Proof of Move Score and Move Information of the Poisson Probability Mass Function

Preliminaries. Write $E_K=\sum_{k=0}^{K}\mu^{k}/k!$ for $K\in\mathbb{N}_0$ and $E_{-1}=E_{-2}=0$. By The Real Exponential Function, $E_K\to\exp(\mu)$ as $K\to\infty$ (the partial sums of the defining series converge to $\exp(\mu)$), and hence also $E_{K-1}\to\exp(\mu)$ and $E_{K-2}\to\exp(\mu)$ as $K\to\infty$: for $\varepsilon>0$, if $|E_L-\exp(\mu)|<\varepsilon$ for all $L\ge L_0$, then $|E_{K-1}-\exp(\mu)|<\varepsilon$ and $|E_{K-2}-\exp(\mu)|<\varepsilon$ for all $K\ge L_0+2$, which is the definition of the limit. By claim 2 of Basic Properties of the Exponential Function, $\exp(-\mu)\exp(\mu)=1$ and $\exp(-\mu)>0$. Since every term $\mu^{k}/k!$ is positive, $E_K$ is nondecreasing in $K$ and $\mu^{k}/k!\le E_K$ for $k\le K$; moreover $E_K\le\exp(\mu)$, by claim 1 of Order Properties of Limits of Real Sequences applied to the constant sequence with value $E_K$ (which converges to $E_K$) and the sequence $(E_L)_{L\ge K}$, which dominates it termwise and converges to $\exp(\mu)$ (a tail of a convergent sequence converges to the same limit, by the argument just given for $E_{K-1}$). Hence $0<p_\mu(k)\le\exp(-\mu)\exp(\mu)=1$ for $k\in\mathbb{N}_0$, so $p_\mu$ takes values in $[0,1]$ and $\{p_\mu>0\}=\mathbb{N}_0$. Finally, each singleton $\{k\}=\bigcap_{j\in\mathbb{N}}(k-1/j,k+1/j)$ is a Borel set (a countable intersection of open intervals), and by Poisson Distribution, $P_\mu(\{k\})=\exp(-\mu)\sum_{j\in\mathbb{N}_0\cap\{k\}}\mu^{j}/j!=\exp(-\mu)\mu^{k}/k!=p_\mu(k)$ for $k\in\mathbb{N}_0$, as asserted in the statement.

(P) Let $f:\mathbb{R}\to[0,\infty)$ vanish on $\mathbb{R}\setminus\mathbb{N}_0$, suppose the partial sums $s_K=\sum_{k=0}^{K}f(k)$ converge to a real number $s$, and let $A$ be a set with $\mathbb{N}_0\subseteq A\subseteq\mathbb{R}$. Then the sum of $f$ over $A$ equals $s$ (note $s\ge0$: each $s_K\ge0$, so claim 1 of Order Properties of Limits of Real Sequences with the constant sequence $0$ gives $s\ge0$). Indeed, for a finite set $F\subseteq A$, either $F\cap\mathbb{N}_0=\emptyset$ and $\sum_{x\in F}f(x)=0\le s$ (if $F\ne\emptyset$ every term vanishes, so the sum is a sum of zeros over $[|F|]$ along an enumeration by Sum over a Finite Index Set, hence $0$ by Finite Sum Notation in a Field; if $F=\emptyset$ the empty-sum convention applies), or $K=\max(F\cap\mathbb{N}_0)$ exists and $\sum_{x\in F}f(x)=\sum_{x\in F\cap\mathbb{N}_0}f(x)\le s_K\le s$, using claims 4 and 3 of Peeling, Splitting, and Interchange for Sums over a Finite Index Set (terms vanishing off $F\cap\mathbb{N}_0$; $F\cap\mathbb{N}_0\subseteq\{0,\dots,K\}$ and the omitted terms are nonnegative, the case $F\cap\mathbb{N}_0=\{0,\dots,K\}$ being an equality) and then claim 1 of Order Properties of Limits of Real Sequences applied to the constant sequence $s_K$ and the nondecreasing sequence $(s_L)_{L\ge K}$, which converges to $s$ as a tail of a convergent sequence (by the argument given for $E_{K-1}$ in the preliminaries). So $s$ is an upper bound of the finite sums. Conversely each $s_K$ is itself a finite sum (over $F=\{0,\dots,K\}\subseteq A$), so any upper bound $b$ satisfies $b\ge s_K$ for all $K$, whence $b\ge s$ by the same claim. Thus $s$ is the least upper bound.

