# Stochastic Extensions of the Elo Rating System
## 3 Stochastic Elo Models
### Continuous Models  
Recall that in the original Elo model, team \(A\) wins if \(X+r_{A}-r_{B}+L>0\), and that should match the odds of \(S_{T}\) being positive, so we can assume \(S_{T}\) is \(X+r_{A}-r_{B}+L\) times some constant. Since the incentives of the teams are the same at every point of the game (score the most points and have the opponent score the least), we can also assume the increments \(S_{t+h}-S_{t}\) are independent (for disjoint intervals) and identically distributed. For this to make sense, \(S_{T}=S_{T/n}+(S_{2T/n}-S_{T/n})+\dots+(S_{T}-S_{T-T/n})\) must be infinitely divisible. Although the logistic distribution is infinitely divisible [14], the increments are not logistic, so we suppose that \(X\) and \(S_{T}\) are normal, and in that case, the increments must also follow the normal distribution by [PERSON]'s decomposition theorem [15], so \(S_{t}\) has i.i.d. Gaussian increments, and it is a Brownian motion with drift:  
\[S_{t}=\mu t+\sigma B_{t}\quad\Rightarrow\quad\mathbb{E}[p_{A}]=\mathbb{P}[N( \mu T,\sigma^{2}T)>0]=\Phi\bigg{(}\frac{\mu T}{\sigma\sqrt{T}}\bigg{)}=\Phi \Big{(}\frac{\mu}{\sigma}\sqrt{T}\Big{)} \tag{19}\]  
where \(B_{t}\) is the standard Brownian motion, i.e., \(B_{0}=0\), and \(B\) has independent and normal increments \(B_{t}-B_{s}\sim N(0,|t-s|)\). Under this model, used by [PERSON] in [4], if we want this expectation to match the one given by the static model, we must have  
\[\Phi\Big{(}\frac{\mu}{\sigma}\sqrt{T}\Big{)}=F_{X}(r_{A}-r_{B}+L)=\Phi\bigg{(} \frac{r_{A}-r_{B}+L}{T}\bigg{)}\quad\Rightarrow\quad\frac{\mu}{\sigma}\sqrt{ T}=\frac{r_{A}-  
Denoting by \(u_{t}:=(T-t)/T\) the fraction of the time that remains, and by \(C\) the constant coefficient \(\frac{c\chi}{\sigma\sqrt{T}}\), we have:  
\[\mathbb{E}[p_{A}|S_{t}=S]=F_{X}\bigg{(}(r_{A}-r_{B}+L)\sqrt{u_{t}}+C\frac{S}{ \sqrt{u_{t}}}\bigg{)} \tag{23}\]  
In the end, we obtain an expression where the term \((r_{A}-r_{B}+L)\sqrt{u_{t}}\) depends on the ratings and decreases to zero as the time \(t\) approaches \(T\), while the term \(C\sqrt{u_{t}}^{-1}S\) depends on the score and becomes larger as the game nears its end, as we would want (unless \(S=0\)). In other words, the behavior or \(\mathbb{E}[p_{A}|S_{t}]\) as \(t\to T\) is exactly what we would expect. Note that setting \(t=0\Rightarrow\sqrt{u_{t}}=1\), and assuming a game starts with a score difference of \(S\) points, the expected score is \(F_{X}(r_{A}-r_{B}+L+CS)\). This allows us to interpret the constant \(C\) as the handicap (measured in rating points) that each goal or point entails at the start of the game. We could in fact use this to estimate \(C\) from game data, although we do not need to pick \(t\approx 0\). For any \(t\in(0,t)\), \(C\) should minimize the mean square error  
\[MSE(t):=\frac{\sum_{k\in M}(p_{a_{k}k}-\mathbb{E}[p_{a_{k}k}|S_{t,k}])^{2}}{|M| }=\frac{1}{|M|}\sum_{k\in M}\bigg{(}p_{a_{k}k}-F_{X}\bigg{(}\sqrt{u_{t}}(r_{a_ {k}}-r_{b_{k}}+L)+C\frac{s_{t,k}}{\sqrt{u_{t}}}\bigg{)}\bigg{)}^{2} \tag{24}\]  
as long as the model is correct, and the minimum should be the same for every \(t\). Conversely, if the function above is minimized for \(C=C^{*}\) for every \(t\), our model is optimal among a certain class, as the following result showcases. Lemma 1: _If there are functions \(f\colon\mathbb{R}\to\mathbb{R}\setminus\{0\}\) and \(V\colon\mathbb{R}^{2}\to\mathbb{R}\), and there exists a \(C^{*}\in\mathbb{R}\) such that \(V(x,C^{*}f(x))\leq V(x,Cf(x))\) for every \(x,C\in\mathbb{R}\), then for every \(g\colon\mathbb{R}\to\mathbb{R}\),_  
\[V(x,C^{*}f(x))\leq V(x,g(x))\qquad\forall x\in\mathbb{R} \tag{25}\]  
Proof: \(V(x,g(x))=V\Big{(}x,\frac{g(x)}{f(x)}f(x)\Big{)}\geq V(x,C^{*}f(x))\)  
In particular, for \(V(t,g)=\frac{1}{|M|}\sum_{k\in M}\big{(}p_{a_{k}k}-F_{X}(\sqrt{u_{t}}(r_{A}-r _{B}+L)+gS_{t,k})^{2}\) and the function \(f:t\mapsto 1/\sqrt{u_{t}}\), this means that our model is optimal among the ones with a prediction of the form \(\mathbb{E}[p_{A}]=F_{X}(\sqrt{u_{t}}(r_{A}-r_{B}+L)+Sg(t))\).