$\lim_\limits{\sigma_A, \sigma_B \to \infty }e^{-\sigma_A-\sigma_B} \sum_{k=0}^{\infty} \frac{{\sigma_A}^k}{k!}\cdot\frac{{\sigma_B}^k}{k!}$

Question

I am currently writing my Bachelor thesis in economics and for one proof idea I would need to compute this limit (if it exists). I am not sure how to approach this due to the double limit.

Clearly, the limit of every summand for finite k is zero but that doesn't really help, does it? (EDIT: if you put the exponential term in the sum, that is.) Adding up zeros infinitely often doesn't imply the sum is zero, right?

Thanks in advance, rm

Answer 1

The limit is zero.

Stream Lined Proof Fix $m \geq 1$, then splitting the summation: \begin{align} 0&\leq e^{-\sigma_A-\sigma_B} \sum_{n=0}^{m-1} \frac{(\sigma_A\sigma_B)^n}{(n!)^2} + e^{-\sigma_A-\sigma_B} \sum_{n=m}^{\infty} \frac{(\sigma_A\sigma_B)^n}{(n!)^2}\\ % &<m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\sum_{n=m}^{\infty} \underbrace{\binom{2n}{n}}_{< \,4^n/\sqrt{3n+1}}\frac{(\sigma_A\sigma_B)^n}{(2n)!}\\ % &< m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\frac{1}{\sqrt{3m+1}} \sum_{n=0}^{\infty} \frac{(\sqrt{4\sigma_A\sigma_B})^{2n}}{(2n)!}\\ % &= m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\frac{1}{\sqrt{3m+1}} \cosh(\sqrt{4\sigma_A\sigma_B})\\ % &< m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B} \frac{1}{\sqrt{3m+1}}e^{\sqrt{4\sigma_A\sigma_B}}\\ % &= m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + \frac{1}{\sqrt{3m+1}} e^{-(\sqrt{\sigma_A}-\sqrt{\sigma_B})^2}\\ % &< m \underbrace{\frac{(\sigma_A+\sigma_B)^{m-1}}{2^{m-1}e^{\sigma_A+\sigma_B}}}_{\text{$\to 0$ as $\sigma_1,\sigma_2 \to \infty$}} + \frac{1}{\sqrt{3m+1}} \end{align} then let $m\to \infty$ and the right hand side approaches $0$.

Detailed Proof Fix $m \geq 1$, then splitting the summation: \begin{align} 0&\leq e^{-\sigma_A-\sigma_B} \sum_{n=0}^{\infty} \frac{(\sigma_A\sigma_B)^n}{(n!)^2}\\ &= e^{-\sigma_A-\sigma_B} \sum_{n=0}^{m-1} \frac{(\sigma_A\sigma_B)^n}{(n!)^2} + e^{-\sigma_A-\sigma_B} \sum_{n=m}^{\infty} \frac{(\sigma_A\sigma_B)^n}{(n!)^2}\\ &< e^{-\sigma_A-\sigma_B}\sum_{n=0}^{m-1} \frac{(\sigma_A\sigma_B)^{m-1}}{(0!)^2} + e^{-\sigma_A-\sigma_B}\sum_{n=m}^{\infty} \frac{(2n)!}{n!\cdot n!} \frac{(\sigma_A\sigma_B)^n}{(2n)!}\\ % &= m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\sum_{n=m}^{\infty} \underbrace{\binom{2n}{n}}_{< \,4^n/\sqrt{3n+1}}\frac{(\sigma_A\sigma_B)^n}{(2n)!}\\ % &< m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\sum_{n=m}^{\infty} \frac{1}{\sqrt{3n+1}}\frac{(4\sigma_A\sigma_B)^n}{(2n)!}\\ % &< m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\frac{1}{\sqrt{3m+1}} \sum_{n=m}^{\infty} \frac{(4\sigma_A\sigma_B)^n}{(2n)!}\\ % &< m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\frac{1}{\sqrt{3m+1}} \sum_{n=0}^{\infty} \frac{(\sqrt{4\sigma_A\sigma_B})^{2n}}{(2n)!}\\ % &= m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\frac{1}{\sqrt{3m+1}} \cosh(\sqrt{4\sigma_A\sigma_B})\\ % &= m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B}\frac{1}{\sqrt{3m+1}} \frac{e^{\sqrt{4\sigma_A\sigma_B}} + e^{-\sqrt{4\sigma_A\sigma_B}}}{2}\\ &< m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + e^{-\sigma_A-\sigma_B} \frac{1}{\sqrt{3m+1}}\frac{2e^{\sqrt{4\sigma_A\sigma_B}}}{2}\\ % &= m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + \frac{1}{\sqrt{3m+1}} e^{-(\sqrt{\sigma_A}-\sqrt{\sigma_B})^2}\\ % &< m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} + \frac{1}{\sqrt{3m+1}} \end{align}