Claim 1. Apply (P) to $f=p_\mu$ and $A=\mathbb{R}$: $s_K=\exp(-\mu)E_K\to\exp(-\mu)\exp(\mu)=1$ by claim 2 of Arithmetic of Limits of Real Sequences. So $\sum_{x\in\mathbb{R}}p_\mu(x)=1$, and $p_\mu$ is a discrete probability mass function by Discrete Probability Mass Function on Euclidean Space; the identities $p_\mu(k)=P_\mu(\{k\})$ and $\{p_\mu>0\}=\mathbb{N}_0$ were shown in the preliminaries.

Claim 2. Let $k\in\mathbb{N}_0$. If $k\ge1$, then $k!=k\cdot(k-1)!$ by Factorial of a Natural Number (with $0!=1$ when $k=1$) and $\mu^{k}=\mu\cdot\mu^{k-1}$, so

$$\frac{p_\mu(k-1)}{p_\mu(k)}=\frac{\mu^{k-1}/(k-1)!}{\mu^{k}/k!}=\frac{k}{\mu}.$$

If $k=0$, then $k-1=-1\notin\mathbb{N}_0$, so $p_\mu(k-1)=0=k/\mu$. In both cases $\rho_{p_\mu,a,w}(k)=w_1\bigl(1-p_\mu(k-a_1)/p_\mu(k)\bigr)=w_1(1-k/\mu)$ by Move Score of a Discrete Probability Mass Function. Finally $k+a_1=k+1\in\mathbb{N}\subseteq\mathbb{N}_0$ for every $k\in\mathbb{N}_0$ by Natural Numbers, so no $k\in\mathbb{N}_0$ has $k+a_1\notin\mathbb{N}_0$.

Claim 3. By Move Information of a Discrete Probability Mass Function and claim 2, $\mathsf{J}(p_\mu;a,w)$ is the sum over $\mathbb{N}_0$ of $f(k)=p_\mu(k)\,w_1^{2}(1-k/\mu)^{2}=\frac{w_1^{2}}{\mu^{2}}\,p_\mu(k)(k-\mu)^{2}$, and we extend $f$ by $0$ to $\mathbb{R}\setminus\mathbb{N}_0$; by (P) with $A=\mathbb{N}_0$ it suffices to show that the partial sums $\sum_{k=0}^{K}p_\mu(k)(k-\mu)^{2}$ converge to $\mu$. Using $(k-\mu)^{2}=k(k-1)+k-2\mu k+\mu^{2}$ and, for $k\ge1$, $k\,\mu^{k}/k!=\mu\,\mu^{k-1}/(k-1)!$, and for $k\ge2$, $k(k-1)\mu^{k}/k!=\mu^{2}\mu^{k-2}/(k-2)!$ (the terms with $k=0$, respectively $k\le1$, vanish),

$$\sum_{k=0}^{K}k\frac{\mu^{k}}{k!}=\mu E_{K-1},\qquad\sum_{k=0}^{K}k(k-1)\frac{\mu^{k}}{k!}=\mu^{2}E_{K-2},$$

so that

$$\sum_{k=0}^{K}p_\mu(k)(k-\mu)^{2}=\exp(-\mu)\bigl(\mu^{2}E_{K-2}+\mu E_{K-1}-2\mu^{2}E_{K-1}+\mu^{2}E_K\bigr).$$

By the preliminaries and claims 1 and 3 of Arithmetic of Limits of Real Sequences, the right side converges to $\exp(-\mu)\exp(\mu)(\mu^{2}+\mu-2\mu^{2}+\mu^{2})=\mu$; multiplying by the constant $w_1^{2}/\mu^{2}$ (claim 3 there), the partial sums of $f$ converge to $w_1^{2}/\mu$. Hence $\mathsf{J}(p_\mu;a,w)=\frac{w_1^{2}}{\mu^{2}}\cdot\mu=\frac{w_1^{2}}{\mu}$. $\blacksquare$