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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

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    $\begingroup$ This seems like the probability that two Poisson variables have the same value, as their means increase without bound. Is that the intended interpretation? $\endgroup$ – Brian Tung Dec 1 '17 at 18:55
  • $\begingroup$ Yeah, that is completely right. $\endgroup$ – user509037 Dec 1 '17 at 20:34
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EDITED.


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.


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 $0$.

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  • $\begingroup$ It might be worth expanding quite a few of these steps (such as after "(by CS)" which is probably 3-4 extra steps using hyperbolic functions). You also completely skipped over where $\binom{2k}{k}$ comes from, and use a weaker bound on it than I provided. This is essentially the same as my answer except I split the sum instead of using Cauchy-Schwartz and showed every single step. $\endgroup$ – adfriedman Dec 1 '17 at 22:12
  • $\begingroup$ @adfriedman, Well, I was motivated by your use of central binomial coefficient so the idea here is not that different. Still I decided to add this answer because the answer is now shorter and also demonstrates the 'worst' decaying speed. It can be check that this is sharp up to constant when $\sigma_A = \sigma_B$. $\endgroup$ – Sangchul Lee Dec 1 '17 at 22:17
  • $\begingroup$ I mean 'worst' decaying speed is quite simple if we stick to non-elementary means $$e^{-2\sigma}\sum_{k=0}^{\infty} \frac{\sigma^{2k}}{(k!)^2} = e^{-2\sigma} I_0(2\sigma)$$ you can derive the worst decay (with constant) to be $$\frac{1}{\sqrt{4\pi \sigma}} + O(\sigma^{-3/2})$$ for $\sigma = \sigma_A = \sigma_B$ by expanding around $\sigma=\infty$, but I figured that was beyond the scope of the OP who mentions that they aren't a mathematician, so detailed steps seem more appropriate than short cryptic ones. $\endgroup$ – adfriedman Dec 1 '17 at 22:42
  • $\begingroup$ @adfriedman, You are right. I was simply thinking that adding more context may be good if OP is interested, but if you don't like it I will delete it. $\endgroup$ – Sangchul Lee Dec 1 '17 at 22:58
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    $\begingroup$ @adfriedman, Accepting your suggestions, I tried to clarify my ideas and some hidden steps. :) $\endgroup$ – Sangchul Lee Dec 1 '17 at 23:56
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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.

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  • $\begingroup$ thanks so much for your help! $\endgroup$ – user509037 Dec 1 '17 at 20:34

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