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

Question

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?

Answer 1

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.