Dealing with the left part of this upperbound, we let $\sigma_A = r\cos\theta$ and $\sigma_B = r\sin\theta$, where $r>0$ and $\theta\in[0,\pi/2]$ (as $0<\sigma_A,\sigma_B$). Additionally note that $\cos\theta + \sin\theta \geq 1$ on $[0,\pi/2]$, so \begin{align} 0&\leq m \frac{(\sigma_A\sigma_B)^{m-1}}{e^{\sigma_A+\sigma_B}} = m \frac{(r\cos\theta \cdot r\sin\theta)^{m-1}}{e^{r\cos\theta+r\sin\theta}} % = m \frac{(r^2\cdot\tfrac12 \sin(2\theta))^{m-1}}{e^{r(\cos\theta+\sin\theta})}\\ % &< m \frac{(r^2 (1))^{m-1}}{2^{m-1}e^{r(1)}} % = \frac{m}{2^{m-1}}\frac{r^{2m-2}}{e^{r}} % \to 0\quad \text{as $r\to\infty$} \end{align}

Therefore

$$ 0\leq \lim_{\sigma_A,\sigma_B\to \infty} e^{-\sigma_A-\sigma_B} \sum_{n=0}^{\infty} \frac{(\sigma_A\sigma_B)^n}{(n!)^2} < \frac{1}{\sqrt{3m+1}}$$ As this holds for all fixed $m$, we let $m\to\infty$ and hence $$\lim_{\sigma_A,\sigma_B\to \infty} e^{-\sigma_A-\sigma_B} \sum_{n=0}^{\infty} \frac{(\sigma_A\sigma_B)^n}{(n!)^2} = 0$$ by the squeeze theorem.

Answer 2

EDITED.

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1. Probabilistic heuristics. Let $N_A \sim \operatorname{Poisson}(\sigma_A)$ and $N_B \sim \operatorname{Poisson}(\sigma_B)$ be independent Poisson random variables. As you are already aware of, the quantity in question is

$$\mathbb{P}( N_A = N_B)$$

It is well-known that the law of $Z_A := (N_A - \sigma_A) / \sqrt{\sigma_A}$ is approximated by a standard normal distribution $\mathcal{N}(0,1)$. This is because $\operatorname{Poisson}(n)$ can be split into $n$ i.i.d. $\operatorname{Poisson}(1)$ variables so that we can apply the classical CLT. (A precise statement is that $Z_A$ converges in distribution to $\mathcal{N}(0,1)$ as $\sigma_A \to \infty$.)

To utilize this fact, notice that

$$ \mathbb{P}(N_A = N_B) = \mathbb{P}\left( Z_A = \frac{N_B - \sigma_A}{\sqrt{\sigma_A}} \right) = \sum_{k=0}^{\infty} \mathbb{P}\left( Z_A = \frac{k - \sigma_A}{\sqrt{\sigma_A}} \right) \mathbb{P}(N_B = k). $$

