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I recently created two models for predicting the outcomes of matches for a particular sport, one is a linear regression model that predicts the margin of the match, and the other is a logistic regression model for predicting the probability of the outcome of the match. I noticed that occasionally the logistic model will predict team A to win (e.g. 57% chance), and the linear model will predict a margin in favour of the opposing team (say, team B wins by 1 goal) for the same match. Both models are trained on the same data and features. I'm curious to know why, and how, this difference between the two models arises. Thanks!

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  • $\begingroup$ Because they are different models that work differently (different assumptions - different methods), so you cannot expect them to return similar results, in general. $\endgroup$ – user2974951 Mar 10 at 13:00
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In case I'm not missing anything from your explanation (some data and details of the model would be useful, though), I guess the reason lies in the different error functions the two models use. In the linear model, you are minimizing the squared difference between the observed and the predicted value:

$$ \sum_i (y_i - \beta \textbf{x}_i - \beta_0)^2, $$

while in the logistic model you are minimizing:

$$ -\sum_i \left( y_i \log p_i + (1-y_i) \log (1-p_i) \right). $$

The other way of looking at it is to say that you have an information loss in the logistic model. It doesn't care by which margin a team has won. If the prediction matches the true outcome, the error will likely be low. In the linear model, on the other hand, if you predict a win by a margin of 1, but the true win was by 100, this will cause a big error.

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  • $\begingroup$ I really appreciate this answer. Do you think a good metric for expressing this discrepancy between these two models would be to compare the accuracy of match outcome (win/loss) using both the linear and logistic models? $\endgroup$ – charl1e Mar 10 at 13:13
  • $\begingroup$ You can do that. Without the data I can only guess, but my guess is that the linear model will be more accurate. $\endgroup$ – Igor F. Mar 10 at 13:15
  • $\begingroup$ Interesting that you would guess that. So, if I want to determine a probability of winning using the linear regression model, am I safe to assume the distribution is normal about my model and calculate the standard deviation in order to measure the probability of winning (i.e. margin greater than zero) $\endgroup$ – charl1e Mar 10 at 13:35

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