- Get link
- Google+
- Other Apps
Let's say we have a game that can be reasonably modeled as two independent Poisson processes with team \(i\) having parameter \(\lambda_i\). If one team wins in regulation with team \(i\) scoring \(n_i\), then it's well-known we have the MLE estimate \(\hat{\lambda_i}=n_i\). But what if the game ends in a tie in regulation with each team scoring \(n\) goals and we have sudden-death overtime? How does this affect the MLE estimate for the winning and losing teams?
Assuming without loss of generality that team \(1\) is the winner in sudden-death overtime. As we have two independent Poisson processes, the probability of this occurring is \(\frac{\lambda_1}{\lambda_1 + \lambda_2}\). Thus, the overall likelihood we'd like to maximize is \[L = e^{-\lambda_1} \frac{{\lambda_1}^n}{n!} e^{-\lambda_2} \frac{{\lambda_2}^n}{n!} \frac{\lambda_1}{\lambda_1 + \lambda_2}.\] Letting \(l = \log{L}\) we get \[l = -{\lambda_1} + n \log{\lambda_1} - {\lambda_2} + n \log{\lambda_2} - 2 \log{n!} + \log{\lambda_1}-\log({\lambda_1 + \lambda_2}).\] This gives \[\begin{equation}
\eqalign{
\frac{\partial l}{\partial \lambda_1} &= -1+\frac{n}{\lambda_1}+\frac{1}{\lambda_1}+\frac{1}{\lambda_1 + \lambda_2}\\
\frac{\partial l}{\partial \lambda_2} &= -1+\frac{n}{\lambda_2}+\frac{1}{\lambda_1 + \lambda_2}.
}
\end{equation}\] Setting both partials equal to \(0\) and solving, we get \[\begin{equation}
\eqalign{
(n-\hat{\lambda_1})(\hat{\lambda_1}+\hat{\lambda_2})+\hat{\lambda_2} &= 0\\
(n-\hat{\lambda_2})(\hat{\lambda_1}+\hat{\lambda_2})-\hat{\lambda_2} &= 0,
}
\end{equation}\] and so \[\begin{equation}
\eqalign{
\hat{\lambda_1} &= (n+1) \frac{2n}{2n+1}\\
\hat{\lambda_2} &= n \frac{2n}{2n+1}.
}
\end{equation}\] For example, if both teams score \(3\) goals in regulation and team \(1\) wins in sudden-death overtime, our MLE estimates are \(\hat{\lambda_1} = 3\frac{3}{7}, \hat{\lambda_2} = 2\frac{4}{7}\).
Intuitively this makes sense, because \(2n\) goals were scored in regulation time, hence we "expect" that the overtime goal occurred around a fraction \(\frac{1}{2n}\) of regulation, so team \(1\) scored \(n+1\) goals in about \(\frac{2n+1}{2n}\) regulation periods and team \(2\) scored \(n\) goals in about \(\frac{2n+1}{2n}\) regulation periods. The standard Poisson process MLE estimates here coincide with the estimates we derived above.
Does this work in practice? Yes! I tested it on my NCAA men's lacrosse model, and it increased the out-of-sample testing accuracy by 0.5%. Surprisingly large for such a small change!
Assuming without loss of generality that team \(1\) is the winner in sudden-death overtime. As we have two independent Poisson processes, the probability of this occurring is \(\frac{\lambda_1}{\lambda_1 + \lambda_2}\). Thus, the overall likelihood we'd like to maximize is \[L = e^{-\lambda_1} \frac{{\lambda_1}^n}{n!} e^{-\lambda_2} \frac{{\lambda_2}^n}{n!} \frac{\lambda_1}{\lambda_1 + \lambda_2}.\] Letting \(l = \log{L}\) we get \[l = -{\lambda_1} + n \log{\lambda_1} - {\lambda_2} + n \log{\lambda_2} - 2 \log{n!} + \log{\lambda_1}-\log({\lambda_1 + \lambda_2}).\] This gives \[\begin{equation}
\eqalign{
\frac{\partial l}{\partial \lambda_1} &= -1+\frac{n}{\lambda_1}+\frac{1}{\lambda_1}+\frac{1}{\lambda_1 + \lambda_2}\\
\frac{\partial l}{\partial \lambda_2} &= -1+\frac{n}{\lambda_2}+\frac{1}{\lambda_1 + \lambda_2}.
}
\end{equation}\] Setting both partials equal to \(0\) and solving, we get \[\begin{equation}
\eqalign{
(n-\hat{\lambda_1})(\hat{\lambda_1}+\hat{\lambda_2})+\hat{\lambda_2} &= 0\\
(n-\hat{\lambda_2})(\hat{\lambda_1}+\hat{\lambda_2})-\hat{\lambda_2} &= 0,
}
\end{equation}\] and so \[\begin{equation}
\eqalign{
\hat{\lambda_1} &= (n+1) \frac{2n}{2n+1}\\
\hat{\lambda_2} &= n \frac{2n}{2n+1}.
}
\end{equation}\] For example, if both teams score \(3\) goals in regulation and team \(1\) wins in sudden-death overtime, our MLE estimates are \(\hat{\lambda_1} = 3\frac{3}{7}, \hat{\lambda_2} = 2\frac{4}{7}\).
Intuitively this makes sense, because \(2n\) goals were scored in regulation time, hence we "expect" that the overtime goal occurred around a fraction \(\frac{1}{2n}\) of regulation, so team \(1\) scored \(n+1\) goals in about \(\frac{2n+1}{2n}\) regulation periods and team \(2\) scored \(n\) goals in about \(\frac{2n+1}{2n}\) regulation periods. The standard Poisson process MLE estimates here coincide with the estimates we derived above.
Does this work in practice? Yes! I tested it on my NCAA men's lacrosse model, and it increased the out-of-sample testing accuracy by 0.5%. Surprisingly large for such a small change!
Comments
Post a Comment