# Autocorrelation in Elo ratings

FiveThirtyEight uses the following formula for their NFL Elo ratings: $$R_i^{k+1} = R_i^k + K \cdot M(z) \cdot A(x) \cdot (S_{ij} - \sigma(x))$$ where $$z$$ is the game's margin of victory, $$x=R_i^k - R_j^k$$, and \begin{align*} M(z) &= \ln (|z|+1) \\ A(x) &= \frac{2.2}{2.2-0.001(-1)^{S_{ij}} x} = \frac{1}{1-(-1)^{S_{ij}}\frac{x}{2200}} \\ S_{ij} &= \begin{cases} 1 & i \text{ wins} \\ 0 & i \text{ loses}\end{cases} \\ \sigma(x) &= \frac{1}{1+10^{-x/400}} \end{align*} Let's ignore ties.

My question is the specific justification for the $$A(x)$$ term: why this form, and why those numbers. (More than a layman's explanation, as found elsewhere already.)

What I (think I) understand already:

I understand that $$A(x)$$ is intended to maintain stationarity in the model. (This, to my understanding is not the same as correcting autocorrelation, but perhaps I misunderstand what 538 is doing.)

Intuitively, we see the function is designed so that

• If Team $$i$$ is the favorite ($$x>0$$), a loss is upweighted ($$A(x)>0$$) and a win is downweighted ($$A(x)<0$$).
• If Team $$i$$ is the underdog ($$x<0$$), the opposite.

However, this leaves the questions: why achieve this in this way? And why use the denominator $$d=2200$$?

Statistically speaking, we want $$\mathbb{E}[R_i^{k+1}] = R_i^k$$, i.e. we shouldn't expect Team $$i$$'s rating to increase --- if we did, "we should have rated them higher to begin with". I believe this issue arises because of the $$M(z)$$ term, because without it, we have \begin{align*} \mathbb{E}[R_k^{k+1}] &= \mathbb{E}[R_i^k] + \mathbb{E}[k(S_{ij}-\sigma(x))] \\ &= R_i^k + k(\mathbb{E}[S_{ij}] - \sigma(x)) \\ &= R_i^k \end{align*} which is fine, I guess.

With $$M(z)$$, we need $$\mathbb{E}[M(z)\cdot A(x) \cdot (S_{ij} - \sigma(x))] = 0$$ which, computing expectation over all possible game outcomes as encoded in $$z$$, given $$R_i^k$$ and $$R_j^k$$, implies $$\int_{-\infty}^0 M(z) A(x) (-\sigma(x)) \text{Pr}(z) \ dz + \int_0^{\infty} M(z) A(x) (1-\sigma(x)) \text{Pr}(z) \ dz = 0$$ over some distribution for $$z$$. Rearranging, we get $$\frac{A(x; i \text{ win})}{A(x; i \text{ lose})} = \frac{\sigma(x)}{1-\sigma(x)} \frac{\mathbb{E}[M(z)|i \text{ lose}]}{\mathbb{E}[M(z)|i \text{ wins}]}$$ and we would want some $$A(x)$$ so this would hold for any $$x$$.

We should be able to interpolate some simple function for the expected $$M(z)$$'s, and then if we are satisfied with our functional form for $$A(x)$$, solve for the denominator $$d$$.

Does this analysis seem correct?

Edit. I worked out the empirical $$M(z)$$'s, and found they are decently approximated by the linear functions: \begin{align*} \mathbb{E}[M(z)|i \text{ win}] &\approx \frac{x}{1000} + 2.2 \\ \mathbb{E}[M(z)|i \text{ lose}] &\approx -\frac{x}{1000} + 2.2 \end{align*} If the $$\sigma/(1-\sigma)$$ term weren't there, this would give $$d=2200$$, but since it is there, I'm not sure what gives.