# Stochastic Extensions of the Elo Rating System
## Appendix B Proof That the Discrete Model Is an Elo Model  
Here, we prove that the discrete model proposed in Section 3.4 is a particular type of Elo model, or in other words, that if \(\mathbb{E}[p_{A}]=\mathbb{P}[Sk(\mu_{1},\mu_{2})>0]+\frac{1}{2}\mathbb{P}[Sk( \mu_{1},\mu_{2})=0]\) for  
\[\mu_{1}=\frac{1}{2}\Bigl{(}\Delta_{r}+\sqrt{\Delta_{r}^{2}+H^{2}}\Bigr{)}\qquad \text{and}\qquad\mu_{2}=\frac{1}{2}\Bigl{(}-\Delta_{r}+\sqrt{\Delta_{r}^{2}+H^ {2}}\Bigr{)}\]  
there exists a continuous distribution \(X\) with domain \(\mathbb{R}\) for which \(\mathbb{E}[p_{A}]=F_{X}(\Delta r)\).  
Proof.: We denote  
\[F(x):=\mathbb{P}[Sk(\mu_{1}(x),\mu_{2}(x))>0]+\frac{1}{2}t\mathbb{P}[Sk(\mu_{1 }(x),\mu_{2}(x))=0]=\]\[=\mathbb{P}\Bigg{[}Sk\Bigg{(}\frac{\sqrt{x+H^{2}}-x}{2},\frac{\sqrt{x+H^{2}}-x}{2} \Bigg{)}>0\Bigg{]}+\frac{1}{2}\mathbb{P}\Bigg{[}Sk\Bigg{(}\frac{\sqrt{x+H^{2}}-x}{2 },\frac{\sqrt{x+H^{2}}-x}{2}\Bigg{)}=0\Bigg{]}\]  
Here, we assume \(H^{2}>0\), since \(\frac{1}{2}H^{2}\) is the expectation of a non-negative quantity (the product of the goals scored by each team). This means that \(\mu_{1}\) is strictly increasing in \(\Delta r\), and \(\mu_{2}=\frac{H^{2}}{4\mu_{1}}\) is strictly decreasing in \(\Delta r\).  
On the other hand, \(g_{k}(a,b):=\mathbb{P}[Sk(a,b)\geq k]\) is increasing in \(a\) and decreasing in \(b\), because for any \(\varepsilon>0\), we have  
\[g_{k}(a+\varepsilon,b)=\mathbb{P}[Sk(a+\varepsilon,b)\geq k]=\mathbb{P}[Sk(a,b)+P(\varepsilon)\geq k]>\mathbb{P}[Sk(a,b)\geq k]=g_{k}(a,b)\]  
\[g_{k}(a,b)=\mathbb{P}[Sk(a,b)\geq k]>\mathbb{P}[Sk(a,b)-P(\varepsilon)\geq k] =g_{k}(a,b+\varepsilon)\]  
where \(P(\varepsilon)\) is a Poisson variable with mean \(\varepsilon\).  
This two propositions imply that \(F(\Delta r)=\frac{1}{2}\mathbb{P}[Sk(\mu_{1},\mu_{2})\geq 1]+\frac{1}{2} \mathbb{P}[Sk(\mu_{1},\mu_{2})\geq 0]=\frac{g_{1}(\mu_{1},\mu_{2})+g_{0}(\mu_{1},\mu_{2})} {2}\) is strictly increasing in \(\mu_{1}\) and decreasing in \(\mu_{2}\), and therefore increasing as a function of \(\Delta r\). To finish, we only need to show that \(\mathbb{E}[p_{A}]\) goes to \(1\) as \(\Delta r\) approaches \(+\infty\), and \(0\) when \(\Delta r\) goes to \(-\infty\), since any monotone function with that asymptotic behavior is a distribution function.  
As \(\Delta r\) goes to infinity, \(\mu_{1}\) goes to infinity and \(\mu_{2}\) goes to zero, so  
\[F(\Delta r)=\mathbb{P}[P(\mu_{1})>P(\mu_{2})]+\frac{1}{2}\mathbb{P}[P(\mu_{1} )=P(\mu_{2})]\geq\mathbb{P}[P(\mu_{1})>0=P(\mu_{2})]=\]  
\[=(1-e^{-\mu_{1}})e^{-\mu_{2}}\xrightarrow[\Delta r\to\infty]{}(1-e^{-\infty} )e^{0}=1\]  
and of course, \(F(\Delta r)=\frac{1}{2}(\mathbb{P}[Sk(\mu_{1},\mu_{2})\geq 1]+\mathbb{P}[Sk(\mu_{1}, \mu_{2})\geq 0])\geq 1\), so by the sandwich rule,  
\[F(\Delta r)\xrightarrow[\Delta r\to\infty]{}1\]  
A similar argument shows that when \(\Delta r\) goes to \(-\infty\),  
\[F(\Delta r)\leq\mathbb{P}[P(\mu_{1})>0]+\frac{1}{2}\mathbb{P}[P(\mu_{2})=0]+ \frac{1}{2}\mathbb{P}[P(\mu_{1})>0]\leq\]  
\[\leq\frac{3}{2}(1-e^{-\mu_{1}})+\frac{1}{2}e^{-\mu_{2}}\xrightarrow[\Delta r \to\infty]{}\frac{3}{2}(1-e^{0})+\frac{1}{2}e^{-\infty}=0\]  
and \(F\) is bounded below by zero, so both limits hold. Therefore, \(F\) is the distribution function of some variable \(X\).