Let me first establish what CBoW and skip-gram are supposed to do. You can skip to the next section if you think this is unnecessary.


My understanding is that Word2Vec is a suite of 2 algorithms, continuous bag-of-words (CBoW) and skip-gram neural-network language model (SGNNLM or simply skip-gram), which are both two-layer neural networks. That is, given a vocabulary $V$ and embedding size $H$, they take vectors of size $|V|$ and pass them through one hidden layer and one output layer, with weight matrices $W_1\in \mathbb{R}^{H\times |V|}$ and $W_2\in \mathbb{R}^{|V|\times H}$, much like a small auto-encoder. The output layer is always cited as having softmax activation, and according to this Coursera course, the hidden layer has ReLU activation. I've seen it cited as having "no" activation ($\varphi(x) = x$), though.

The difference between CBoW and skip-gram is supposedly that CBoW "predicts the current word based on the context", and the skip-gram "predicts surrounding words given the current word". That's a literal quote from the original Word2Vec paper, Mikolov 2013. Here's their figure, which looks intuitive, but in reality is notoriously confusing:

Mikolov figure

Since both are used to move from a one-hot encoding to a smaller embedding (the weights in one or both of the matrices), their input and output format is still one-hot. I know that per training example, CBoW starts out with $C$ one-hot vectors (one for each context word) and just averages them to get a single $|V|$-sized vector. (This is why it's called "bag of words": because averaging is commutative, you lose the order of the context words.) Apart from this tricky many-to-one transformation, the rest of the network makes sense.

Skip-gram has no issue encoding the input, but it has a one-to-many transformation at the end. This is a problem.

The issue

What does skip-gram do in its final layer? I have been scouring various StackExchange sites to find a consensus on this question, and there seem to be fourdistinct camps.

  1. Skip-gram predicts $C$ softmaxes of size $|V|$ using $C$ different matrices $W_{2}^{(1)} \dots W_{2}^{(C)}$. Examples: this article (at least the figures), this article stating "each with its own weight matrix", and this question.

  2. Skip-gram predicts $C$ softmaxes of size $|V|$ using the same matrix $W_2$. Examples: this question and this answer.

  3. Skip-gram predicts $1$ softmax of size $|V|$ using one matrix $W_2$, and the dataset consist of (center word, context word) pairs instead of (center word, all context words). Examples: this tutorial.

  4. Skip-gram is a binary classifier predicting whether a given word is "in context" vs. "out of context" for another word. The dataset consists of pairs of words that either appeared close to each other (positive) or were randomly paired (negative). Examples: this article, this article, and this answer.

Skip-gram is usually explained as having one output matrix, which is hypothesis (2) or (3), yet producing $C$ different words in the output layer, from the same hidden representation, which is hypothesis (1). See this article and the figure in this question. That's a magic trick. Pure fiction.

I find (3) the most reasonable, since it only has two weight matrices and doesn't produce useless copies. In that case, skip-gram actually doesn't predict any words, but rather produces a single $|V|$-sized vector which represents the average one-hot vector of the context words. This single vector can then account for one term in the loss per context word.

Yet, there are some major issues with (3):

  • You can never reach $0$ loss. Let's say the loss function is log likelihood, and compute the loss for one context window: $$ \ell(\text{word}, \text{context}) = \sum_{i=1}^C \ln P(\text{context}_i \mid \text{word}) $$ Clearly, that probability cannot be $1$ for each context word. In the perfect case, it is $1/C$.

  • It cannot handle negative examples like (4). Let $y$ be a binary variable that is $1$ when a pair of words is positive and $0$ when it is negative. What is $P(y \mid w_1,w_2)$ given the above model? It can't just be $P(w_2 \mid w_1)$. Why? Look at this binary loss: $$\begin{aligned} \ell(w_1, w_2, y) &= y\ln P(y = 1 \mid w_1, w_2) + (1-y)\ln (1 - P(y = 1 \mid w_1,w_2)) \\ &= y\ln P(w_2 \mid w_1) + (1-y)\ln (1 - P(w_2 \mid w_1)) \end{aligned}$$ As we saw before, best-case scenario, $P(w_2 \mid w_1) = 1/C$ for a positive example, which means $P(y = 0 \mid w_1,w_2) = 1 - 1/C$, clearly way higher. For example: with a context of $C = 4$, the model predicts that a word is context by giving it 25% probability, and accidentally assigns the remaining 75% to all other words, so it always predicts that a word is out of context. You simply cannot combine softmax likelihood with negative sampling. This answer specifically says that softmax likelihood isn't used at all, and instead we use "noise-contrastive estimation".

So, once and for all: does Word2Vec's skip-gram NNLM even produce context words?


1 Answer 1


The answer is 3, but keep in mind that the original algorithm iterates over all available context words for each center word, thus all possible pairs are included for each center word (some frequent center words that may be stop words can be discarded depending on parameters). Also the original work uses a faster alternative algorithm to replace the standard softmax.

Depending on the negative sampling parameter, the training set can also have more pairs, where the second word is chosen at random.

See section 2 of another paper of Mikolov et al https://arxiv.org/pdf/1310.4546.pdf and the original source code https://code.google.com/archive/p/word2vec/.


OK, I've looked at the paper and the code I suggested above. I think they contain all the information to correctly answer your question.

My (wrong) answer above was based on a simplification of the method. Actually softmax is only discussed as a simpler and slower alternative to hierarchical softmax. Pure softmax over all dictionary words does not seem to be used in the original word2vec implementation.

I think the correct answer is that Skip-gram learns to maximize the log-likelihood of the "Hierarchical Softmax" probability given in formula (3) of the paper (which I gave a link to) over the pairs of nearby words.

For a single data point the "Hierarchical Softmax" loss can be expressed as a sum of terms over nodes on the tree path leading from the root to the center word. Each of those terms looks like ordinary log loss (binary cross entropy), as in your item 4., but because the summing over the path in a tree, overall it is certainly more complicated than your 4.

If negative sampling is turned on, the method just adds (4) from the paper to the loss. This would be almost exactly like your item 4.

  • $\begingroup$ Iterating over all $C$ context words is what the summation at the bottom does. Anyway: how are negative samples included in the loss if (3) is the answer? The binary loss I wrote out doesn't seem to work. $\endgroup$
    – Mew
    Jan 21 at 16:07
  • $\begingroup$ You are right. My answer is incomplete. $\endgroup$
    – Valentas
    Jan 23 at 16:27
  • $\begingroup$ See my revised answer @Mew. $\endgroup$
    – Valentas
    Jan 26 at 16:05
  • $\begingroup$ Where does this "tree" come from? It's unclear from the paper how it is constructed or what the significance of the root is. Also, the two equations you point to (3 and 4 in the paper) seem to measure different things: one measures $p(\text{context} \mid \text{center})$, the other measures some kind of similarity $p(\text{similar} \mid \text{context, center})$. Both use some kind of "representation" $v$ of the relevant words, but this is never explained; is it a weight vector? Is it the output of the plain softmax? $\endgroup$
    – Mew
    Jan 27 at 13:04
  • $\begingroup$ The tree is a Huffman tree of words, constructed using just word frequencies. The idea is described (without much detail) in the referenced paper of Morin and Bengio (2005). I found the word2vec.c source code easier to understand. $v$s are just the word vectors. And very good point on measuring different quantities. I'd guess using their sum is an engineering trick to make use of both? Mathematically you can view them as optimising two different objective functions at the same time. How much weight each of them has is chosen by them empirically via parameter $k$. $\endgroup$
    – Valentas
    Jan 27 at 14:57

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