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It's common short-hand in neural networks literature to refer to categorical cross-entropy loss as simply "cross-entropy." However, this terminology is ambiguous because different probability distributions have different cross-entropy loss functions.

So, in general, how does one move from an assumed probability distribution for the target variable to defining a cross-entropy loss for your network? What does the function require as inputs? (For example, the categorical cross-entropy function for one-hot targets requires a one-hot binary vector and a probability vector as inputs.)

A good answer will discuss the general principles involved, as well as worked examples for

  • categorical cross-entropy loss for one-hot targets
  • Gaussian-distributed target distribution and how how this reduces to usual MSE loss
  • A less common example such as a gamma distributed target, or a heavy-tailed target
  • Explain the relationship between minimizing cross entropy and maximizing log-likelihood.
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2 Answers 2

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Suppose that we are trying to infer the parametric distribution $p(y|\Theta(X))$, where $\Theta(X)$ is a vector output inverse link function with $[\theta_1,\theta_2,...,\theta_M]$.

We have a neural network at hand with some topology we decided. The number of outputs at the output layer matches the number of parameters we would like to infer (it may be less if we don't care about all the parameters, as we will see in the examples below).

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In the hidden layers we may use whatever activation function we like. What's crucial are the output activation functions for each parameter as they have to be compatible with the support of the parameters.

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Some example correspondence:

  • Linear activation: $\mu$, mean of Gaussian distribution
  • Logistic activation: $\mu$, mean of Bernoulli distribution
  • Softplus activation: $\sigma$, standard deviation of Gaussian distribution, shape parameters of Gamma distribution

Definition of cross entropy:

$$H(p,q) = -E_p[\log q(y)] = -\int p(y) \log q(y) dy$$

where $p$ is ideal truth, and $q$ is our model.

Empirical estimate:

$$H(p,q) \approx -\frac{1}{N}\sum_{i=1}^N \log q(y_i)$$

where $N$ is number of independent data points coming from $p$.

Version for conditional distribution:

$$H(p,q) \approx -\frac{1}{N}\sum_{i=1}^N \log q(y_i|\Theta(X_i))$$

Now suppose that the network output is $\Theta(W,X_i)$ for a given input vector $X_i$ and all network weights $W$, then the training procedure for expected cross entropy is:

$$W_{opt} = \arg \min_W -\frac{1}{N}\sum_{i=1}^N \log q(y_i|\Theta(W,X_i))$$

which is equivalent to Maximum Likelihood Estimation of the network parameters.

Some examples:

$$\mu = \theta_1 : \text{linear activation}$$ $$\sigma = \theta_2: \text{softplus activation*}$$ $$\text{loss} = -\frac{1}{N}\sum_{i=1}^N \log [\frac{1} {\theta_2(W,X_i)\sqrt{2\pi}}e^{-\frac{(y_i-\theta_1(W,X_i))^2}{2\theta_2(W,X_i)^2}}]$$

under homoscedasticity we don't need $\theta_2$ as it doesn't affect the optimization and the expression simplifies to (after we throw away irrelevant constants):

$$\text{loss} = \frac{1}{N}\sum_{i=1}^N (y_i-\theta_1(W,X_i))^2$$

$$\mu = \theta_1 : \text{logistic activation}$$ $$\text{loss} = -\frac{1}{N}\sum_{i=1}^N \log [\theta_1(W,X_i)^{y_i}(1-\theta_1(W,X_i))^{(1-y_i)}]$$ $$= -\frac{1}{N}\sum_{i=1}^N y_i\log [\theta_1(W,X_i)] + (1-y_i)\log [1-\theta_1(W,X_i)]$$

with $y_i \in \{0,1\}$.

