3
$\begingroup$

Most of the literature on scoring rules that I know of deals with discrete measurable spaces, but in this paper by Gneiting and Raferty a very general definition of a scoring rule is given. I don't see in the paper, however, any concrete examples of (strictly) proper scoring rules for general measurable spaces. I'm hoping someone can provide such an example.

Let me recall the setup (with some simplifications and minor modifications to suit my purposes). Let $(\Omega, \mathcal A)$ be a measurable space and let $\mathcal P$ be the set of probability measure on this space (Gneiting and Rafferty actually allow $\mathcal P$ to be any convex set of probability measures). A scoring rule $S: \mathcal P \times \Omega \to [-\infty, \infty)$ is a function that is measurable in its second argument. Write $$S(P,Q) = \int S(P, \omega)Q(d\omega).$$ Say that $S$ is (strictly) proper if $$S(P,P) \geq S(Q,P)$$ holds for all $P,Q \in \mathcal P$ (with equality iff $P=Q$).

A scoring rule $S$ is regular if $S(P,P) > -\infty$ for all $P \in \mathcal P$. Gneiting and Rafferty prove the following representation theorem.

A regular scoring rule is (strictly) proper if and only if there exists a (strictly) convex, real-valued function $G$ on $\mathcal P$ such that $$S(P, \omega) = G(P) - \int G^*(P, \omega')P(d\omega') + G^*(P,\omega)$$ for all $P \in \mathcal P$ and $\omega \in \Omega$, where the function $G^*(P, \cdot): \Omega \to [-\infty, \infty]$ (the subtangent of $G$ at $P$) is measurable and satisfies $$G(Q) \geq G(P) + \int G^*(P, \omega')(Q-P)(d\omega')$$ for all $Q \in \mathcal P$.

I'm looking for an example of a strictly proper scoring rule that illustrates the theorem and doesn't depend on $\Omega$ being countable. Since the strict properness fo $S$ implies that $S$ is regular, I suppose this amounts to choosing a strictly convex function $G$ that has a subtangent at every $P$. Is there an obvious or natural choice of such a function?

$\endgroup$

3 Answers 3

3
$\begingroup$

Turns out that Gneiting and Raftery give an example in Section 4.2 of the continuous ranked probability score (CRPS), which is strictly proper for $\mathcal{P}$ equal to the Borel probability measures on $\mathbb{R}$ with finite first moment. Writing a forecast as a cumulative distribution function $F$, the score is

$$ CRPS(F, x) = -\int_{-\infty}^{\infty} \Big(F(y) - \mathbf{1}[y \geq x] \Big)^2 dy . $$

It "corresponds to the integral of the Brier scores for the associated probability forecasts at all real-valued thresholds." Apparently this is a popular scoring rule in statistics.

$\endgroup$
4
  • $\begingroup$ It looks like they say it's strictly proper only for the subclass of $\mathcal P$ consisting of those probability measures with finite first moment. $\endgroup$
    – aduh
    Commented Oct 18, 2020 at 6:37
  • 2
    $\begingroup$ @aduh ah, thanks for catching that! I assume it's because the expected score becomes $-\infty$ for many (any?) forecasts, making it weakly proper as they say.... But at least on $[0,1]$ it will be unrestricted. $\endgroup$
    – usul
    Commented Oct 18, 2020 at 13:59
  • $\begingroup$ Is there an easy way to see that CRPS is indeed strictly proper on probabilities over $[0,1]$? I'm having trouble showing it. $\endgroup$
    – aduh
    Commented Oct 19, 2020 at 0:09
  • $\begingroup$ @aduh, Hmm, I don't know. It must have to do with $F$ being required to be weakly increasing, implying somehow that it's not possible to modify $F$ on a set of measure zero. $\endgroup$
    – usul
    Commented Oct 19, 2020 at 15:38
2
$\begingroup$

Well, it might be important to limit $\mathcal{P}$ here. If we consider the space $\Omega = \mathbb{R}$ with Lebesgue measure, we might take $\mathcal{P}$ to be the set of distributions with a continuous density function (no point masses). Then I believe the log scoring rule works: $S(f, \omega) = \log f(\omega)$ where $f$ is a density function. $G$ should be the negative of differential entropy. If we in addition limit $\mathcal{P}$ to be square integrable, quadratic score should work: something like $S(f, \omega) = 2 f(\omega) - \|f\|_2^2$. Here $G(f) = \|f\|_2^2$ or something similar.

