|\to \frac{\pi}{4}$$ a.s.

My ideas:

Define $$D_{n}=|\{i\in\{1,...,n\}|X_{i}\in K\}|$$

$$(D_{n})_{n}$$ are IID random variables as a transformation of the random variables $$(X_{n})_{n}$$

It is clear that I need to use the Strong Law of Large Numbers, nonetheless I assume that $$\mathbb E[D_{n}]=\frac{\pi}{4}$$ but how do I know what $$D_{n}$$ looks like?

• I agree with @Mindlack. I am guess that the set should be a disk of radius $1$ instead of the whole $[-1,1]^2$. – Ankitp Feb 21 at 4:28
• I have corrected this @Ankitp – SABOY Feb 22 at 16:53
• Why post a bounty on this when the question is not even stated correctly? Once you get the notation right (as Ankitp suggests), this is literally just a straightforward application of the SLLN. – Shalop Feb 22 at 23:35

1 Answer

The trick here is to consider the indicator function $$Y_i = 1_K(X_i) = \begin{cases} 1, & \mbox{if X_i in K, and} \\ 0, & \mbox{otherwise.} \end{cases}$$ Using this notation, the $$Y_i$$ are i.i.d. random variables with $$\sum_{i=1}^n Y_i = D_n$$ and $$\mathbb{E}(Y_i) = 0 P(Y_i = 0) + 1 P(Y_i = 1) = P(X_i \in K) = \pi/4$$. Applying the strong law of large numbers to these random variables, you get $$\frac1n D_n = \frac1n \sum_{i=1}^n Y_i \longrightarrow \mathbb{E}(Y_1) = \pi/4$$ as $$n\to\infty$$, almost surely.