  • Regression: Gamma response

$$\alpha \text{(shape)} = \theta_1 : \text{softplus activation*}$$ $$\beta \text{(rate)} = \theta_2: \text{softplus activation*}$$

$$\text{loss} = -\frac{1}{N}\sum_{i=1}^N \log [\frac{\theta_2(W,X_i)^{\theta_1(W,X_i)}}{\Gamma(\theta_1(W,X_i))} y_i^{\theta_1(W,X_i)-1}e^{-\theta_2(W,X_i)y_i}]$$

Some constraints cannot be handled directly by plain vanilla neural network toolboxes (but these days they seem to do very advanced tricks). This is one of those cases:

$$\mu_1 = \theta_1 : \text{logistic activation}$$ $$\mu_2 = \theta_2 : \text{logistic activation}$$ ... $$\mu_K = \theta_K : \text{logistic activation}$$

We have a constraint $\sum \theta_i = 1$. So we fix it before we plug them into the distribution:

$$\theta_i' = \frac{\theta_i}{\sum_{j=1}^K \theta_j}$$

$$\text{loss} = -\frac{1}{N}\sum_{i=1}^N \log [\Pi_{j=1}^K\theta_i'(W,X_i)^{y_{i,j}}]$$

Note that $y$ is a vector quantity in this case. Another approach is the Softmax.

*ReLU is unfortunately not a particularly good activation function for $(0,\infty)$ due to two reasons. First of all it has a dead derivative zone on the left quadrant which causes optimization algorithms to get trapped. Secondly at exactly 0 value, many distributions would go singular for the value of the parameter. For this reason it is usually common practice to add a small value $\epsilon$ to assist off-the shelf optimizers and for numerical stability.

As suggested by @Sycorax Softplus activation is a much better replacement as it doesn't have a dead derivative zone.

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Summary:

  1. Plug the network output to the parameters of the distribution and take the -log then minimize the network weights.
  2. This is equivalent to Maximum Likelihood Estimation of the parameters.
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  • $\begingroup$ What is the input to the network in this case? $\endgroup$
    – theQman
    Commented Feb 14, 2022 at 18:54
  • $\begingroup$ The answer you yourself linked (i.e this one.) @Sycorax is using X as a random variable, i think you have to review your knowledge; you downvoted my question because I said X was a random variable. $\endgroup$
    – user306937
    Commented Aug 17 at 15:29
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I'm going to answer for targets whose distribution family is an exponential family. This is typically justified as the minimum assumptive distribution. Let us denote the observed distributions to be $X_1, X_2, \dots$, the predictive distributions produced by the model to be $Y_1, Y_2, \dots$.

Every exponential family admits two important parametrizations: natural and expectation. Let the expectation parameters of the observed distributions be $\chi_i$, and the natural parameters of the predictive distributions be $\eta_i$.

  • How does one move from an assumed probability distribution for the target variable to defining a cross-entropy loss for your network?

The cross entropy of an exponential family is

$$H^\times(X; Y) = -\chi^\intercal \eta + g(\eta) - E_{x\sim X}\left(h(x)\right). $$

where $h$ is the carrier measure and $g$ the log-normalizer of the exponential family. We typically just want the gradient of the cross entropy with respect to the predictions, which is is just

$$\frac{dH^\times(X; Y)}{d\eta} = g'(\eta)-\chi. $$

$g'(\eta)$ is just the expectation parameters of the prediction.

  • What does the function require as inputs?

We require the pair $(\eta_i, \chi_i)$.

Let's go through your examples:

Categorical cross-entropy loss for one-hot targets. The one-hot vector (without the final element) are the expectation parameters. The natural parameters are log-odds (See Nielsen and Nock for a good reference to conversions). To optimize the cross entropy, you let the gradient be the difference of one-hot vectors.

Gaussian-distributed target distribution (with known variance). The cross entropy is simply a paraboloid, and therefore corresponds to MSE. Its gradient is linear, and is simply the difference of the observed and predicted means.

A less common example such as a gamma distributed target, or a heavy-tailed target. Same thing: the optimization is done as a difference of expectation parameters. For the gamma distribution, the expectation parameters are $(\frac{k}{\lambda}, \psi(k) - \log \lambda)$ where $k$ is the shape and $\lambda$ is the rate.

The relationship between minimizing cross entropy and maximizing log-likelihood is a good question. Minimizing log-likelihood is the special case where the target is a sample $x$ (or delta distribution) rather than a distribution $X$. I think for the optimization you do the same thing as above except you just use $\chi=x$. The log-likelihood calculation is just the log-density of the predictive distribution evaluated at $x$.

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