Absolute continuity of stopped sequence of absolutely continuous random variables

Question

Suppose I have a list of $n$ iid random variables $X_i$ absolutely continuous (say $X_i\sim U[0,1]$ for simplicity) and consider $X_{\tau\wedge n}$ where $\tau$ is a stopping time. Is it correct to assume that in general also $X_{\tau\wedge n}$ is an absolutely continuous random variable? I would say yes since it is always going to be one of the $X_i$, intuitively, but to show it, is something more rigorous than this required? Thanks for any help. EDIT: in particular, is it necessary to prove it as follows or is it trivial?: $\tau\wedge n\in\{1,\ldots,n\}$ so $T_i=\{\tau\wedge n=i\}_{i\in[n]}$ is a partitioning of $\Omega$. Then by the law of total probability $\mathbb{P}(X_{\tau\wedge n}\leq t)=\sum_{i=1}^n\mathbb{P}(X_i\leq t|T_i)\mathbb{P}(T_i)=\sum_{i=1}^nt\mathbb{P}(T_i)=t$. Thus $X_{\tau\wedge n}\sim U[0,1]$ and the result follows. EDIT: following the discussion in the comments, it is clear that in this calculation independence from the stopping time was assumed, but it need not be in general. Still the absolute continuity keeps holding.

Answer 1

Indeed $X_{\tau \wedge n}$ is an absolutely continuous random variable, since for any set $A$ of Lebesgue measure zero, $$P(X_{\tau \wedge n} \in A) \le \sum_j P(X_j \in A)=0 \,.$$

However, even if $ X_i \sim U[0,1]$, the distribution of $X_{\tau \wedge n}$ could be different: Consider $\tau=\min\{j: U_j >1/2\}$.