The cross entropy for a binary class is

$$p \ln \hat p + (1-p) \ln (1-\hat p),$$

where $p$ is the probability of the label A and $\hat p$ is the predicted probability of label A. Since we have binary classes, $p$ is either 1 or 0. However, the predicted probabilities can be any value between $[0,1]$.

Probability

For a very simple case, $\hat p$ might be a sigmoid like expression with exponential in it,

$$p \sim \frac{1}{1 + \exp(-x)},$$

where $x$ is some kind of input or intermediate input.

The problem is, exponentials may blow up if $p\to 0$. To deal with it, we can factor a exponential out,

$$p = e^{a} ( e^{-a}p ).$$

We choose $a$ carefully so that this eliminates the factors that leads to $p\to 0$.

For example, if we have a Gaussian like probability

$$\ln p \sim \ln \left(\sum_i \exp \left( -x_i^2 \right)\right),$$

we know that $x$ can be as large as $1e^6$. One such value in the training data will destroy our loss function as we will have to calculate the exponential then $\ln$. Though we can calculate this manually and it is fine, our computer will treat $e^{-1e6}$ as 0 and $\ln 0$ leads to negative infinity. We do not get the correct answer. To deal with it, we will factor out the exponentials, and rewrite the expression as

\begin{align} p \sim & a + \ln\left(e^{-a} \sum_i \exp \left( -x_i^2 \right)\right) \\ \sim & a + \ln \left(\sum_i \exp \left( -x_i^2 - a \right)\right) \end{align}

where $a$ can be $-1e6$. In this case, we do not hit the overflow problem in the computer.

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cards/machine-learning/neural-networks/log-sum-exp-trick Links to:

LM (2021). 'The log-sum-exp Trick', Datumorphism, 07 April. Available at: https://datumorphism.leima.is/cards/machine-learning/neural-networks/log-sum-exp-trick/.