I have liblinear model file for a classifier learned using logistic regression. In the file, they say, the weight vector and intercept term. But when I simply plot it as $$w^Tx + b$$ on the original data, it does not seems like the proper decision boundary. But in case of SVM, plotting

$$w^Tx+b$$ seems to be the correct way. Is there any difference in case of Logistic Regression with liblinear ?


1 Answer 1


Logistic regression does not have decision boundaries. It is a method to estimate probabilities of events/class membership. Decisions are made in a separate step once you know the estimated risk along with utilities/costs/loss function, which is the way optimum decisions are made.

  • $\begingroup$ but if i am not mistaken, it is among the linear classifiers, and the probability of class membership is decided based on $$\frac{1}{1+exp(-f(x))}$$ where $$f(x) = w^Tx+b$$, which is a linear decision boundary. $\endgroup$
    – Shew
    Dec 4, 2014 at 16:57
  • $\begingroup$ That would be true if "decision" were part of logistic regression. It is not. $\endgroup$ Dec 4, 2014 at 16:58
  • $\begingroup$ what do you mean by "decision" being part of logistic regression ?. In normal settings (without any thresholding), the decision is similar to the case of SVMs, right ?. +ve, $$if p(x) \geq 0.5$$ (similar to $$f(x) >= 0$$ $\endgroup$
    – Shew
    Dec 4, 2014 at 17:03
  • $\begingroup$ I said that "decision" is not part of logistic regression. The construct you wrote which uses a highly arbitrary 0.5 risk threshold is not part of logistic regression. Logistic regression is a direct probability estimator. $\endgroup$ Dec 4, 2014 at 17:10
  • $\begingroup$ Ok. In that case, what is the $$w$$ and $$b$$ in the logistic model file ?. What is the plane $$w^T+x$$ corresponds to ? $\endgroup$
    – Shew
    Dec 4, 2014 at 20:06

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