This already tells us something, since if $Z_A$ where truly normal then each $\mathbb{P}\left( Z_A = \frac{k - \sigma_A}{\sqrt{\sigma_A}} \right)$ would have been zero and thus the right-hand side is truly zero. Or a bit more precisely, since

\begin{align*} \mathbb{P}\left( Z_A = \frac{k - \sigma_A}{\sqrt{\sigma_A}} \right) &= \mathbb{P}\left( Z_A \in \left( \frac{k - \frac{1}{2} - \sigma_A}{\sqrt{\sigma_A}}, \frac{k + \frac{1}{2} - \sigma_A}{\sqrt{\sigma_A}} \right) \right) \\ &\approx \int_{\frac{k - \sigma_A}{\sqrt{\sigma_A}} - \frac{1}{2\sqrt{\sigma_A}}}^{\frac{k - \sigma_A}{\sqrt{\sigma_A}} + \frac{1}{2\sqrt{\sigma_A}}} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \, dx \\ &\lesssim \frac{1}{\sqrt{2\pi \sigma_A}}, \end{align*}

we can expect that $\mathbb{P}(N_A = N_B) \lesssim \frac{1}{\sqrt{2\pi \sigma_A}}$ which converges to $0$.

Although this is nothing but a heuristics, it is in fact not far from truth. Indeed this idea can be directly used to prove the convergence to $0$ using some well-known results such as local limit theorem.

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2. Analytical proof. Let us first accept the following fact:

Fact. there exists $C> 0$ satisfying

$$ \forall k \geq 0: \qquad \binom{2k}{k} \leq C \frac{4^k}{\sqrt{2k+1}}. $$

This is easily proved once we know the Stirling's approximation, but let us simply accept this and proceed. Also, the reason why we use this is to replace the hard-to-understand term $k!^2$ by a more familiar-looking term. We will see that this certainly benefits our computation.

Now write

\begin{align*} \sum_{k=0}^{\infty} \frac{(\sigma_A \sigma_B)^k}{k!k!} &= \sum_{k=0}^{\infty} \underbrace{ \frac{(2k)!}{k!k!} }_{=\binom{2k}{k}} \cdot \frac{(\sigma_A \sigma_B)^k}{(2k)!} \leq \sum_{k=0}^{\infty} \frac{C}{\sqrt{2k+1}} \frac{(4\sigma_A \sigma_B)^k}{(2k)!}. \end{align*}

Let us write $\alpha = 2\sqrt{\sigma_A \sigma_B}$ for simplicity. (This choice is not arbitrary, but at this point let's simply say we want to save some hand-labor by shortening the notation.) Proceeding by writing

$$ C \sum_{k=0}^{\infty} \frac{\alpha^{2k}}{(2k)!\sqrt{2k+1}} = C \sum_{k=0}^{\infty} \sqrt{ \frac{\alpha^{2k}}{(2k)!} } \cdot \sqrt{ \frac{\alpha^{2k}}{(2k+1)!} } $$

and applying Cauchy-Schwarz inequality (CS for short), we obtain

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

Although utilizing CS is not an intuitive step, certain explanations are possible. First, this allows to resolve the pesky square-root term $\sqrt{2k+1}$. Second, since $\frac{\alpha^{2k}}{(2k+1)!}$ and $\frac{\alpha^{2k}}{(2k)!}$ does not differ too much, we does not lose too much under CS. (Recall the equality condition for CS!) We are now good to go, by bounding each sum using the Taylor expansion of $e^{\alpha}$:

\begin{align*} \sum_{k=0}^{\infty} \frac{\alpha^{2k}}{(2k+1)!} = \frac{1}{\alpha} \sum_{k=0}^{\infty} \frac{\alpha^{2k+1}}{(2k+1)!} \leq \frac{e^{\alpha}}{\alpha} \qquad\text{and}\qquad \sum_{k=0}^{\infty} \frac{\alpha^{2k}}{(2k)!} \leq e^{\alpha}. \end{align*}

Plugging this back, our long journey ends and we obtain

$$ \sum_{k=0}^{\infty} \frac{(\sigma_A \sigma_B)^k}{k!k!} \leq C \frac{e^{\alpha}}{\sqrt{\alpha}} = \frac{C}{\sqrt{2} (\sigma_A \sigma_B)^{1/4}} e^{2\sqrt{\sigma_A\sigma_B}}. $$

Therefore, multiplying both sides by $e^{-\sigma_A-\sigma_B}$, we obtain

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

By the squeezing theorem, as $\sigma_A, \sigma_B \to 0$ the sum converges to $0$.