Yesterday I asked about $\lim_\limits{\sigma_A, \sigma_B \to \infty }e^{-\sigma_A-\sigma_B} \sum_{k=0}^{\infty} \frac{{\sigma_A}^k}{k!}\cdot\frac{{\sigma_B}^k}{k!}$, and some fellow users helped me understand that this expression goes to zero. I hoped that I could use the logic of this proof to also understand this, slightly different, limit and ideally prove that it goes to zero as well. It didn't work out.

I see two possibilities to prove that this second limit goes to zero. Firstly, one could show that the ratio of the two probabilities is bounded. Secondly, one could autonomously prove that the second expression goes to zero. Unfortunately, it seems there is no clear result as to that one of the two expression is bounded by the other, or at least I can't find one, as the exponent for $\sigma_B$ is higher for one, but it also has $(k+1)!$ instead of $k!$.

The interpretation of the expressions are that we have two independent poisson distributed random variables, one with mean $\sigma_A$ (say X) and one with mean $\sigma_B$ (say Y). One of the expressions is the probability that X=Y, as the means go to infinity, the other one that X=Y+1 for the means going to infinity.

TL;DR: Ideally, I need to show that both expressions are zero. Here are two proofs that the one not in the heading is zero.

Thanks so much in advance, rm


I fixed the error in my previous answer.

Similar to my previous answer, notice by Stirling's approximation that

$$ \frac{(2k+1)!}{k!(k+1)!} \leq C \frac{2^{2k+1}}{\sqrt{2k+2}} $$

for some $C > 0$ and for all $k \geq 0$. So if we let $\alpha = 2(\sigma_A \sigma_B)^{1/2}$, then

\begin{align*} \sum_{k=0}^{\infty} \frac{\sigma_A^k}{k!}\cdot\frac{\sigma_B^{k+1}}{(k+1)!} &\leq C\sum_{k=0}^{\infty} \frac{2^{2k+1} \sigma_A^k \sigma_B^{k+1}}{(2k+1)!\sqrt{2k+2}} \\ &= C\sqrt{\frac{\sigma_B}{\sigma_A}} \sum_{k=0}^{\infty} \frac{\alpha^{2k+1}}{(2k+1)!\sqrt{2k+2}} \\ &\leq C\sqrt{\frac{\sigma_B}{\sigma_A}} \left( \sum_{k=0}^{\infty} \frac{\alpha^{2k+1}}{(2k+1)!} \right)^{1/2} \left( \sum_{k=0}^{\infty} \frac{\alpha^{2k+1}}{(2k+2)!} \right)^{1/2} \\ &\leq C \sqrt{\frac{\sigma_B}{\sigma_A}} \cdot \frac{e^{\alpha}}{\sqrt{\alpha}}. \end{align*}

Now let $\delta = |\sqrt{\sigma_A} - \sqrt{\sigma_B}|$ and notice the following lemma:

Lemma. We have $\sup_{x,y \geq 1} \frac{x}{y}e^{-(x-y)^2} < \infty$.

Proof of Lemma. May assume that $x > y \geq 1$. Then by the mean value theorem, we can easily check that $\log x - \log y \leq x - y$. Therefore

$$ (\log x - \log y) - (x - y)^2 \leq (x-y) - (x-y)^2 \leq \frac{1}{4}. $$

Therefore we have $\sup_{x,y \geq 1} \frac{x}{y}e^{-(x-y)^2} \leq e^{1/4} < \infty$. ////

Applying this lemma to $(x, y) = (\sqrt{\sigma_A}, \sqrt{\sigma_B})$, we obtain

$$ \sum_{k=0}^{\infty} \frac{\sigma_A^k}{k!}\cdot\frac{\sigma_B^{k+1}}{(k+1)!} e^{-\sigma_A -\sigma_B} \leq \frac{C}{\sqrt{\alpha}} \sqrt{\frac{\sigma_B}{\sigma_A}} \cdot e^{-\left(\sqrt{\sigma_A} - \sqrt{\sigma_B}\right)^2} \leq \frac{C'}{(\sigma_A \sigma_B)^{1/4}} $$

for some absolute constant $C' > 0$. This is enough to prove that the sum converges to $0$.

  • $\begingroup$ THANK YOU! I tried to modify the proof as well but my version was too weak. $\endgroup$ – user509037 Dec 3 '17 at 21:54
  • $\begingroup$ I have one more question: Did you maybe lose the $4^k$ on your way? Otherwise I don't understand where it's gone. $\endgroup$ – user509037 Dec 4 '17 at 10:17
  • $\begingroup$ @user509037, You are right, I must have been simply lost in the soup of notations. Let me try to fix the proof. $\endgroup$ – Sangchul Lee Dec 4 '17 at 11:15

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