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I'm currently reading the book Deep Learning (Goodfellow et al., 2015) and had a question regarding the calculation of a gradient when explaining backpropagation for a certain example. For anyone who's curious, this is from section 6.5.9: Differentiation outside the Deep Learning Community.

Suppose we have variables $p_1, p_2, ... , p_n$ representing probabilities and variables $z_1, z_2, ... , z_n$ representing unnormalized log probabilities. Suppose we define

$$q_i = \frac{e^{z_i}}{\sum_i e^{z_i}}$$

where we build the softmax function out of exponentiation, summation and division operations, and construct a cross-entropy loss $J = -\sum_i p_i \log{q_i}$. A human mathematician can observe that the derivateive of $J$ with respect to $z_i$ takes a very simple form: $q_i - p_i$.

I don't know how this result was derived, and was hoping that someone could give me some tips or advice. What I have so far is

$$\log{q_i} = \log{e^{z_i}} - \log({\sum_i e^{z_i}})$$

$$ \begin{align} p_i\log{q_i} & = p_i \log{e^{z_i}} - p_i \log({\sum_i e^{z_i}}) \\ & = p_iz_i - p_i\log(\sum_i e^{z_i}) \end{align}$$

If we take the derivative of $J = p_i\log{q_i}$ then I can understand that $d/dz_i (p_i z_i) = p_i$, but how do we differentiate the second term that contains the logarithm of the summation?

Thank you.

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Your derivation of $p_i\log q_i$ is fine. Based upon it we obtain for $J$:

\begin{align*} J&=-\sum_{j=1}^np_jz_j+\sum_{j=1}^np_j\log\left(\sum_{k=1}^ne^{z_k}\right)\\ &=-\sum_{j=1}^np_jz_j+\log\left(\sum_{k=1}^ne^{z_k}\right)\tag{1} \end{align*}

In the last line we use the sum of the probabilities $p_j,1\leq j\leq n$ is equal to $1$.

From (1) we obtain the derivation of $J$ with respect to $z_i$ as: \begin{align*} \color{blue}{\frac{d}{dz_i}J} &=\frac{d}{dz_i}\left(-\sum_{j=1}^np_jz_j\right)+\frac{d}{dz_i}\left(\log\left(\sum_{k=1}^ne^{z_k}\right)\right)\\ &=-p_i+\frac{e^{z_i}}{\sum_{k=1}^ne^{z_k}}\\ &\,\,\color{blue}{=-p_i+q_i} \end{align*} in accordance with the claim.

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Well, $$\frac{\partial J}{\partial z_i} = \frac{\partial}{\partial z_i} (-\sum_j p_j\log q_j )=\frac{\partial}{\partial z_i}( -p_i\log q_i) \\= \frac{\partial}{\partial z_i} (-p_i \log (e^{z_i}/\sum_k e^{z_k}))\\ =\frac{\partial}{\partial z_i} (-p_iz_i\log e + p_i\log(\sum_k e^{z_k})) \\= \frac{\partial}{\partial z_i} (-p_iz_i) + \frac{\partial}{\partial z_i}p_i\log(\sum_k e^{z_k}) \\= -p_i + p_ie^{z_i}/\sum_k e^{z_k} \\= -p_i+p_iq_i.$$

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  • $\begingroup$ But didn't the op claim the partial derivative to be $q_i - p_i$? $\endgroup$ – Viktor Glombik Jul 19 at 9:00
  • $\begingroup$ Needs checking by a human mathematician. $\endgroup$ – Wuestenfux Jul 19 at 9:07
  • $\begingroup$ Hello, thanks for the answer. In the first line why (or how) did you get rid of the symbol for summation? Is this a rule in calculus? $\endgroup$ – Seankala Jul 20 at 1:42
  • $\begingroup$ @Seankala: Note $p_1$ is a function of $z_1,$ but $p_2$ is not related to $z_1.$ $p_i$ for $i \neq 1 $ are all constant with respect to $z_1.$ The same argument works for the $q_i.$ The derivative is linear, and the derivative of a constant is $0.$ $\endgroup$ – Artimis Fowl Jul 21 at 21:49
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    $\begingroup$ @Wuestenfux I think there is a mistake in the third to last line - note $z_i = \log p_i \implies p_i = e^{z_i},$ so we can't treat $p_i$ as a constant there. Likewise for $q_i.$ $\endgroup$ – Artimis Fowl Jul 21 at 21:55

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