Let $\{X_i \}_{i=1}^n$ be i.i.d. from a Poisson random variable with mean $\lambda$ and let $$M_n = \max_{1 \le i \le n} X_i.$$ What is a tight upper bound on $\mathbb{E}[M_n]$? I can prove that \begin{equation} \mathbb{E}[M_n] \le \frac{(\lambda+1) \log n}{\log \log n} + O \left( \frac{1}{\log \log n} \right) \end{equation} but numerically this bound is not tight. Can someone give a tighter analysis or point to a reference with one?

Proof of my bound:

Let $s > 0$. Then by Jensen's inequality, $$ e^{s \mathbb{E}[M_n]} \le \mathbb{E}[e^{sM_n}] \le \sum_{i-1}^n \mathbb{E}[e^{sX_i}] = ne^{\lambda(e^s-1)}$$ from the moment generating function. Then taking the logarithm of both sides gives $$ \mathbb{E}[M_n] \le \frac{\log n}s + \frac{\lambda(e^s-1)}{s}.$$ Finally, the minimum of the function on the right hand side should be around $s = \log \log n$ which gives the bound above.

  • $\begingroup$ Some back of the envelope calculations lead me to believe that is the right asymptotic growth. What numerics do you have to show it's not tight? Is it the $\log n/ \log \log n$ asymptotic or just the constant $\lambda+1$? $\endgroup$ – zoidberg Dec 7 '18 at 2:24
  • $\begingroup$ Just the constant $\lambda + 1$. I do think its $O(\log n / \log \log n)$. $\endgroup$ – Sandeep Silwal Dec 7 '18 at 2:27
  • $\begingroup$ Does the constant 1 match better? You're clearly losing something by going from max to sum. $\endgroup$ – zoidberg Dec 7 '18 at 2:35
  • $\begingroup$ 1 is definitely too small. Also the constant should depend on $\lambda$. $\endgroup$ – Sandeep Silwal Dec 7 '18 at 2:37
  • $\begingroup$ Are you sure? Take a look at this paper: arxiv.org/pdf/0903.4373.pdf Apparently $M_n$ becomes super concentrated as $n \to \infty$ and with high probability lies in an interval of length 1 located $~ \log n /\log \log n$ independent of $\lambda$. $\endgroup$ – zoidberg Dec 7 '18 at 2:48

The paper https://arxiv.org/pdf/0903.4373.pdf mentions that as $n \to \infty$, $M_n$ becomes concentrated at two adjacent values located asymptotically at $\sim \log n/ \log \log n$ independent of $\lambda$ with probability approaching 1. Even tighter estimates are given there.

  • $\begingroup$ Hmm interesting, can you clarify how tight the bounds are exactly? The paper does not give much detail on the rate of convergence $\endgroup$ – Sandeep Silwal Dec 7 '18 at 2:57
  • $\begingroup$ You can see from the Figure 3. It's really tight apparently. They claim the formula they give has error less than 1 (at least over the values of $n$ and $\lambda$ in that figure). They don't seem to offer much in the way of proofs. $\endgroup$ – zoidberg Dec 7 '18 at 2:59

After the fact, but a quick observation: in your proof, set instead $$ s \stackrel{\rm def}{=} \log\left(\frac{\log n}{\lambda}+1\right) $$ Then $$ \mathbb{E}[M_n] \leq \frac{\log n}{s} + \frac{\lambda(e^s-1)}{s} = \frac{2\log n}{s} \operatorname*{\sim}_{n\to\infty} 2\frac{\log n}{\log\log n} $$ which is better than your bound for $\lambda >1$.


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