# Free parameters in logistic regression

When applying logistic regression, one is essentially applying the following function $$1/(1 + e^{\beta x})$$ to provide a decision boundary, where $$\beta$$ are a set of parameters that are learned by the algorithm, and $$x$$ is an input feature vector. This appears to be the general framework provided by widely available packages such as Python's sklearn.

This is a very basic question, and can be manually implemented by normalization of the features, but shouldn't a more accurate decision boundary be given by: $$1/(1 + e^{\beta (x - \alpha)})$$, where $$\alpha$$ is an offset? Of course an individual can manually subtract a pre-specified $$\alpha$$ from the features ahead of time and achieve the same result, but wouldn't it be best for the logistic regression algorithm to simply let $$\alpha$$ be a free parameter that is trained, like $$\beta$$? Is there a reason this is not routinely done?

You get the same effect from including a bias term, i.e. $$\frac{1}{1+\exp(-\beta x + bias)}$$, and the defaults for most software is to include such bias.
To see why, you can do the math, expanding $$\beta(x-a)$$ and seeing that you can express the sum of those differences as a single variable.
• Fair enough, but these parameters have physical significance that is then lost, no? When plotting the decision boundary, if there is more than one feature, then the shifts resulting from a particular bias term are not unique. For example, $0.5x_1 + 0.5x_2 + 10$ is equivalent to $0.5(x_1 - 10) + 0.5(x_2 + 30)$ or $0.5(x_1 + 30) + 0.5(x_2 - 10)$. But what if one wants to essentially find the marginalized decision boundary for a single feature and find the shift for only $x_1$, for example? – Mathews24 Oct 13 '18 at 22:53