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If $X_1, \ldots X_n$ are independent real-valued random variables such that $E[X_k] = 0$ and $E[X_k^2]$ is finite for each $k$, Kolmorogov's inequality gives an upper bound on $P[\max_{1\le k \le n}|S_k| < \lambda]$, where $S_k = X_1 + \ldots X_k$. In some work relating to Dvoretzky's stochastic approximation theorem, a colleague and I are looking for a bound where the sums of prefixes $S_k$ are replaced by sums of contiguous subsequences $S_k - S_j$. I.e., under the same assumptions on the $X_k$, we would like a bound, tending to $0$ as $\lambda \to 0$, on $P[\max_{1\le j < k \le n}[|S_k - S_j| < \lambda]$. This seems like a natural thing to ask for, but we haven't been able to find anything in the literature that we can adapt to give such a bound. We would be very grateful for any pointers or hints. Alternatively, we would be very interested in a counter-example showing that $P[\max_{1\le j < k \le n}|S_k - S_j| < \lambda]$ need not tend to $0$ as $\lambda \to 0$.

Edit: apologies, there is a repeated and highly significant typo in the above: "${} < \lambda$" should read "${} > \lambda$" throughout. What we are looking for is an upper bound, tending to $0$ as $\lambda \to \infty$ on $P[\max_{1\le j < k \le n}[|S_k - S_j| > \lambda]$.

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  • $\begingroup$ Kolmorogov's maximal inequality provides a lower bound on $P(\max_{1\le k \le n}|S_k| < \lambda)$. $\endgroup$ Commented Feb 19 at 20:48
  • $\begingroup$ It is not a good idea to use $\Bbb P$ and $\Bbb E$ to denote the probability and the expectation; use $P,E$ or $\mathsf P,\mathsf E$ instead. Blackboard bold is reserved for $\Bbb R$, $\Bbb C$, $\Bbb Z$, and other similar sets. $\endgroup$ Commented Feb 19 at 20:52
  • $\begingroup$ @IosifPinelis: I have changed the notation in the question. $\endgroup$
    – Rob Arthan
    Commented Feb 19 at 20:55
  • $\begingroup$ In fact, I think the amended question actually follows easily from the Kolmogorov inequality and the triangle inequality. So I think I'lll delete this question as I don't think it adds much value to MO users. $\endgroup$
    – Rob Arthan
    Commented Feb 21 at 13:50
  • $\begingroup$ You should not change the question, especially so drastically, especially after an answer was given (which you gratefully accepted). $\endgroup$ Commented Feb 21 at 14:22

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If $p_i:=P(X_i=0)>0$ for all $i$, then $$P_n(\lambda):=P(\max_{1\le j < k \le n}|S_k - S_j| < \lambda)\ge P(X_1=0,\dots,X_n=0)=p_1\cdots p_n>0$$ for all $\lambda>0$, so that $P_n(\lambda)\not\to0$ as $\lambda \downarrow 0$.

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  • $\begingroup$ Many thanks! That will save us a lot of time exploring a dead-end. $\endgroup$
    – Rob Arthan
    Commented Feb 19 at 20:52
  • $\begingroup$ I am glad this was of help. $\endgroup$ Commented Feb 19 at 20:54
  • $\begingroup$ Sorry, I made a significant error in the statement of the question. See the edit. $\endgroup$
    – Rob Arthan
    Commented Feb 21 at 13:14

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