Considering a binary classification problem with data $D = \left\{ (x_i, y_i) \right\}_{i=1}^n$, $x_i \in \mathbb{R}^d$ and $y_i \in \{0,1\}$. Given the following definitions,

$$f(x) = x^T \beta$$

$$p(x) = \sigma(f(x)) \quad \text{with} \quad\sigma(z) = 1/(1 + e^{-z})$$

$$L(\beta) = \sum_{i=1}^n \Bigl[ y_i \log p(x_i) + (1 - y_i) \log [1 - p(x_i)] \Bigr]$$

where $\beta \in \mathbb{R}^d$ is a vector. $p(x)$ is a short-hand for $p(y = 1\ |\ x)$. The task is to compute the derivative $\frac{\partial}{\partial \beta} L(\beta)$. A tip is to use the fact $$\frac{\partial}{\partial z} \sigma(z) = \sigma(z) (1 - \sigma(z))$$

So here is my approach so far:

\begin{align*} L(\beta) & = \sum_{i=1}^n \Bigl[ y_i \log p(x_i) + (1 - y_i) \log [1 - p(x_i)] \Bigr]\\ \frac{\partial}{\partial \beta} L(\beta) & = \sum_{i=1}^n \Bigl[ \Bigl( \frac{\partial}{\partial \beta} y_i \log p(x_i) \Bigr) + \Bigl( \frac{\partial}{\partial \beta} (1 - y_i) \log [1 - p(x_i)] \Bigr) \Bigr]\\ \end{align*}

\begin{align*} \frac{\partial}{\partial \beta} y_i \log p(x_i) &= (\frac{\partial}{\partial \beta} y_i) \cdot \log p(x_i) + y_i \cdot (\frac{\partial}{\partial \beta} p(x_i))\\ &= 0 \cdot \log p(x_i) + y_i \cdot (\frac{\partial}{\partial \beta} p(x_i))\\ &= y_i \cdot (p(x_i) \cdot (1 - p(x_i))) \end{align*}

\begin{align*} \frac{\partial}{\partial \beta} (1 - y_i) \log [1 - p(x_i)] &= (1 - y_i) \cdot (\frac{\partial}{\partial \beta} \log [1 - p(x_i)])\\ & = (1 - y_i) \cdot \frac{1}{1 - p(x_i)} \cdot p(x_i) \cdot (1 - p(x_i))\\ & = (1 - y_i) \cdot p(x_i) \end{align*}

$$\frac{\partial}{\partial \beta} L(\beta) = \sum_{i=1}^n \Bigl[ y_i \cdot (p(x_i) \cdot (1 - p(x_i))) + (1 - y_i) \cdot p(x_i) \Bigr]$$

So basically I used the product and chain rule to compute the derivative. I am afraid, that my solution is wrong, because on page 120 of The Elements of Statistical Learning it says the gradient is

$$\sum_{i = 1}^N x_i(y_i - p(x_i;\beta))$$

I don't know what could have possibly gone wrong. Any advice on this?

  • $\begingroup$ Did you mean $p(x)=\sigma(p(x))$ ? Because I don't see you using $f$ anywhere. $\endgroup$ Apr 28, 2012 at 7:08
  • $\begingroup$ Also, note your final line can be simplified to: $\sum_{i=1}^n \Bigl[ p(x_i) (y_i - p(x_i)) \Bigr]$. $\endgroup$ Apr 28, 2012 at 7:12
  • $\begingroup$ Yes, absolutely, thanks for pointing out, it is indeed $p(x) = \sigma(p(x))$. But isn't the simplification term: $\sum_{i=1}^n [p(x_i) ( 1 - y \cdot p(x_i)]$ ? $\endgroup$
    – Mahoni
    Apr 28, 2012 at 7:16
  • $\begingroup$ Maybe, but I just noticed another mistake: when you compute the derivative of the first term in $L(\beta)$. $\endgroup$ Apr 28, 2012 at 7:18
  • $\begingroup$ Ah, are you sure about the relation being $p(x)=\sigma(f(x))$? Because if that's the case, then I can see why you don't arrive at the correct result. $\endgroup$ Apr 28, 2012 at 7:21

3 Answers 3


So, if $p(x)=\sigma(f(x))$ and $\frac{d}{dz}\sigma(z)=\sigma(z)(1-\sigma(z))$, then

$$\frac{d}{dz}p(z) = p(z)(1-p(z)) f'(z) \; .$$

This changes everyting and you should arrive at the correct result this time.

In particular,

$$\frac{d}{dz}\log p(z) = (1-p(z)) f'(z)$$


$$\frac{d}{dz}\log (1-p(z)) = -p(z) f'(z) \; .$$

  • $\begingroup$ Of course, I ignored the chain rule for that one! $\endgroup$
    – Mahoni
    Apr 28, 2012 at 7:26
  • $\begingroup$ Also be careful because your $\beta$ is a vector, so is $x$. So you should really compute a gradient when you write $\partial/\partial \beta$. $\endgroup$ Apr 28, 2012 at 7:29

The classification problem data can be captured in one matrix and one vector, i.e. $\{X,y\}$.

Then the relevant quantities are the vectors $$\eqalign{ f &= X^T\beta \cr p &= \sigma(f) \cr }$$ and their differentials and logarithmic differentials $$\eqalign{ df &= X^Td\beta \cr dp &= p\circ(1-p)\circ df \cr\cr d\log(p) &= \frac{dp}{p} \,=\, (1-p)\circ df \cr d\log(1-p) &= \frac{-dp}{1-p} \,=\, -p\circ df \cr }$$ where $(g\circ h)$ and $\big(\frac{g}{h}\big)$ denote element-wise (aka Hadamard) multiplication and division.

The likelihood function is a scalar which can be written in terms of Frobenius products $$\eqalign{ L &= y:\log(p) + (1-y):\log(1-p) \cr }$$ whose differential is $$\eqalign{ dL &= y:d\log(p) + (1-y):d\log(1-p) \cr &= y:(1-p)\circ df - (1-y):p\circ df \cr &= (y-p):df \cr &= \big(y-p\big):X^Td\beta \cr &= X\,\big(y-p\big):d\beta \cr }$$ Yielding the gradient as $$\eqalign{ \frac{\partial L}{\partial\beta} &= X\,(y-p) \cr }$$

  • $\begingroup$ This is the matrix form of the gradient, which appears on page 121 of Hastie's book. $\endgroup$
    – greg
    Nov 14, 2015 at 19:27
  • $\begingroup$ Why did the transpose of X become just X? How did you remove the transpose by moving the order to the front? $\endgroup$
    – Sticky
    Jan 29, 2018 at 3:19

In your third line, while differentiating you missed out $1/p(x_i)$ which is the derivative of $\log(p(x_i))$.

\begin{eqnarray} d/db(y_i \cdot \log p(x_i)) &=& \log p(x_i) \cdot 0 + y_i \cdot(d/db(\log p(x_i))\\ &=& y_i \cdot 1/p(x_i) \cdot d/db(p(x_i)) \end{eqnarray}

Note that $d/db(p(xi)) = p(x_i)\cdot {\bf x_i} \cdot (1-p(x_i))$ and not just $p(x_i) \cdot(1-p(x_i))$.

Also in 7th line you missed out the $-$ sign which comes with the derivative of $(1-p(x_i))$.


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