Suppose we had some data points $\{(x_i,y_i)\}$ where $x_i$ is a real number and $y_i$ is zero or one. We want to fit the logistic function to the data where the logistic function is $\hat{y}=\frac{1}{1+e^{-(\beta x +\beta_0)}}$. To do this we would like to choose $\beta_1$ and $\beta_0$ in order to minimize the residual sum of squares, $RSS = \sum_i (y_i-\hat{y}_i)^2$.[nevermind that cross-entropy is a better cost function.] I believe there is only a single minimum. How can I know? Does anyone know of a proof for this or a counter example?


1 Answer 1


Sigmoid itself is not a convex function (see this) and square loss based on sigmoid, such as $ \left( A - \frac{1}{1 + e^{-z}} \right)^2 $ is not convex.

Simply plotting the square loss function with A = 5 shows that it's convex for x > 0 and concave, otherwise

enter image description here

If you want mathematical proof, take the second derivative and you'll see that it not strictly positive nor negative.

  • $\begingroup$ What is the point? What does the residual squared being neither convex nor concave have to do with whether there is only one local minimum, the global minimum? $\endgroup$ Commented Mar 22, 2016 at 2:30
  • $\begingroup$ sorry if it's not clear, the function is convex meaning local minima is global minima, strictly convex means only one global minima. The fact that a function is convex helps a lot in optimization procedure. If it's not convex, then it's difficult to find minima. $\endgroup$
    – Tu N.
    Commented Mar 22, 2016 at 4:33
  • $\begingroup$ Ah, so I believe you are saying that we cannot use convexity to prove whether a minimum is unique or not. $\endgroup$ Commented Mar 22, 2016 at 6:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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