# How to prove the logistic loss function is strongly convex?

The logistic loss function is: $$\mathcal{L}=\frac{1}{n}\sum_{i=1}^n\log(1+\exp(-y_ix_i^T\theta))$$ in which $$y_i\in\{-1,+1\},x\in \mathbb{R}^d$$. How to show that $$\mathcal{L}$$ is strongly convex.

My thinkings: Can we get the $$\nabla^2 \mathcal{L}(\theta)$$ and show $$\nabla^2 \mathcal{L}(\theta)-mI$$ is PSD for some $$m$$?

• Yes, you can compute the hessian wrt $\theta$ and it will be straightforward to see it is PSD. Computing it is a little bit of a hassle but does have a nice closed form expression Mar 15, 2019 at 12:33
• The computation of the gradient and Hessian of logistic loss function is given here math.stackexchange.com/questions/3098910/… Mar 16, 2019 at 8:37

It is not strongly convex. Take $$n=d=1$$. You are getting a function of the form $$f(x)=\log(1+\exp( a x))$$. Its second derivative is $$f''(x) = \frac{a^2 \exp( a x) } { (1 + \exp(ax))^2}$$ Assuming $$a > 0$$, you have $$\lim_{x \to -\infty} f''(x) = 0$$. Thus, there is no positive constant which bounds $$f''$$ from below. A similar argument shows the same if $$a < 0$$.
• You're absolutely right. And yet: I think it may be possible to prove strong convexity in the multidimensional case with certain conditions on $(y_i,x_i)$. I don't know this for sure. But this is certainly the case that's most interesting in machine learning applications. Mar 18, 2019 at 15:36
• @MichaelGrant intuitively, i do not believe so. Since each term of the log-loss behaves, approximately, like $\max(0, -y_i x_i^T \theta)$, and therefore any $\alpha \|x\|^2$ eventually 'curves up' faster. Mar 18, 2019 at 18:03