0
$\begingroup$

Given a function $CE=-\sum_i y_i log(\hat{y}_i)$ and $\hat{y}_i=\frac{e^{\theta_i}}{\sum_j e^{\theta_j}}$, where $y$ and $\theta$ are vectors. The question asks to compute $\frac{\partial CE}{\partial \theta}$.

(Hint from the original question: $y$ is the one-hot label vector, you might want to consider the fact many elements of $y$ are zeros, and assume that only the $k$-th dimension of $y$ is one.)

My solution: $$\frac{\partial CE(y,\hat{y})}{\partial \theta_i}= -\sum_i y_i \frac{1}{\hat{y}_i} \frac{e^{\theta_i}\sum_j e^{\theta_i} -e^{2\theta_i}}{(\sum_j e^{\theta_j})^2}= -\sum_i y_i \frac{\sum_j e^{\theta_j}}{e^{\theta_i}} \frac{e^{\theta_i}\sum_j e^{\theta_j} -e^{2\theta_i}}{(\sum_j e^{\theta_j})^2}=\sum_i y_i (\hat{y}_i-y_i)$$

thus, $\frac{\partial CE(y,\hat{y})}{\partial \theta_i}$=\begin{cases} \hat{y}_i-1, \quad i=k\\ 0, \quad \text{otherwise} \end{cases}

Solution given:

$\frac{\partial CE(y,\hat{y})}{\partial \theta}=\hat{y}-y$

or equivalently, $\frac{\partial CE(y,\hat{y})}{\partial \theta_i}$=\begin{cases} \hat{y}_i-1, \quad i=k\\ \hat{y}_i, \quad \text{otherwise} \end{cases}

The difference shows on calculations of otherwise, did I miss anything in my solution?

Update (Solved):

$\endgroup$
0
$\begingroup$

\begin{align*} \frac{\partial CE(y,\hat{y})}{\partial \theta_k} &= \sum_i \frac{\partial CE(y,\hat{y})}{\partial \hat{y}_i} \frac{\partial \hat{y}_i}{\partial \theta_k}\\ &= -y_k \frac{1}{\hat{y}_k} \frac{\partial \hat{y}_k}{\partial \theta_k}- \sum_{i,i \neq k} y_i \frac{1}{\hat{y}_i} \frac{\partial \hat{y}_i}{\partial \theta_k}\\ &= - y_k \frac{\sum_k e^{\theta_k}}{e^{\theta_k}} \frac{e^{\theta_k}\sum_j e^{\theta_j} -e^{2\theta_k}}{(\sum_j e^{\theta_j})^2}+\sum_{i,i\neq k}y_i \frac{\sum_j e^{\theta_j}}{e^{\theta_i}} \frac{e^{\theta_i} e^{\theta_k}}{(\sum_j e^{\theta_j})^2}\\ &=-y_k(1-\hat{y}_k)+ \sum_{i,i\neq k}y_i \hat{y}_k\\ &= -y_k + \hat{y}_k\sum_i y_i\\ &= \hat{y_k}-y_k \end{align*}

| cite | improve this answer | |
$\endgroup$

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.