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Let $X_i \sim \mathrm{Pois}(1)$ be a sequence of $n$ i.i.d. random variables (with Poisson distribution with parameter 1). I'm interested in the asymptotic behavior of $$\mathbb E[\max_{i \in \{1\ldots n\}}X_i],$$ i.e., the expected maximum value of the sequence for large $n$.

The exact answer is $$\sum^\infty_{k = 0}\left[1-\left(\sum_{i=0}^k \frac{e^{-1}}{i!}\right)^n\right],$$ but I'm not really sure how to massage this into something that's easy to work with. I think the results I'm looking for are in a paper called "A note on Poisson maxima," but I can't find a copy of it online.

RobPratt
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Alf
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  • Not a rigorous comment but here I go. Does that the expected value exist here? If you take infinitely many iid random variables with Poisson distribution, wouldn't you expect their maximum to tend to infinity also? I may be totally wrong but just wanted to share my view. – Calculon Jul 15 '14 at 19:20
  • The expectation does tend to infinity, but I'm interested in its asymptotic growth. For instance, does it grow as $\Theta(\lg n)$, $\Theta(\lg \lg n)$, etc.? – Alf Jul 15 '14 at 19:25
  • Just speculation, I have not calculated. Is anything useful to be obtained from hoping that the inner sum is sufficiently close to $1-\frac{e^{-1}}{(k+1)!}$? Or else use the better geometric series approximation for the tail? – André Nicolas Jul 15 '14 at 19:30
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    It seems that it grows as $\log n /\log \log n$, according to http://arxiv.org/pdf/0903.4373.pdf – leonbloy Jul 15 '14 at 20:09
  • Im little confused on your formula for expectation should letting F(t) represent the CDF of Pois(1) the probability that Max=t would be $F(t)^{n}-F(t-1)^{n}$ thus wouldn't expectation look like $\sum_{t=0}^{\infty} t(F(t)^{n}-F(t-1)^{n})$ – Kamster Jul 15 '14 at 21:10
  • This paper talks about approximating maximum of poisons http://arxiv.org/pdf/0903.4373.pdf – Kamster Jul 15 '14 at 21:11
  • It seems like to me that your exact answer is actually calculating $\sum_{t=0}^{\infty}Pr(Max>t)$ – Kamster Jul 15 '14 at 21:16
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    @user159813 For an integer-valued random variable, $\sum_{t=0}^{\infty} \mathrm{Pr}(X > t)= \mathbb E X$. – Alf Jul 15 '14 at 23:10
  • @Jon yea I think my friend just told me that property yesterday, is that property related to survival functions?http://thirdorderscientist.org/homoclinic-orbit/2013/6/25/the-darth-vader-rule-mdash-or-computing-expectations-using-survival-functions – Kamster Jul 15 '14 at 23:12
  • @user159813 Yes. The proof is easy (and is left as an exercise at the end of the blog post you linked to). – Alf Jul 15 '14 at 23:14
  • @Jon ok sorry about my confusion on that – Kamster Jul 15 '14 at 23:16
  • @user159813 No problem! It's a useful fact. – Alf Jul 15 '14 at 23:23
  • https://link.springer.com/content/pdf/10.1007/BF00533727.pdf – zhoraster Oct 07 '22 at 15:53

1 Answers1

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For $M_n=\max_{i \in \{1\ldots n\}}X_i$, from 1 it is known that

$$ \lim_{n \rightarrow \infty}\mathbb{P} \left \{ M_n= I_n \text{ or } M_n=I_n+1 \right \}=\mathbb{P} \left \{ I_n \le M_n \le I_n+1 \right \}=1.$$

where $I_n$ is a sequence with $I_n \sim \frac{\log n}{\log \log n}$.

Hence, if the left-side limit below exists, from $ \mathbb E[M_n] \ge I_n \times \mathbb{P} \left \{ I_n \le M_n \le I_n+1 \right \} $, one can see that:

$$ \lim_{n \rightarrow \infty} \frac {\mathbb E[M_n]}{\frac{\log n}{\log \log n}} \ge1.$$

One may suspect that $ \lim_{n \rightarrow \infty} \frac {\mathbb E[M_n]}{\frac{\log n}{\log \log n}} =c$; however, my numerical study shows that the ratio does not converge.

Instead, I could find the following well-fitted linear relation based on the exact values of $\mathbb E[M_n]$ for $n=10^2,…,10^{16}$:

$$\log \mathbb E[M_n] \sim \small 0.347553651849616 + 0.788253819406444 \times \log\log n -0.253241028492585 \times \log\log\log n$$

with Multiple R: 0.999981, R Square: 0.999962, Adjusted R Square: 0.999956, Standard Error: 0.002858, and F P-value: 3.066E-27.

Hence, I guess something like the following may hold for some small $ \epsilon_1,\epsilon_2,\epsilon_3 $:

$$ \lim_{n \rightarrow \infty} \frac {\mathbb E[M_n]}{e^{0.347+ \epsilon_1}\frac{(\log n) ^{0.788+ \epsilon_2}}{(\log \log n)^{0.253+ \epsilon_3}}} =1.$$

For $n=10^2,…,10^{16}$, the absolute deviations of the above rate for $ \epsilon_1=\epsilon_2=\epsilon_3=0 $ from 1 is less than 0.008 and is 0.00219 on average; in fact, for $n=10^2,…,10^{16}$

$$0.997 \le \frac {\mathbb E[M_n]}{e^{0.347}\frac{(\log n) ^{0.788}}{(\log \log n)^{0.253}}} \le 1.008,$$

which seems sufficiently accurate for most practical cases considering that $\mathbb E[M_n]$ is less than 18 for $n\le 10^{16}$.

An alternative, inspired by the approximation of $I_n$ given in 2 as $x_0$ and utilizing the approximation $\log x -\log \log x$ of the principal branch of the Lambert function, can be the following for some $ \theta_1,\theta_2,\theta_3>0 $:

$$ \frac {\mathbb E[M_n]}{\frac{\theta_1 \log n}{\log \log n-\log \log (\log n-\theta_2)-\theta_3}} \sim 1.$$

For $n=10^2,…,10^{16}$ and $ \theta_1=1,\theta_2=0.15,\theta_3=0.22 $, which are set by manual try and error, the above ratio is between 0.993 and 1.073. The first approximation above seems better because it is more accurate and the parameters can be easily obtained by multiple linear regression for any desired value of $\lambda$, which is 1 here.

Hope these observations help to find a complete answer.

Amir
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