What is the argument for running maximum likelihood estimation on Bradley-Terry models?

I have thought about estimating ratings for teams in a sport competition under the BT probability model. Some of the following may lack rigor, or possibly be sloppy, but I hope the key idea gets through...

We have two teams $i$ and $j$, probability of $i$ winning is \begin{align} p_{ij} = \frac{r_i}{r_i+r_j} \end{align} and thus, if $\hat{p}_{ij}$ is our best estimate of $p_{ij}$, we get \begin{align} r_i \approx \frac{\hat p_{ij}r_j}{1-\hat p_{ij}} \end{align} but because this applies to all teams we get (possibly with some additional weight factor required on each term) \begin{align} r_i \approx \sum_{j\neq i} \frac{\hat p_{ij}}{1-\hat p_{ij}}r_j. \end{align} This is an easy eigenvalue problem in the form $\boldsymbol r = A\boldsymbol r$.

What are the upsides of doing maximum likelihood estimation of $\boldsymbol r$?

EDIT: Actually the following is probably the correct expression: \begin{align} r_i \approx \frac{1}{n_i}\sum_{j\neq i} \frac{\hat p_{ij}n_{ij}}{1-\hat p_{ij}}r_j. \end{align}

where $n_{ij}$ is the total games played between $i$ and $j$.

Take a look at

David R. Hunter. MM-Algorthims for generalized Bradley-Terry models. The Annals of Statistics, 32:384–406, 2004

Basically estimating "true" rankings is combinatorial problem. Furthermore it involves some non-linearity. Hence, you cannot simply apply a linear model or eigenvalues - unless you assume some linear relationship between the rankings. The general upside of maximum likelihood estimation is that your objective is a monotonically increasing function and every interation increases the likelihood (however, you don't know what the best likelihood is; that's the drawback).

• But the model IS linear, according to the derivation above. There is no assumptions except for the basic probability model. What am I missing here? Oct 18 '16 at 11:44
• Using SVD to determine the rankings is a legitimate method to aggregate rankings. Maximum likelihood estimation just makes sense from a Baysian point of view - what is a most likely configuration/ranking that is responsible for the scores. If you impose a linear relationship you question would be how your coefficients in the linear model produce such rankings: here you assume a linear influence of one item towards another. From Baysian point of view the interaction of probabilties is not linear. I hope that makes sense; I'm not very sure if this answers your comment.
– Drey
Oct 18 '16 at 12:29
• Sort of, I suspect what you're saying is that just because $\hat p_{ij}$ is the best estimator of $p_{ij}$, it is not necessarily true that $r_{i}$ is best estimated by algebraically solving for $r_i$ after replacing all instances of $p_{ij}$ with its best estimate $\hat p_{ij}$...? Oct 18 '16 at 12:36
• Mhh, I think simply put if you know for sure that $i$ is in linear relationship with $\forall j \neq i$ then surely you can solve the algebraically (assuming that you have complete rankings). If you doubt you can look for a ranking that maximizes the probability of observing scores $r_i$ (see citation above). The resulting rankings (of the $r_i$) may be the same (and usually will be given enough data and low noise). Just there are many different ways to compute the scores.
– Drey
Oct 18 '16 at 12:52