Let $\xi_1, \xi_2,\cdots$ be i.i.d. random elements with distribution $\mu$ in some measurable space $(S,\mathcal S)$, fix a set $A\in \mathcal S$ with $\mu A >0$, put $\tau$ = $\inf\{k; \xi_k \in A \}$ (the first hitting time). Show that $\xi_\tau$ has distribution $\mu[\cdot|A]=\mu[\cdot\cap A]/\mu A$.

Intuitively, $\xi_\tau\in B \Leftrightarrow \xi_\tau \in A\cap B$, since $\xi_i$ are all i.i.d., we can intuitively call them $\xi$, its distribution is then $P\{\xi_\tau \in B\} = P\{\xi\in B|\xi \in A\}=\mu[B\cap A]/\mu A$.

But I don't know how to argue it rigorously, by definition $\xi_i$ and $\xi_j$ have identical distributions only means $P\{\xi_i\in B\} = P\{\xi_j\in B\}$.

(I also accept answers that totally unrelated to my argument.)

BTW, is it by definition $\xi_\tau(w)=\lim_{k\rightarrow \infty} \xi_k(w)$ if $\tau(w) = \infty$? But it becomes strange as $S$ is just a space hence we don't know what the $\lim$ means. (Though in this problem we could ignore it because it simply makes $\xi_\tau(w) \notin B$, so basically anyone answering this question could ignore this minor bug unless it's important in the answer.)


First things first, luckily enough $\tau$ is a.s. finite. Indeed $\tau$ is geometrically distributed:

$$\mathbb{P}(\tau = k) = pq^{k-1}$$ with $p = \mathbb{P}(\xi \in A)$ and $q = 1-p.$ For this reason $\tau <{+}\infty$ holds a.s. (essentially you are doing successive Bernoulli trials). Now let us compute the distribution of $\xi_{\tau}$. Note that $\tau$ assumes only a countable set of values, thus we can write for any $B \in \mathcal{S}$: $$ \mathbb{P}(\xi_{\tau} \in B ) = \sum_k \mathbb{P}(\xi_{k} \in B, \tau = k) = \sum_k \mathbb{P}(\xi_k \in B \cap A, \ \ \xi_l \in A^c, \forall l < k ) $$ and now we can use independence to compute this last term: $$ \mathbb{P}(\xi_{\tau} \in B ) = \sum_k \mathbb{P}(\xi_k \in B \cap A)q^{k-1} = \mathbb{P}(\xi \in B | \xi \in A)\cdot p\sum_{k \ge 1}q^{k-1}. $$

Now evaluating the geometric series gives the required result.


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