# Logistic Regression is convex proof

I am trying to make sense of this paper qwone.com/~jason/writing/convexLR.pdf

"Regularized Logistic Regression is Strictly Convex" by Jason D. M. Rennie.

I am following the proof and formula (1) is a given:

$$-\ln(P(\vec{y}\mid X,\vec{w})) = \sum_{i=1}^N \ln(1+e^{(-y_{i}\vec{w}^T\vec{x}_i)}$$

Assuming:

$$g(z) = \frac{1}{1+e^{-z}}$$

I also see how

$$1-g(z) = \frac{e^{-z}}{1+e^{-z}}$$

$$\frac{\partial g(z)}{\partial z} = -g(z)(1-g(z))$$

If I differentiate g(z) w.r.t. z I get:

$$\frac{\partial g(z)}{\partial z} = \frac{e^{-z}}{(1+e^{-z})^2}$$

which is $g(z)(1-g(z))$ not $-g(z)(1-g(z))$

Also, when doing (2) I get the negative of what is expressed there (taking into account it is performing the partial differential of - L.H.S. of (1)):

$$\frac{\partial (-\text{L.H.S. (1)} )}{\partial w_j}$$

Here is the graph of $\displaystyle g(z)=\frac{1}{1+e^{-z}}$:
which is clearly increasing, hence the derivative should be positive and as your calculations show $g'(z)=g(z)(1-g(z))>0$. This could be a typo in the paper you mentioned. Further equation (2) in the "paper" is correct since it is the derivative of $\displaystyle\log P(.)$ not $\displaystyle-\log P(.)$. The second derivative in the paper is also correct.