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Let $z_1, \dots, z_n$ be iid standard Normal, and let $x \in \mathbb{R}^n$. Put $\|u\|_\infty = \max_i |u_i|$.

Define $$ F(x) = \mathbb{E}\Big[\|x + z\|_\infty - \|z\|_\infty\Big] $$ If $\|x\|_\infty \geq 2 \mathbb{E} \|z\|_\infty = \Theta(\sqrt{\log n})$, then we clearly have $F(x) = \Theta(\|x\|_\infty)$. In fact, we easily see: $$ \frac{1}{2}\|x\|_\infty \leq F(x) \leq \|x\|_\infty \quad \mbox{if} \quad \|x\|_\infty \geq 2 \mathbb{E} \|z\|_\infty. $$ Thus, the question is now what happens when $\|x\|_\infty \ll \mathbb{E} \|z\|_\infty$. In this case, can one develop a sharp, up to multiplicative constants independent of dimension, approximation to $F(x)$?

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  • $\begingroup$ I suspect no good general bounds are possible without saying something about the structure of $x$. If all entries of $x$ are the same, then $F(x) \asymp \|x\|_\infty$, but if $x$ has a single large entry then $F(x) \ll \|x\|_\infty$. $\endgroup$
    – Alf
    Commented Dec 14 at 19:17
  • $\begingroup$ Is that true? I would still expect $F$ to behave quadratically near the origin, so I am wondering if the relation $F(x \mathbf{1}_n) \asymp |x|$ is really true for all $x$? $\endgroup$
    – Drew Brady
    Commented Dec 14 at 20:23
  • $\begingroup$ There's two ways to see that such quadratic behavior has to arise. First, consider the one-dimensional case, then it is clear that $F(x) \asymp \min\{|x|, |x|^2\}$. Secondly, for general $n$, note that if $\|x\|_\infty \ll (\log n)^{-1/2}$, then with high probability $\|x + g\|_\infty = |x_i + g_i|$ and $\|g\|_\infty = |g_i|$--the maximal coordinate is the same, and hence such quadratic behavior comes from the one-dimensional case again. $\endgroup$
    – Drew Brady
    Commented Dec 14 at 22:07
  • $\begingroup$ Fair enough, I was considering the regime where $1 \lesssim \|x\|_\infty \ll \sqrt{\log n}$. Do you agree that in that (wide) range, the behavior will depend on the structure of $x$? $\endgroup$
    – Alf
    Commented Dec 14 at 23:16

1 Answer 1

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Since we know that $E|Z|_\infty\sim \sqrt{\log n}$, the first easy case is if $|x|_\infty=o( \sqrt{\log n})$. In that case, $E |x+z|_\infty \sim E|z|_\infty$ in the sense that the ratio goes to $1$ asymptotically. This follows from the bounds $$ |z|_\infty-|x|_\infty\leq |z+x|_\infty\leq |z|_\infty+|x|_\infty.$$ In particular, trivially $\Omega(|x|_\infty/n)\leq |F(x)|\leq|x|_\infty$, which can’t be improved (consider the case of all coordinates constant for the upper bound, and of $x$ possessing a single non-zero coordinate for the lower bound).

So the only interesting regime (which maybe you do not care about) is when $|x|_\infty\sim c\sqrt{\log n}$ where $c\in (0,\infty)$. In that case, $F(x)/E|z|_\infty$ depends on the empirical measure $n^{-1}\sum_{i=1}^n \delta_{x_i/\sqrt{\log n}}$: If it is close to a non-degenerate probability measure, say $\mu$, then the ratio may be non-zero asymptotically. One can write a reasonably accurate expression as function of $\mu$. On the other hand if $\mu=\delta_0$ then only the infinity norm of $x$ enters in the ratio, and only if it larger than $E|z|_\infty$.

If one wants more accurate bounds on $F$, one needs to make assumptions on the empirical measure of the entries of $x$, with appropriate scaling.

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  • $\begingroup$ In the case that $\|x\|_\infty \ll \mathbb{E} \|z\|_\infty$, from your argument, is it clear what the additive gap $\mathbb{E}\|x + z\|_\infty - \mathbb{E} \|z\|_\infty$ looks like? I am not sure that I understood. $\endgroup$
    – Drew Brady
    Commented yesterday
  • $\begingroup$ Sorry, there was a stupid typo, corrected. The absolute value of the difference you ask about is bounded by $|x|_\infty$, which is of lower order than the leading term. It is not clear from your question what kind of result you look for. $\endgroup$ Commented 17 hours ago
  • $\begingroup$ When you say the bound "cannot be improved", what you mean, I guess is that if you have a bound depending only on the infinity norm of $x$, then this cannot be improved. My question is whether or not it is possible to have a more precise characterization of $F(x)$, for instance near the origin. A Taylor expansion would indicate that for $x$ close enough to $0$ we should have $F(x) \asymp \|x\|_2^2/n$. On the other hand we know for $\|x\|_\infty \gtrsim \sqrt{\log n}$, that $F(x) \asymp \|x\|_\infty$. Can a more precise characterization be obtained? $\endgroup$
    – Drew Brady
    Commented 16 hours ago

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