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This is related to my earlier question, which I'm trying to break down into parts (this being the first) since it seemed too large. I'm reading notes on word vectors here. Specifically, I'm referring to section 4.2 on page 7. First, regarding points 1 to 6 - here's my understanding:

If we have a vocabulary $V$, the naive way to represent words in it would be via one-hot-encoding, or in other words, as basis vectors of $R^{|V|}$ - say $e_1, e_2,\ldots,e_{|V|}$. We want to map these to $\mathbb{R}^n$, via some linear transformation such that the images of similar words (more precisely, the images of basis vectors corresponding to similar words) have higher inner products. Assuming the matrix representation of the linear transformation given the standard basis of $\mathbb{R}^{|V|}$ is denoted by $\mathcal{V}$, then the "embedding" of the $i$-th vocab word (i.e. the image of the corresponding basis vector $e_i$ of $V$) is given by $\mathcal{V}e_i$.

Now suppose we have a context "The cat ____ over a", CBoW seeks to find a word that would fit into this context. Let the words "the", "cat", "over", "a" be denoted (in the space $V$) by $x_{i_1},x_{i_2},x_{i_3},x_{i_4}$ respectively. We take the image of their linear combination (in particular, their average): $$\hat v=\mathcal{V}\bigg(\frac{x_{i_1}+x_{i_2}+x_{i_3}+x_{i_4}}{4}\bigg)$$

We then map $\hat v$ back from $\mathbb{R}^n$ to $\mathbb{R}^{|V|}$ via another linear mapping whose matrix representation is $\mathcal{U}$: $$z=\mathcal{U}\hat v$$

Then we turn this score vector $z$ into softmax probabilities $\hat y=softmax(z)$ and compare it to the basis vector corresponding to the actual word, say $e_c$. For example, $e_c$ could be the basis vector corresponding to "jumped".

Here's my interpretation of what this procedure is trying to do: given a context, we're trying to learn maps $\mathcal{U}$ and $\mathcal{V}$ such that given a context like "the cat ____ over a", the model should give a high score to words like "jumped" or "leaped", etc. Not just that - but "similar" contexts should also give rise to high scores for "jumped", "leaped", etc. For example, given a context "that dog ____ above this" wherein "that", "dog", "above", "this" are represented by $x_{j_1},x_{j_2},x_{j_3},x_{j_4}$, let the image of their average be

$$\hat w=\mathcal{V}\bigg(\frac{x_{j_1}+x_{j_2}+x_{j_3}+x_{j_4}}{4}\bigg)$$

This gets mapped to a score vector $z'=\mathcal{U}\hat w$. Ideally, both score vectors $z$ and $z'$ should have similarly high magnitudes in their components corresponding to similar words "jumped" and "leaped".

Is my above understanding correct? Consider the following quote from the lectures:

We create two matrices, $\mathcal{V} \in \mathbb{R}^{n\times |V|}$ and $\mathcal{U} \in \mathbb{R}^{|V|\times n}$, where $n$ is an arbitrary size which defines the size of our embedding space. $\mathcal{V}$ is the input word matrix such that the $i$-th column of $\mathcal{V}$ is the $n$-dimensional embedded vector for word $w_i$ when it is an input to this model. We denote this $n\times 1$ vector as $v_i$. Similarly, $\mathcal{U}$ is the output word matrix. The $j$-th row of $\mathcal{U}$ is an $n$-dimensional embedded vector for word $w_j$ when it is an output of the model. We denote this row of $\mathcal{U}$ as $u_j$.

It's not obvious to me why $v_i=\mathcal{V}e_i$ should be the same as or even similar to $u_i$. How does the whole backpropagation procedure above ensure that?

Also, how does the procedure ensure that basis vectors corresponding to similar words $e_i$ and $e_j$ are mapped to vectors in $\mathbb{R}^n$ that have high inner product? (In other words, how is it ensured that if words no. $i_1$ and $i_2$ are similar, then $\langle v_{i_1}, v_{i_2}\rangle$ and $\langle u_{i_1}, u_{i_2}\rangle$ have high values?)

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  • $\begingroup$ The term you're looking for is "distributional hypothesis." $\endgroup$ – Sycorax says Reinstate Monica Aug 16 at 14:07
  • $\begingroup$ @Sycorax: I've read up on it - at least the wikipedia page. That's the underlying hypothesis but my question is about the mechanism of the CBoW method - on how it ensures embedded representations are set up according to the distributional hypothesis. $\endgroup$ – Shirish Kulhari Aug 16 at 19:31
  • $\begingroup$ @ShirishKulhari, I believe it is 'ensured' by the loss function definition in the backpropagation design. It will be data driven as the inner product is made to be large by definition, and in the case above, the transpose inner product set will be high $\endgroup$ – Vass Aug 27 at 20:15

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