3
$\begingroup$

In the book here: http://neuralnetworksanddeeplearning.com/chap3.html

If you scroll down to Exercise 2 in the Softmax Section, it says

Show that $\partial a^L_{j}/\partial z^L_{k}$ is positive if $j=k$ and negative if $j \neq k$. As a consequence, increasing $z^L_j$ is guaranteed to increase the corresponding output activation, $a^L_j$, and will decrease all the other output activations.

Here, $$a_j = \frac{e^{z^L_{j}}}{\sum_{k}{e^{z^L_{k}}}}$$

I managed to prove the part when $j \neq k$ by differentiating as normal to get $$-\frac{e^{z^L_{j}}}{\left(\sum_{k}{e^{z^L_{k}}}\right)^2}$$

which is obviously always negative. However I'm having trouble with when $j=k$. When I differentiated I got an inequality which simplified to proving $$\sum_{k}{e^{z^L_k}}>1$$ I am unsure of how to do this.

$\endgroup$
2
$\begingroup$

Let's remove the $L$ superscripts. The derivative with respect to $z_L$ is $$ \frac{\partial a_j}{\partial z_j} = \frac{e^{z_j} \sum_k e^{z_k} - e^{z_j} e^{z_j}}{\left(\sum_k e^{z_k}\right)^2} = \frac{e^{z_j} \sum_{k \neq j} e^{z_k}}{\left(\sum_k e^{z_k}\right)^2} > 0. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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