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Set-up. Consider a random walk $S_n=\sum_{i=1}^n X_i$, where $\{ X_i, 1\leq i < \infty \} $ is a sequence of i.i.d. random variables with distribution $\mu$, $\mathbb{E}X_1 = 0$. Let $a > 0$. Given $|S_n| \leq a $, the random variables $X_1, X_2, ..., X_n$ are still identically distributed (because of the invariance under permutations), but they are not independent.

Question. Is it always true that the convergence in distribution takes place:

$$ X_1 \Big| \{|S_n| \leq a\} \Rightarrow X_1 \ \ \ \text{ as } n\to \infty \ ? \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) $$

Some thoughts. On the left hand side of (1) is the conditional distribution of $X_1$ given $\{|S_n| \leq a\}$. For instance if $\mu = \frac 12 \delta _1 + \frac 12\delta _{-1}$, then (1) is not difficult to prove. What about the general case? This question is remotely related to these two questions: one and two.

Proof idea (based on Thomas Kojar's comment) Set $R_n = S_n - X_1$. For a Borel set $B\subset \mathbb{R}$ we have

$$ \mathbb{P} \{ X_1 \in B, -a \leq S_n \leq a \} = \mathbb{E} \Big( \mathbf{I}\{ X_1 \in B \} \mathbb{E} \big[ \mathbf{I}\{ -a - X_1 \leq R_n \leq a - X_1 \} \big| X_1 \big]\Big). $$ Therefore if for all $M > 0$ $$ \sup\limits _{b \in [-M,M]} \big|\ln \mathbb{P} \{ -a - b \leq R_n \leq a - b \} - \ln \mathbb{P} \{ -a \leq R_n \leq a \}\big| \to 0, \ \ \ n \to \infty \ \ \ \ \ \ \ (2) $$ then $\mathbb{E} \big[ \mathbf{I}\{ -a - X_1 \leq R_n \leq a - X_1 \} \big| X_1 \big]$ is approximately $\mathbb{P} \{ -a \leq R_n \leq a \} $ and (1) follows. Now, (2) is a consequence of Stone's local limit theorem, at least for a wide class of non-lattice distributions $\mu$.

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    $\begingroup$ How about doing it by hand by writing $$S_{n+1}=X_1+R_n$$ and then applying strong law to show that the effect of X1 is negligible in the second event in $$P(X_1>t, -a/n<X_1/n+R_n/n<a/n)\approx P(X_1>t) P( -a/n-c_n<R_n/n<a/n+c_n)+o(g_n)$$ for some $g_n,c_n\to 0$ at least when if we assume that X1 is upper/lower bounded? $\endgroup$
    – Thomas Kojar
    Commented Oct 6, 2022 at 19:28
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    $\begingroup$ I think you are correct and for a wide class of $\mu$ (1) is going to follow a local limit theorem. I modify the post. $\endgroup$
    – Viktor B
    Commented Oct 6, 2022 at 20:01
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    $\begingroup$ You also have to assume that $P(|S_n|\le a)\ne0$. Then, I think, this should be true for all zero-mean $\mu$ -- but the proof is probably quite nontrivial. $\endgroup$
    – Iosif Pinelis
    Commented Oct 6, 2022 at 20:36
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    $\begingroup$ The real challenge here is to show that this holds for all zero-mean $\mu$ (or maybe to construct a counterexample). $\endgroup$
    – Iosif Pinelis
    Commented Oct 9, 2022 at 0:06
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    $\begingroup$ In general that should follow from the strong ratio limit theorem, see Section 4.3 in arxiv.org/abs/1511.01721 for a recent reference. $\endgroup$
    – user8965
    Commented Nov 26 at 8:33

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