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Let $\{X_n\}_{n \in \mathbb{N}}$ be i.i.d. real random variables with $\mathbb{E}[X_i] = \mu \in \mathbb{R}$. Let $S_n = X_1 + X_2 + \cdots + X_n$.

Let $\nu \leq \mu$ be such that $\mathbb{P}[S_n < \nu \cdot n] = e^{-\gamma n + o(n)}$ for some $\gamma > 0$.

I wonder if the following is true / known / published: $$\mathbb{P}[S_k < \nu \cdot k \mbox{ for all } k \leq n] = \frac{1}{f(n)} \cdot \mathbb{P}[S_n < \nu \cdot n]$$ for some sub-exponential $f \colon \mathbb{N} \to \mathbb{R}$. Equivalently, $$\mathbb{P}[S_k < \nu \cdot k \mbox{ for all } k \leq n] = e^{-\gamma n + o(n)}.$$

I'm not sure if the first moment assumption on $X_n$ is enough, but if possible I would like to prove this without extra assumptions.

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    $\begingroup$ I think it should be true that $$ \frac{1}{n}P[S_n<\nu n] \leq P[S_k<\nu k \text{ for all }k\leq n] \leq P[S_n<\nu n], $$ which is, probably, what you need. The 2nd inequality is evident; as for the 1st one, note that $S_n<\nu n$ implies that there is a cyclic shift of $X_1,\ldots,X_n$ such that $S'_k<\nu k \text{ for all }k\leq n$ for the ``new'' partial sums (just shift to the point where $S_j - \nu j$ is maximized). $\endgroup$
    – Serguei Popov
    Commented Oct 20, 2015 at 22:09
  • $\begingroup$ That would be great. $\endgroup$
    – Vladimir
    Commented Oct 20, 2015 at 22:17
  • $\begingroup$ I'm not sure, though, if what I wrote is formally true; however, if you substitute all $<$'s to $\leq$'s in the statement, then it should hold. $\endgroup$
    – Serguei Popov
    Commented Oct 20, 2015 at 22:23
  • $\begingroup$ You mean k < n? $\endgroup$
    – Vladimir
    Commented Oct 20, 2015 at 22:33
  • $\begingroup$ I mean $S_k\leq \nu k$ etc. $\endgroup$
    – Serguei Popov
    Commented Oct 20, 2015 at 22:35

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