0
$\begingroup$

I'm new to machine learning and currently working on logistic regression. but i don't know how to deal this problem. let us consider the logistic regression for a dataset $(x_n,y_n)\ (x_i \in \mathbb R^d, y_i \in \{+1,-1\})$, let $\Phi(x)=(\phi_p(x))^T$ be a p-dimensional vector of functions and $x\in\mathbb R^d$ be a parameter vector. A probabilistic model is defined as $$p(y|x)=\frac{1}{1+\exp(-y\theta^T\Phi(x))}$$ $$(p(+1|x)=\frac{1}{1+\exp(-\theta^T\Phi(x))}=\frac{\exp(\theta^T\Phi(x))}{1+\exp(\theta^T\Phi(x))},p(-1|x)=\frac{1}{1+\exp(\theta^T\Phi(x))})$$ and the logistic regression estimates the parameter $\theta$ by maximizing $L(\theta)=\sum_{i=1}^n\log p(y_i|x_i,\theta)$.

Then, how can I found an equation representing a decision boundary which satisfy $p(+1|x)=p(-1|y)$?

$\endgroup$
0
$\begingroup$

The decision boundary follows from your last sentence

Then, how can I found an equation representing a decision boundary which satisfy $p(+1|x)= p(−1|x)$?

(correcting $p(-1|y)$ to $p(-1|x)$). The decision boundary is given by the set $$\left\{x\in\mathbb{R}^d:p(+1|x) = p(-1|x)\right\}.$$ Then expanding the condition, \begin{align*} \frac{1}{1+\exp(-\theta^T\Phi(x))} &= \frac{1}{1+\exp(\theta^T\Phi(x))} \\ 1+\exp(-\theta^T\Phi(x)) &= 1+\exp(\theta^T\Phi(x)) \\ -\theta^T\phi(x) &= \theta^T\Phi(x)\\ \Rightarrow \theta^T\Phi(x) &= 0. \end{align*} And so the decision boundary is given by the set, $$\{x\in\mathbb{R}^d:\theta^T\Phi(x) = 0\}.$$

$\endgroup$
2
  • $\begingroup$ Thank you for the the answer. For $x\in\mathbb R^2$ and $\Phi (x)=(x_1,x_2)^T$, is the decision boundary $\theta_1x_1=-\theta_2x_2$? $\endgroup$ – J.Maisel Jun 2 '20 at 0:43
  • $\begingroup$ Yes, that would follow from $\theta^T\Phi(x) = 0$ . $\endgroup$ – ztkpat001 Jun 2 '20 at 6:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.