Let $\{X_n\}_{n\ge1}$ be a sequence of i.i.d. random variables with distribution $\mu$. For $A\in\mathcal{B}(\mathbb{R}):=\text{Borel $\sigma$-algebra of $\mathbb{R}$}$ with $\mu(A)\in(0,1)$, define
\begin{align*}
\tau=\inf{\{k\ge1, X_k\in A\}}
\end{align*}
(1) Prove that $P(\tau<\infty)=1$.
(2) Prove that $X_\tau$ has distribution given by
\begin{align*}
P(X_\tau\in H)=\frac{\mu(H\cap A)}{\mu(A)},\quad H\in\mathcal{B}(\mathbb{R}).
\end{align*}
I know that $\tau$ is a stopping time with respect to $F_n:=\sigma(X_1,…,X_n)$, in particular $\tau$ is a hitting time but I cannot figure out how to use this information to solve the problem. I feel like this is a pretty standard question and I have overthought myself into frustration with it, any help here would be greatly appreciated.
Best Answer
Let $\{X_n\}_{n \in \mathbb N}$ be a family of i.i.d r.vs with distribution $\mu$. Let $A \in \mathcal B(\mathbb R)$ be such that $\mu(A) \in (0,1)$. Define $\tau_A := \inf \{ n \ge 1 : X_n \in A \}$.
a) Note that $\tau_A$ has geometric distribution with parameter $\mu(A)$. Indeed, for $n \ge 1$: $$ \mathbb P(\tau_A = n) = \mathbb P(X_1,...,X_{n-1} \not \in A, X_n \in A) = \mu(A)(1-\mu(A))^{n-1} $$ So (since $\mu(A) \in (0,1)$ !) $$ \mathbb P(\tau_A < \infty) = \sum_{n=1}^\infty \mathbb P(\tau_A = n) = \mu(A)\sum_{n=1}^\infty (1-\mu(A))^{n-1} = \frac{\mu(A)}{1-(1-\mu(A))} = 1 $$
b) Let $X_{\tau_A}(\omega) = X_{\tau_A(\omega)}(\omega)$ (it is well defined up to the set of measure $0$)
Let any $H \in \mathcal B(\mathbb R)$. We have: $$ \mathbb P(X_{\tau_A} \in H) = \sum_{n=1}^\infty \mathbb P(X_n \in H \cap \{ \tau_A = n \}) = \sum_{n=1}^\infty \mathbb P(\{X_1,...,X_{n-1} \not \in A\} \cap \{X_n \in H \cap A \}) $$
Last equality, during to the fact, that since $\tau_A = n$, we need every $X_1,...,X_{n-1}$ not to fall into $A$, and $X_n$ must fall into $A$ and $H$ simultaneously. Since they are independent, we get it is equal to $(1-\mu(A))^{n-1} \cdot \mu(H \cap A)$, so: $$ \mathbb P(X_{\tau_A} \in H) = \mu(A\cap H) \sum_{n=1}^\infty (1-\mu(A))^{n-1} = \frac{\mu(A \cap H)}{\mu(A)}$$