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?

  • 2
    $\begingroup$ 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

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 $\ge\frac1{2e}>0$.


(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 $1/(1+\epsilon)$.


Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.