The Angry Statistician

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation . Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc. Let our Pythagorean win expectation formula be $$w=\frac{P^e}{P^e+1},$$ where $w$ is the win fraction expectation, $P$ is runs/allowed (or similar) and $e$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $g$ games is $$W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$$ so for one estimate we only need to compute the value of the partial derivative $\frac{\partial W}{\partial P}$ at $P=1$. Note that $$W = g\left( 1-\frac{1}{P^e+1}\right),$$ and so $$\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$$ and it follows \[ \frac{\partial W}{\partial P}(P=1) ...

Given the number of runs scored and runs allowed by a baseball team, what's a good estimate for that team's win fraction? Bill James famously came up with what he called the " Pythagorean expectation " $$w = \frac{R^2}{R^2 + A^2},$$ which can also be written as $$w = \frac{{(R/A)}^2}{{(R/A)}^2 + 1}.$$ More generally, if team $i$ scores $R_i$ and allows $A_i$ runs, the Pythagorean estimate for the probability of team $1$ beating team $2$ is $$w = \frac{{(R_1/A_1)}^2}{{(R_1/A_1)}^2 + (R_2/A_2)^2}.$$ We can see that the estimate of the team's win fraction is a consequence of this, as an average team would by definition have $R_2 = A_2$. Now, there's nothing magical about the exponent being 2; it's a coincidence, and in fact is not even the "best" exponent. But what's a good way to estimate the exponent? Note the structural similarity of this win probability estimator and the Bradley-Terry estimator $$w = \frac{P_1}{P_1+P_2}.$$ Here ...