We have two random variables $X$, $Y$ with respectively given probability distributions $P_X, P_Y$.
We construct a sequence of random variables in the following way:
- We toss a fair coin and if the result is heads we insert as the first element of the sequence X, if it's tails Y.
- We repeat this for the second element of the sequence and then proceed inductively
Call this sequence $\{Z_n\}_{n \in N}$ then would we have that
$$\frac{1}{n} \sum_{i = 1}^n \mathbb{1}_{Z_i \in A } \rightarrow \frac{ P_X(X \in A) + P_Y(Y \in A) }{2}$$
almost surely? Where $A$ is a borel set s.t. $P_X(X \in A) \ne 0$ and $P_Y(Y \in A) \ne 0$ and I have used $\mathbb{1}$ as the symbol for the indicator function.
Yes. Your $Z_i$ are drawn from the distribution which is the mixture of the distributions $P_X$ and $P_Y$, namely from $P_Z = (P_X+P_Y)/2$, defined by $$P_Z(S) = \frac 1 2 P_X(S) + \frac 1 2 P_Y(S)$$ for all measurable S. The SLLN applied to the sequence $\mathbb 1_{Z_i\in A}$ is your result.