I'm much more used to the finite or countable setting - sorry I don't have more general examples! Only one paper[1] comes to my mind in this setting, and it makes the restriction above. Maybe others will know of more references.

I can mention some nontrivial weakly proper scoring rules. Let's take $\Omega$ to be the interval $[0,1]$. We know that a scoring rule for the mean of a distribution is $s: [0,1] \times [0,1] \to \mathbb{R}$ defined by $s(\mu, \omega) = -(\mu - \omega)^2$. We can lift this to a weakly proper scoring rule for distributions, $S(p, \omega) = -(\mu_p - \omega)^2$ where $\mu_p$ is the mean of $p$. Here I guess $G(p) = \mu_p^2$, ish. This scoring rule is also weakly proper on $\Omega = \mathbb{R}$ as long as $\mathcal{P}$ only contains distributions with finite mean. You can also lift scoring rules for other "properties" of distributions, e.g. obtaining $S(p,\omega) = -|m_p - \omega|$ where $m_p$ is a median of $p$.

[1] Proper Local Scoring Rules. Parry, Dawid, and Lauritzen. Annals of Statistics, 2012. https://arxiv.org/abs/1101.5011

$\endgroup$
3
  • $\begingroup$ Thanks for this. You're right that restricting $\mathcal P$ can be important. I suppose I should clarify, then, that my question is about the case where $\mathcal P$ is the set of all probability measures on some space. To make this more concrete, let $(\Omega, \mathcal A)$ be the unit interval with its Borel algebra. What's an explicit example of a strictly proper scoring rule on the set $\mathcal P$ of all probability measures on $(\Omega, \mathcal A)$? $\endgroup$
    – aduh
    Commented Oct 17, 2020 at 2:55
  • 1
    $\begingroup$ @aduh, yeah, it's a great question. I can give the above nontrivial weakly proper rules, but I wonder if there might not exist a strictly proper one. $\endgroup$
    – usul
    Commented Oct 17, 2020 at 10:46
  • $\begingroup$ Never mind - see my other answer; G&R give an example of the continuous ranked probability score (CRPS). $\endgroup$
    – usul
    Commented Oct 18, 2020 at 3:37
1
$\begingroup$

Suppose $\scr F$ is a countably generated $\sigma$-algebra on any $\Omega$ (in particular, the Borel $\sigma$-algebra on any Polish space will work) and $\scr P$ is all countably additive probabilities on $\scr F$. Suppose $A_1,A_2,...$ generate $\scr F$. Let $S_n(P,\omega) = -(1_{A_n}(\omega)-P(A_n))^2$ be a Brier accuracy score with respect to $A_n$ and let $S=\sum_n 2^{-n} S_n$. Then it's elementary that $S_n$ has the following strict propriety condition: $S_n(P,P) \ge S_n(Q,P)$ with equality iff $P(A_n)=Q(A_n)$. Hence, $S$ is proper, and $S(P,P) > S(Q,P)$ if $P$ and $Q$ differ on at least one of the $A_n$. But since the $A_n$ generate $\scr F$, if $P$ and $Q$ differ anywhere, they differ on at least one of the $A_n$, so $S$ is strictly proper. Moreover, it's regular, and in fact it's uniformly bounded, and $P\mapsto S(P,\omega)$ is continuous for each fixed $\omega$ in the $\ell^\infty(\scr F)$ metric. Finding $G$ should then be an easy exercise.

$\endgroup$

You must log in to answer this question.

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