3
$\begingroup$

For word2vec with negative sampling, the cost function for a single word is the following according to word2vec: $$ E = - log(\sigma(v_{w_{O}}^{'}.u_{w_{I}})) - \sum_{k=1}^K log(\sigma(-v_{w_{k}}^{'}.u_{w_{I}})) $$

$v_{w_{O}}^{'}$ = hidden->output word vector of the output word

$u_{w_{I}}$ = input->hidden word vector of the output word

$v_{w_{k}}^{'}$ = hidden->output word vector of the negative sampled word

$\sigma$ is the sigmoid function

And taking the derivative with respect to $v_{w_{O}}^{'}.u_{w_{j}}$ is:

$ \frac{\partial E}{\partial v_{w_{j}}^{'}.u_{w_{I}}} = \sigma(v_{w_{j}}^{'}.u_{w_{I}}) * (\sigma(v_{w_{j}}^{'}.u_{w_{I}}) - 1) $ $ if w_j = w_O $

$ \frac{\partial E}{\partial v_{w_{j}}^{'}.u_{w_{I}}} = \sigma(v_{w_{j}}^{'}.u_{w_{I}}) * \sigma(-v_{w_{j}}^{'}.u_{w_{I}}) $ $ if w_j = w_k \ for \ k = 1...K$

Then we can use chain rule to get

$ \frac{\partial E}{\partial v_{w_{j}}^{'}} = \frac{\partial E}{\partial v_{w_{j}}^{'}.u_{w_{I}}} * \frac{\partial v_{w_{j}}^{'}.u_{w_{I}}}{\partial v_{w_{j}}^{'}} $

Is my reasoning and derivative correct? I am still new to ML so any help would be great!

$\endgroup$
0
$\begingroup$

Looks good to me. This derivative is also presented in the paper (equations 56-58).

The paper you're linking to is the most advanced attempt - at least to best of my knowledge - to explain how word2vec works, but there is also a lot of other resources on the topic (just search for word2vec on arxiv.org). If you're interested in word2vec, you may find GloVe interesting too (see: Linking GloVe with word2vec).

$\endgroup$
2
  • $\begingroup$ And the GloVe vectors are significantly faster to compute than word2vec $\endgroup$ – Vladislavs Dovgalecs Jul 29 '15 at 15:25
  • $\begingroup$ Please advise which paper has the equation 56-58. I think the original paper is arxiv.org/pdf/1310.4546.pdf which does not have 56-58. $\endgroup$ – mon Apr 11 at 2:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.