Doubt on formulating cost function for GloVe

I'm reading the notes here and have a doubt on page 2 ("Least squares objective" section). The probability of a word $$j$$ occurring in the context of word $$i$$ is $$Q_{ij}=\frac{\exp(u_j^Tv_i)}{\sum_{w=1}^W\exp(u_w^Tv_i)}$$

Training proceeds in an on-line, stochastic fashion, but the implied global cross-entropy loss can be calculated as $$J=-\sum_{i\in corpus}\sum_{j\in context(i)}\log Q_{ij}$$ As the same words $$i$$ and $$j$$ can occur multiple times in the corpus, it is more efficient to first group together the same values for $$i$$ and $$j$$: $$J=-\sum_{i=1}^W\sum_{j=1}^WX_{ij}\log(Q_{ij})$$

where $$X_{ij}$$ is the total number of times $$j$$ occurs in the context of $$i$$ and the value of co-occuring frequency is given by the co-occurence matrix $$X$$. This much is clear. But then the author states that the denominator of $$Q_{ij}$$ is too expensive to compute, so the cross entropy loss won't work.

Instead, we use a least square objective in which the normalization factors in $$P$$ and $$Q$$ are discarded: $$\hat J=\sum_{i=1}^W\sum_{j=1}^WX_i(\hat P_{ij}-\hat Q_{ij})^2$$ where $$\hat P_{ij}=X_{ij}$$ and $$\hat Q_{ij}=\exp(u_j^Tv_i)$$ are the unnormalized distributions.

$$X_i=\sum_kX_{ik}$$ is the number of times any word appears in the context of $$i$$. I don't understand this part. Why have we introduced $$X_i$$ out of nowhere? How is $$\hat P_{ij}$$ "unnormalized"? Is there a tradeoff in switching from softmax to MSE?

(As far as I know, softmax made total sense in skip gram because we were calculating scores corresponding to different words (discrete possibilities) and matching the predicted output to the actual word - similar to a classification problem, so softmax makes sense.)

• In the notes, the author specifically says " it requires the distribution Q to be properly normalized, which involves the expensive summation over the entire vocabulary", so I think you should also have said this, because it was not clear what the denominator is in your context. So, the denominator is the normalization factor. – nbro Sep 17 at 17:20
• I think I have the answer to your question: How is $\hat P_{ij}$ "unnormalized"?. In section 1.2 of the notes, the author defines $P_{ij}$ as the fraction between $X_{ij}$ and $X_i$, so I suppose that $\hat P_{ij}$ is just $X_{ij}$ (only the numerator), which is number of times word $j$ occur in the context of word $i$. – nbro Sep 17 at 17:25
• @nbro: Ah you're absolutely right. Careless of me not to see that. But I'm still not sure why $X_i$ comes in – Shirish Kulhari Sep 17 at 17:36