1
$\begingroup$

The minimax equation for generative adversarial networks

$$\min_G \max_D V(D,G) = \mathbb{E}_{\boldsymbol{x}\sim p_{data}(\boldsymbol{x})}[\log D(\boldsymbol{x})] + \mathbb{E}_{\boldsymbol{z}\sim p_{\boldsymbol{z}}(\boldsymbol{z})}[\log(1 - D(G(\boldsymbol{z}))] $$

Why do we use logarithms instead of just

$$\min_G \max_D V(D,G) = \mathbb{E}_{\boldsymbol{x}\sim p_{data}(\boldsymbol{x})}[ D(\boldsymbol{x})] + \mathbb{E}_{\boldsymbol{z}\sim p_{\boldsymbol{z}}(\boldsymbol{z})}[(1 - D(G(\boldsymbol{z}))] $$

$\endgroup$

1 Answer 1

1
$\begingroup$

It's common in machine learning to do this log-trick, i.e. rather than optimizing $f(\mathbf{x})$, you optimize $\log f(\mathbf{x})$.

There are 3 main reasons why we (can) do this.

  1. When your objective function is the product of multiple probabilities (or, more generally, small numbers), i.e. $f(\mathbf{x}) = \prod_{i=1}^N p(x_i)$, then $\log f(\mathbf{x}) = \log \left( \prod_{i=1}^N p(x_i) \right) = \sum_{i=1}^N \log p(x_i)$ (see this), which is more numerical stable because we got rid of multiplication of possibly very small numbers, which can lead to underflow.

  2. For the same reason, we can also compute derivatives more easily, as we can simply compute the derivatives of each component (because derivatives are linear)

  3. The logarithm is monotonically increasing, so $\log f(\mathbf{x})$ has the same optima as $f(\mathbf{x})$ (simple proof here).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .