Gradient and Hessian of $\sum_i \log \left(1 + \exp\left\{ -t_i \left(w^T x_i\right)\right\} \right) + \mu \|w \|_2^2$?

Is my Gradient and Hessian of the following correct? \begin{align} f &= \sum_i \log \left(1 + \exp\left\{ -t_i \left(w^T x_i\right)\right\} \right) + \mu \|w \|_2^2 \ , \end{align} where $$t_i \in \mathbb{R}$$, $$w, x_i \in \mathbb{R}^n$$, and $$\mu \in \mathbb{R}$$.

I want to find the gradient and Hessian w.r.t. $$w$$, that is $$\frac{\partial f}{\partial w}$$ and $$\frac{\partial^2 f}{\partial w^2}$$.

Partial attempt

\begin{align} \frac{\partial f}{\partial w} &= \sum_i \left( \frac{-t_i x_i \exp\left\{ -t_i \left(w^T x_i\right)\right\}}{1 + \exp\left\{ -t_i \left(w^T x_i\right)\right\} } \right) + 2\mu w \ \ \\ &= \sum_i \left( \frac{-t_i x_i }{1 + \exp\left\{ +t_i \left(w^T x_i\right)\right\} } \right) + 2\mu w \ . \end{align}

Hessian

\begin{align} \frac{\partial^2 f}{\partial w^2} &= \frac{\partial}{ \partial w} \left[ \sum_i \left( \frac{-t_i x_i }{1 + \exp\left\{ +t_i \left(w^T x_i\right)\right\} } \right) + 2\mu w \right] \\ &= \sum_i \left( \frac{t_i^2 x_i x_i \ \exp\left\{ +t_i \left(w^T x_i\right)\right\}}{\left(1 + \exp\left\{ +t_i \left(w^T x_i\right)\right\}\right)^2 } \right) + 2\mu I \ . \end{align}

I Think my Hessian is for sure incorrect, isn't it? because I am getting in the numerator of the first part as $$x_i x_i$$... how would two vectors just multiply :( ...

• The last part should have the outer product $x_i x_i^\top$ instead of $x_i x_i$. – angryavian Feb 3 at 18:35
• ok, can you derive that? – learning Feb 3 at 18:38

In order to write the function in purely matrix form, first note that the $$\{x_i\}$$ vectors are columns of a single matrix $$X$$. Next use $$(\circ)$$ to denote the elementwise/Hadamard product and (:) to denote the trace/Frobenius product, i.e. $$A:B = {\rm Tr}(A^TB)$$
Define the following variables. \eqalign{ a &= t\circ X^Tw &\implies da = t\circ X^Tdw \cr b &= \exp(-a) &\implies db = -b\circ da \cr p &= \exp(a) &\implies dp = p\circ da \implies 1=b\circ p \cr c &= \log(1+b) &\implies dc = \frac{db}{1+b} \cr } Write the function in terms of these variables. Then calculate its differential and and back-substitute variables until we arrive at the gradient with respect to $$w$$. \eqalign{ f &= \mu\,w:w + 1:c \cr df &= 2\mu\,w:dw + 1:dc \cr &= 2\mu\,w:dw + \frac{1}{1+b}:db \cr &= 2\mu\,w:dw - \frac{1}{1+b}:b\circ da \cr &= 2\mu\,w:dw - \frac{b}{1+b}:t\circ X^Tdw \cr &= 2\mu\,w:dw - X\Big(\frac{t\circ b}{1+b}\Big):dw \cr &= \bigg(2\mu\,w - X\Big(\frac{t}{p+1}\Big)\bigg):dw \cr g = \frac{\partial f}{\partial w} &= 2\mu\,w - X\Big(\frac{t}{1+p}\Big) \cr } Now find the differential and gradient of $$g$$. \eqalign{ dg &= 2\mu\,dw + X\Big(\frac{t\circ dp}{(1+p)\circ(1+p)}\Big) \cr &= 2\mu\,dw + X\Big(\frac{t\circ p\circ da}{1+2p+p\circ p}\Big) \cr &= 2\mu\,dw + X\Big(\frac{t\circ p\circ t\circ X^Tdw}{1+2p+p\circ p}\Big) \cr } Replace the Hadamard products with diagonal matrices, e.g. \eqalign{ P &= {\rm Diag}(p),\,\, T &= {\rm Diag}(t),\,\, I &= {\rm Diag}(1) \cr Px &= p\circ x \cr } Therefore \eqalign{ dg &= \Big(2\mu I + X(I+2P+P^2)^{-1}T^2PX^T\Big)\,dw \cr H = \frac{\partial g}{\partial w} &= 2\mu I + X(I+2P+P^2)^{-1}T^2PX^T \cr }