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?



The minimum of the reciprocals of some Poisson random variables
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 $\ge\frac1{2e}>0$. 


