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Let $\epsilon_1, \dots, \epsilon_n$ be random signs, equiprobably in $\{-1, 1\}$, independently. Let $S_k = \sum_{j=1}^k \epsilon_j$. I am wondering what is known about the expectation $$ \mathbb{E}\Big[\max_{k \leq n} |S_k| \Big]. $$

It can be seen as the maximum distance from the origin over $n$ steps of a simple random walk which moves left or right from the origin with equal probability.

A naive bound is via Lévy's maximal inequality, which implies that the quantity above is bounded above by $2 \sqrt{n}$. Can the constant $2$ be improved?

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  • $\begingroup$ A natural conjecture would be that the constant can be improved to $\sqrt{\pi/2}$ since this the constant for the Brownian motion. See: math.stackexchange.com/questions/1137263/…. However I am unsure if this calculation carries over to the discrete symmetric simple random walk. $\endgroup$
    – Drew Brady
    Commented Feb 23, 2023 at 6:09
  • $\begingroup$ The calculation by Sangchul Lee in that answer should work by embedding in the natural way. The random walk and coupled Brownian motion never differ by more than $1$ (in the natural sense) and $\vert T_n - n\vert=O(\sqrt{n\ln n})$ (or something like that) with overwhelming probability as the increments $T_k-T_{k-1}$ are subexponential and i.i.d. so applying Bernstein. That means you can lower and upper bound the desired expectation by the expectation of the max abs value of the Brownian motion up to time $n-O(\sqrt{n\ln n})$ and $n+O(\sqrt{n\ln n})$ up to negligible terms. $\endgroup$ Commented Feb 23, 2023 at 7:14
  • $\begingroup$ @Drew Brady The calculation on MSE deals with $\max_{0 \le k \le n} |S_k|$, which is greater than $\max_{0 \le k \le n} S_k$. In the present situation, the right constant is $\sqrt{2/\pi}$ in the asymptotic regime $n \to +\infty$. $\endgroup$ Commented Feb 24, 2023 at 18:39

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If you just care about the asymptotics, it is indeed just $(1+o(1))\sqrt{\pi n/2}$, where the $o(1)$ term decays like $\tilde{O}(1/n^{1/4})$; this can be done using the approach that I suggested of using the natural embedding to compare to Brownian motion at the obvious stopping times, the fact that the Brownian motion does not fluctuate by more than 1 between stopping times, and the fact that $\vert T_n - n\vert=\tilde{O}(\sqrt{n})$ with very high probability by Bernstein's inequality applied to the (subexponential) increments in stopping times. I can make this more rigorous if needed.

If you really do care about the best constant $C^*$ such that $\mathbb{E}[\max_{k\leq n} S_k]\leq C\sqrt{n}$ for every $n$, not just for large enough $n$, there's a couple of approaches; it's true that $C^*$ is at most twice the optimal constant when you take the absolute values outside the max because $\max_{k\leq n}\vert \sum_{i=1}^k \varepsilon_i\vert\leq \vert\max_{k\leq n} \sum_{i=1}^k \varepsilon_i\vert+\vert\min_{k\leq n} \sum_{i=1}^k \varepsilon_i\vert$, which can be treated exactly via the reflection principle. This should (at least morally) get you something like $2\sqrt{2/\pi}\approx 1.596$. Just for variety, here's a slightly different approach that gives a slightly better constant of $\pi/2\approx 1.571$.

Simply note that if $g_1,\ldots,g_n$ are i.i.d. standard Gaussians, then by Jensen's inequality, \begin{align*} \mathbb{E}\left[\max_{k\leq n} \left\vert \sum_{i=1}^k \varepsilon_i \right\vert\right] &= \sqrt{\frac{\pi}{2}} \mathbb{E}\left[\max_{k\leq n} \left\vert \sum_{i=1}^k \varepsilon_i \mathbb{E}[\vert g_i\vert] \right\vert\right]\\ &\leq \sqrt{\frac{\pi}{2}}\mathbb{E}\left[\max_{k\leq n} \left\vert \sum_{i=1}^k \varepsilon_i\vert g_i\vert \right\vert\right]\\ &=\sqrt{\frac{\pi}{2}}\mathbb{E}\left[\max_{k\leq n} \left\vert \sum_{i=1}^k g_i \right\vert\right]\\ &\leq \sqrt{\frac{\pi}{2}}\mathbb{E}\left[\max_{t\leq n} \left\vert B_t\right\vert\right]\\ &=\pi/2\cdot \sqrt{n}, \end{align*} using the symmetry of standard Gaussians and upper bounding by a coupled Brownian motion and appealing to the MSE solution.

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