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Let $X_1,\dots,X_k$ denote a collection of independent samples of a Poisson random variable whose mean also happens to be equal to $k$. Does the quantity $$k\boldsymbol{E}\min\left\{ \frac{1}{1+X_{1}},\dots,\frac{1}{1+X_{k}}\right\} $$ have a strictly positive limit as $k$ becomes large?

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    $\begingroup$ Yes and using a large deviation principle one can proves that this goes to 1. $\endgroup$ – RaphaelB4 Jan 18 at 14:57
  • $\begingroup$ @RaphaelB4 : Can you expand your comment into a formal answer? From my answer, it seems clear that, to get the value of the limit, one needs to have the asymptotics of $P(X_1>x)$ uniformly in the zone $x=O(k)$ (which will be given by a certain analytic function of $\lfloor x\rfloor$, rather than by an analytic function of $x$). On the other hand, a large deviation principle only gives the asymptotics of $\ln P(X_1>x)$ -- which is much, much less informative than the needed asymptotics of $P(X_1>x)$. $\endgroup$ – Iosif Pinelis Jan 20 at 0:29
  • $\begingroup$ Previous comment continued: Cf. e.g. the asymptotics $\int_0^1(1-ct)^k\,dt\sim1/(ck)$ for $c\in(0,1)$ as $k\to\infty$, which of course depends on $c$, whereas $\ln(ct)\sim\ln t$ as $t\downarrow0$. So, I don't see how a large deviation principle by itself could be enough here. $\endgroup$ – Iosif Pinelis Jan 20 at 0:32
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Let \begin{equation} Y_i:=\frac1{1+X_i}, \quad Y:=\min(Y_1,\dots,Y_k). \end{equation} Then \begin{equation} EY=\int_0^\infty P(Y>y)\,dy=\int_0^\infty P(Y_1>y)^k\,dy. \end{equation} Next, for $y\in(0,\frac1{1+2k})$ and $x:=\frac1y-1>2k$, we have \begin{equation} P(Y_1>y)=1-P(X_1>x),\quad P(X_1>x)\le P(X_1>2k)=P(X_1-k>k)\le1/k, \end{equation} by Chebyshev's inequality, whence $P(Y_1>y)\ge1-1/k$ and \begin{equation} EY\ge\int_0^{1/(1+2k)} P(Y_1>y)^k\,dy\ge\int_0^{1/(1+2k)} (1-1/k)^k\,dy\sim\frac1{2ek} \end{equation} as $k\to\infty$. So, if the limit of $k\,EY$ exists, it must be $\ge\frac1{2e}>0$.

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(To answer Losif Pinelis, this was a bit too long for a comment)

For $X$ a Poisson of mean $k$, $$\mathbb{P}(X\geq (1+\epsilon)k) =\mathbb{P}(e^{X\epsilon/4}\geq e^{(1+\epsilon)k\epsilon/4}) \\ \leq \frac{\mathbb{E}(e^{X\epsilon / 4})}{e^{(1+\epsilon)k\epsilon/4}} =\frac{\exp(k(e^{\epsilon /4}-1))}{e^{(1+\epsilon)k\epsilon/4}} \\ \approx\exp(-k \epsilon^2(\frac{1}{4}-\frac{1}{2*4^2}) $$ for small $\epsilon$. One can then finish the proof of Pinelis and get that the limit is larger than $1/(1+\epsilon)$.

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