Suppose we want to predict context words $w_{i-h}, \dots, w_{i+h}$ given a target word $w_i$ for a window size $h$ around the target word $w_i$. We can represent this as: $$p(w_{i-h}, \dots, w_{i+h}|w_i) = \prod_{-h \leq k \leq h, \ k \neq 0} p(w_{i+k}|w_i)$$ where we model the probabilities of a word $u$ given another word $v$ as $$p(u|v) = \frac{\exp(\left<\phi_u, \theta_v \right>)}{\sum_{u' \in W} \exp(\left<\phi_{u'}, \theta_v \right>)}$$ where $\phi_u, \theta_v$ are some vector representations for words $u$ and $v$ respectively and $\left<\phi_u, \theta_v \right>$ is the dot product between these vector representations (which represents some sort of similarity between the words) and $W$ is a matrix of all the words.
In Skip-Gram Negative Sampling, we want to learn the embeddings $\phi_u, \theta_v$ that maximize the following: $$\sum_{u \in W} \sum_{v \in W} n_{uv} \log \sigma(\left<\phi_u, \theta_v \right>) +k \mathbb{E}_{\bar{v}} \log \sigma(-\left<\phi_u, \theta_{\bar{v}} \right>)$$
Question. How exactly does this work? For example, suppose $k=5$, the target word $w_i$ is $\text{apple}$ and we want to find $p(\text{pie}| \text{apple})$. Let $n_{uv} = 10$ (number of times pie co-occurs with apple). Then we sample $5$ random words $\bar{v}$ that did not occur with $\text{apple}$ and whichever term in the sum is bigger is the one we predict? For example, if the first term in the sum is larger than the second term then we would predict that $p(\text{pie}| \text{apple}) \approx 1$? Otherwise we predict that $p(\text{pie}| \text{apple}) \approx 0$? Is this the correct intuition?
Source. Here at around the 10:05 mark.
