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(Cross post from MSE)

Let $\xi_i$ be iid random variables, and define:

$$S_{(k)} = \sum_{i=1}^N \xi_{i+k}$$

Now, define:

$$\tau = \min \left\{ k : S_{(k)} \notin (a,b) \right\}$$

How can I find $\mathbb{E}[\tau]$ or even $P(\tau < t)$, and whether there are any conditions under which these do not depend on the specific distribution of $\xi_i$ but on its first two moments?


To be more precise, I would expect to be able to express $\mathbb{E}[\tau]$ in terms of $N$, the statistical properties of $S_{(0)}$ (for instance, in case we knew the initial state of the register and would be interested in larger values of $k$) and the moments of $\xi_i$. This is usually the case for problems in which one new iid is added to the sum when increasing $k$.

However, here $S_{(k)}$ acts as a shift register/buffer/sliding window/etc., where the oldest value is discarded every time a new one is added. Moreover, other techniques used in similar situations, like Wald's Identities, cannot be applied here, due to the sum taking place over a fixed number $N$ of iids.

Indeed, it turns out that one can prove that the following is a martingale:

$$M_n=S_n\left(\frac{N}{N-1} \right)^n - \mu \sum_{i=1}^n \left(\frac{N}{N-1} \right)^i$$ but the exponential terms prevent us from using the OST since $M_n$ is not uniformly integrable.

Finally, I find it strange that there seems not to be a lot of research about the statistics of these moving windows/shift registers, but I am not able to find any bibliography on this type of problems.

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    $\begingroup$ Isn't $S_{(k)}$ just a random walk? Specifically, you have $S_{(k + 1)} - S_{(k)} = \xi_{k + N + 1} - \xi_{k}$? This should then let you use all the theory of random walks directly... $\endgroup$ Commented Jul 27 at 20:14
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    $\begingroup$ Yes but the fact that $\xi_k$ is correlated with $S_{(k)}$ makes most results not applicable $\endgroup$ Commented Jul 27 at 20:17
  • $\begingroup$ OK, one more stab - you can translate $S_{(k)}$ into an $N$-state Markov chain (continuous state, discrete time). There appears to be a strong theory on stopping times for Markov chains (that I admit to not really understanding), e.g.: doi.org/10.1215/ijm/1256049786 $\endgroup$ Commented Jul 27 at 20:41
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    $\begingroup$ You can translate to a multidimensional state that depends only on the previous time, i.e. $\bar{S}_{(k)}$ r.v. over $\mathbb{R}^N$ with the transition probability $P(\bar{S}_{(k + 1)} = (a_1, ..., a_N) | \bar{S}_{(k)} = (b_1, ..., b_N)) = P(\xi_{k + N + 1} = a_1)$ when $a_2 = b_1, a_3 = b_2, ..., a_N = b_{N - 1}$ and 0 otherwise. Then - following the intro in Baxter & Chacon (link above) - you need to transform your stopping criterion into a "balayage" $\mu \to \nu$ and determine the potential operator $G$ for this process and you get your expected stopping time as $\int (\mu - \nu) dG$. $\endgroup$ Commented Jul 28 at 20:01
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    $\begingroup$ Sure. Steps are uniform on $[-1,1]$, $a=-N-1, b=N+1$ (note that $N,a,b$ are fixed parameters in your setup!) $\endgroup$
    – fedja
    Commented Aug 9 at 12:59

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