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I am trying to understand the loss function which is used for the word2vec model, but I don't really follow the argumentation behind this video https://www.youtube.com/watch?v=ERibwqs9p38&t=5s, at 29:30.

The formula which is unclear is the following:

$J(\theta) = \displaystyle-\dfrac{1}{T} {\sum_{t=1}^{T} \sum_{-m <= j <=m, \\j\ne0} log(p(w_{t+j}|w_t))}.$

  • $T$ is the number of words in the vocabulary
  • $w_t$ is a given word and we try to calculate the probablity that another word $w_{t+j}$ occurs within a window of +/- $m$ words ahead.
  • $\theta$ is the solution we're after. It is essentially a $2*d*Tx1$ dimensional vector which contains all the columns of the matrices V and U.

At a first sight it looks all pretty clear: we're iterating though the whole vocabulary and for each (fixed word then), we add up all probabilities that another word occurs within a window around that fixed word.

However, it fails apart for me when I consider that a word $w_t$ occurs in many positions in the corpus and might occur multiple times with a different word. E.g, the words 'deep learning' often occur together, which indicates that there's a contextual relation between them. Why would we only count them twice? It seems like the formula above is counting each pair $p(w_{t+j}|w_t)$ just twice (e.g. once for $p(deep|learning)$ and once for $ p(learning|deep)$). IMO we should need a correcting term that adjusts for the missing 'frequency', e.g.

$J(\theta) = \displaystyle-\dfrac{1}{T} {\sum_{t=1}^{T} \sum_{-m <= j <=m, \\j\ne0} log(\lambda(w_t,w_{t+j})p(w_{t+j}|w_t))}.$

In the case where $\lambda(x,y) = 1$, we get the formula above, but we could also be free to chose a function that boosts frequent occurrences (pairwise). The formula above could then be seen as a special case when you don't care that words that occur more often together get a boost.

On the other hand, when the formula above already accounts for multiple occurrences, then where is this visible? The author then continues and defines $p(o|w)$ as $exp(o^T*w)/\sum(exp(u^T*w))$. In this particular case, I don't see that we're counting the dot-product as many times as the word $o$ is in the neighbourhood of the word $w$. Maybe the choice of $p$ is just very simplistic and represents a model where the fact that 2 words are in the neighbourhood (just somewhere in the corpus) is enough (bag of words model???). It's hard to see then that such models deliver good performance in NLP though.

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Not sure what the video said, but $T$ should not be the vocabulary size, but the training corpus size (number of all words).

For example, if your training corpus is

deep learning is popular . i love deep learning . i want to learn more about it.

Then when you sume up over $T$, you will sum up all the word pairs in the corpus including duplicates. The word pair (deep learning) is indeed calculated twice.

For details please refer to the original skip-gram paper, and notice the definitoin (1) on page 2.

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  • $\begingroup$ So if T is the corpus size and not the vocabulary size, then this changes everything. It was not clear from the formula what T was. Thank you for the clarification $\endgroup$ – shaft Sep 21 '18 at 22:02

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