10
$\begingroup$

Assuming a discrete variable $z_j$ with unnormalized probability $\alpha_j$, one way to sample is to apply argmax(softmax($\alpha_j$)), another is to do the Gumbel trick argmax($\log\alpha_j+g_j$) where $g_j$ is gumbel generated noise. This second approach is useful if we want to do something like variational auto encoding (i.e., encode an input $x_j$ into a latent discrete variable $z_j$). Then, if the goal was to have the full distribution over possible outcomes for $z_j$, we can use softmax transformation on top of the perturbation with Gumbel noise: $$\pi_j = \frac{e^{\log \alpha_j+g_j}}{\sum_{k=1}^{k=K}e^{\log \alpha_k+g_k}}\ \ \ \text{where}\ \ g_k=-\log(-\log(\epsilon\sim {U}(0,1))).$$ Why this isn't enough? Why do we need to include the temperature $\tau$ term in this? And rewrite, $$\pi_j = \frac{e^{\frac{\log\alpha_j+g_j}{\tau}}}{\sum_{k=1}^{k=K}e^\frac{\log \alpha_k+g_k}{\tau}}\ \ \ \text{where}\ \ g_k=-\log(-\log(\epsilon\sim {U}(0,1)))$$ I understand that the temperature makes the vector $\pi=[\pi_1, ...,\pi_k]$ smoother or rougher (i.e., high temperature just makes all $\pi_i$s to be the same, and generates a flatter distribution, and $\tau=1$ just makes the two equations identical) but why do we need it in practice? All we want (i.e., in VAE) is to decouple the stochastic aspect of the sampling (i.e, move the stochastic part of it to the input) which is achieved by the Gumbel trick, and then somehow replace the one-hot vector draw with a continuous vector, which we get by doing the softmax($\log\alpha_j+g_j$) which we will get by using the first equation. I am sure I am missing something fundamental, but can't see what it is...

$\endgroup$

1 Answer 1

4
$\begingroup$

one way to sample is to apply argmax(softmax($\alpha_j$))

That is hardly "sampling", given that you deterministically pick the largest $\alpha_j$ every time. (also, you said that $\alpha$ is the unnormalized probability but that doesn't make sense seeing as log probabilities go into the softmax). The correct way to sample would be sample(softmax($x$)), where $x$ are the logits. Indeed, the goal of gumbel-softmax is not to replace the softmax operation as you've written it, but the sampling operation:

We can replace sample($p$) where $p$ are a vector of probabilities with argmax($\log p + g$) where $g$ is the gumbel noise. Of course, this is equivalent to argmax($x + g$) where $x$ are again the logits. To conclude, sample(softmax($x$)) and argmax($x+g)$ are equivalent procedures.

Then, if the goal was to have the full distribution over possible outcomes for $z_j$, we can use softmax transformation on top of the perturbation with Gumbel noise.

In fact you already have a distribution over all possible outcomes.

However, argmax($x+g$) is not differentiable wrt $x$, therefore to backpropagate we replace its gradient with the gradient of softmax($(x+g)\tau^{-1}$). When $\tau \rightarrow 0$, the expression approaches argmax.

Picking a reasonable, small values of $\tau$ will ensure a good estimate of the gradient while ensuring that the gradients are numerically well behaved.

and $\tau=1$ just makes the two equations identical

In fact, there is no special significance to $\tau = 1$. Rather, $\tau \rightarrow 0$ makes the gradient estimate unbiased but high in variance, where as larger values of $\tau$ add more bias to the gradient estimate but lower the variance.

$\endgroup$
7
  • $\begingroup$ Thanks for the comment. But I think this still doesn't answer the question. You are explaining the impact of $\tau$ on the gradient. If we consider the temperature inclusion as a hack to control the gradient variance, then your answer is fine. $\endgroup$ Commented Sep 14, 2018 at 18:53
  • $\begingroup$ @user3639557 You asked why temperature is needed: without temperature (with temperature defaulting to 0), you have the nondifferentiable function argmax, which is a problem for backpropagation. $\endgroup$
    – shimao
    Commented Sep 16, 2018 at 15:34
  • 2
    $\begingroup$ Simple normalization (via softmax) of Gumbel-perturbed values corresponds to the case that doesn't include $\tau$ (or $\tau$=1). The question is why don't we just use that. Based on your answer, it seems you are postulating $\tau$ is included to control the variance of gradient. And I am wondering if this was just a hack after all? Otherwise, there must be a mathematical derivation that starts from something and reaches this particular form of softmax with temperature. $\endgroup$ Commented Sep 17, 2018 at 12:10
  • 2
    $\begingroup$ @user3639557 $\tau = 1$ creates biased gradients. Choosing $\tau = 1$ is as arbitrary as choosing $\tau = \pi$. If having $\tau$ at all is a "hack" then setting $\tau$ to an arbitrary value of 1 is surely also a hack. The "default value" of $\tau$ is not 1, but 0, but we can't really use that because it makes the function non-differentiable. $\endgroup$
    – shimao
    Commented Sep 17, 2018 at 14:35
  • $\begingroup$ shimaos answer also does not satisfy me unfortunately. How does leaving out the temperature make the function non-differentiable it not clear to me $\endgroup$
    – CD86
    Commented May 1, 2021 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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