I usually see a discussion of the following loss functions in the context of the following types of problems:

  • Cross entropy loss (KL divergence) for classification problems
  • MSE for regression problems

However, my understanding (see here) is that doing MLE estimation is equivalent to optimizing the negative log likelihood (NLL) which is equivalent to optimizing KL and thus the cross entropy.


  • Why isn't KL or CE used also for regression problems?
  • What's the relationship between CE and MSE for regresion? Are they one and the same loss under some circumstances?
  • If different, what's the benefit of using MSE for regression instead?

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2 Answers 2


In a regression problem you have pairs $(x_i, y_i)$. And some true model $q$ that characterizes $q(y|x)$. Let's say you assume that your density

$$f_\theta(y|x)= \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left\{-\frac{1}{2\sigma^2}(y_i-\mu_\theta(x_i))^2\right\}$$

and you fix $\sigma^2$ to some value

The mean $\mu(x_i)$ is then e.g. modelled via a a neural network (or any other model)

Writing the empirical approximation to the cross entropy you get:

$$\sum_{i = 1}^n-\log\left( \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left\{-\frac{1}{2\sigma^2}(y_i-\mu_\theta(x_i))^2\right\} \right)$$

$$=\sum_{i = 1}^n-\log\left( \frac{1}{\sqrt{2\pi\sigma^2}}\right) +\frac{1}{2\sigma^2}(y_i-\mu_\theta(x_i))^2$$

If we e.g. set $\sigma^2 = 1$ (i.e. assume we know the variance; we could also model the variance than our neural network had two ouputs, i.e. one for the mean and one for the variance) we get:

$$=\sum_{i = 1}^n-\log\left( \frac{1}{\sqrt{2\pi}}\right) +\frac{1}{2}(y_i-\mu_\theta(x_i))^2$$

Minimizing this is equivalent to the minimization of the $L2$ loss.

So we have seen that minimizing CE with the assumption of normality is equivalent to the minimization of the $L2$ loss

  • 1
    $\begingroup$ Then what's the point of the Gaussian assumption? In linear regression, we make a Gaussian assumption to do parameter inference, which is less important in neural networks. $\endgroup$
    – Dave
    Commented Jul 14, 2020 at 19:19
  • 1
    $\begingroup$ That means approximating the expectation with respect to the true data distribution $p^*(x)$ with a Monte Carlo estimate using a set of samples $S$: $ \int dx \, p^*(x) f(x) \approx \frac{1}{\vert S \vert}\sum_{i \in S} f(x_i), x_i \sim p^*(x)$. Since typically we are unable to evaluate the integral analytically or in reasonable (polynomial) time for most problems. $\endgroup$
    – Eweler
    Commented Jul 15, 2020 at 4:18
  • 1
    $\begingroup$ @Josh as Eweler already pointed out: we imagine that $H(q, f_\theta)$ has some fixed but unknown value (for a fixed $\theta$) by the term $1/n\sum_{i=1}^n-\log(f_\theta(x_i)) \approx \int -\log(f_\theta(x)) q(x)dx$ we approximate this quantity empirically and minimize this quantity as a proxy because we have no way to minimize the term we actually care about, i.e. the underlying real cross-entropy. This is generally referred to as empirical risk minimization (the risk is the theoretical value). Note that we usually drop the $1/n$ because is irrelevant to the optimization. $\endgroup$
    – Sebastian
    Commented Jul 15, 2020 at 6:13
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    $\begingroup$ Yes exactly, they are minimizing the same loss (or to be more precise their solution is the same (CE loss has this additional constant $n\log(\frac{1}{\sqrt{2\pi}})$ that does not matter for the optimization No without a normal distribution this does not hold. If you substitute normal distribution with Laplace distribution this will result in the minimization of the $L1$ loss $\endgroup$
    – Sebastian
    Commented Jul 15, 2020 at 13:43
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    $\begingroup$ Exactly, you got it:) $\endgroup$
    – Sebastian
    Commented Jul 15, 2020 at 16:44

The mean squared error is the cross-entropy between the data distribution $p^*(x)$ and your Gaussian model distribution $p_{\theta}$. Note that the standard MLE procedure is:

$$ \begin{align} \max_{\theta} E_{x \sim p^*}[\log p_{\theta}(x)] &= \min_{\theta} \left(- E_{x \sim p^*}[\log p_{\theta}(x)]\right)\\ &= \min_{\theta} H(p^* \Vert p_{\theta}) \\ &\approx \min_{\theta} \sum_i \frac{1}{2} \left(\Vert x_i - \theta_1\Vert^2/\theta_2^2 - \log 2 \pi \theta_2^2\right) \end{align} $$

Where $H(p^* \Vert p_{\theta})$ denotes the CE and we use a Monte Carlo approximation to the expectation. And as you stated, this is equivalent to minimizing the KL divergence between the data distribution and your model distribution. Commonly the variance $\theta_2$ is fixed and drops out of the objective.

Some people get confused because certain textbooks introduce the cross-entropy in terms of the Bernoulli/Categorical distribution (almost all machine learning libraries are guilty of this!), but it applies more generally than the discrete setting.

  • $\begingroup$ Thanks Eweler. When you said "this is equivalent to minimizing the KL divergence between the data distribution and your model." would you mind elaborating? Sorry not sure I follow that just from the expression that you wrote. $\endgroup$
    – Josh
    Commented Jul 15, 2020 at 3:04
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    $\begingroup$ Sorry, I was a bit loose there. What I meant was that minimization of the cross-entropy is equivalent to minimizing the KL divergence between the true data distribution $p^*$ and your model distribution $p_{\theta}$. Typically $p_{\theta}$ belongs to some restricted class of functions, so this will be nonzero. To show this, note that the cross entropy is: $E_{x \sim p^*}[\log p_{\theta}(x)] = E_{x \sim p^*}[ \log p_{\theta}(x) - \log p^*(x)] + E_{x \sim p^*}[\log p^*(x)] $ $\endgroup$
    – Eweler
    Commented Jul 15, 2020 at 4:22
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    $\begingroup$ Thanks! So to circle back to some of the questions in the OP. Is optimizing the CE loss equivalent to optimizing the MSE loss in regression? (or only under a normality assumptions?). If they are not equivalent in the general case, what's a good way to think about their relationship? $\endgroup$
    – Josh
    Commented Jul 15, 2020 at 14:07

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