Suppose we have following objective function that we want to minimize: $$\mathcal{L} = \frac{1}{2} \sum_{u}\sum_{j \in I(u)}((b_u + b_j + \mu + q_j^T(p_u + \lvert I(u)\rvert ^ {-\frac{1}{2}}\sum_{i \in I(u)}y_i)) - r_{u,j})^2 + \ldots$$

$u$ is set of users, $I(u)$ is the index of items rated by user $u$, $p_u$ is a vector of size $n$, $q_j$ is a vector of size $m$, If I want to use batch gradient descent for learning model parameters how can I compute $\frac{\partial{E}}{\partial{y_i}}$?

I have looked at the paper and author have computed the gradient as follows: $$\forall i \in I(u), \frac{\partial \mathcal{L}}{\partial y_i} = \sum_{j \in I(u)} e_{u,j} \lvert I(u) \rvert^{-\frac{1}{2}} q_j$$

and also they have computed gradient of $p_u$ as follows: $$\frac{\partial{\mathcal{L}}}{\partial{p_u}} = \sum_{j \in I(u)} e_{u,j}q_j$$

What really confuses me is the $\forall i \in I(u)$ term for $y_i$ values, why we have that? why we didn't treat $y_i$ just like $p_u$

Edit: For refrence you can take a look at this link this paper.

• any comment on question? Do I need to clarify anything? – Rathma Jun 23 '17 at 19:51

I'm not sure where the confusion is. Let $\mathcal{L}_u$ be the expression of everything under the $\sum_u$, i.e., $\mathcal{L} = \frac{1}{2}\sum_u \mathcal{L}_u$. I'm assuming you're doing the batch gradient descent over users. So for a single user, under $\mathcal{L}_u$, the $y_i$ term appears within the context of $\sum_{j \in I(u)} \sum_{i \in I(u)}$ whereas $p_u$ appears only in the context of $\sum_{j \in I(u)}$. In other words, for $i \not\in I(u)$, $\mathcal{L}_u$ is not dependent on $y_i$.
• thanks for your answer, what happens if we had some other variable like $\sum_{k \in T(u)} p_k$ was present instead of $y_i$? – Rathma Jun 25 '17 at 12:28
• Assuming these are same $p$ variables as before, then the partial derivative expression would be a little more complex. Look at partial derivative relative to $p_k$, then have $S_1(\dots) + S_2(\dots)$ where $S_1$ is 0 for $k \neq u$ and is $\sum_{j \in I(u)} e_{u,j}q_j$ otherwise, and $S_2$ is 0 for $k \not\in T(u)$ and is $\sum_{j \in I(u)} e_{u,j}|I(u)|^{-\frac{1}{2}}q_j$ otherwise. Hope that's clear, hard to write it inline in comment with no preview – MotiN Jun 25 '17 at 12:38