Maximum a posteriori estimators as a limit of Bayes estimators


Maximum a posteriori and Bayes estimators are two common methods of point estimation in Bayesian statistics. It is commonly accepted that maximum a posteriori estimators are a limiting case of Bayes estimators with 0–1 loss. In this paper, we provide a counterexample which shows that in general this claim is false. We then correct the claim that by providing a level-set condition for posterior densities such that the result holds. Since both estimators are defined in terms of optimization problems, the tools of variational analysis find a natural application to Bayesian point estimation.

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Both authors express their gratitude to Roger J.-B. Wets for his guidance and supervision. This paper is dedicated to him, in honor of his 80th birthday.

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Corresponding author

Correspondence to Robert Bassett.

Additional information

Robert Bassett’s work was supported by the Programme Gaspard Monge pour l’Optimisation et la recherche opérationnelle (PGMO). Julio Deride was partially supported by NSF funding CMMI 1538263.

Proof of Lemma 1

Proof of Lemma 1

In this appendix we provide the proof of Lemma 1.


Let \(\theta \in \mathbb {R}^{n}\). To show hypo-convergence, we must show that for each sequence \(\theta ^{\nu } \rightarrow \theta \), \(\limsup _{\nu } f^{\nu }(\theta ^{\nu }) \le f(x)\) and that there exists a sequence \(\theta ^{\nu } \rightarrow \theta \) with \(\liminf _{\nu } f^{\nu }(\theta ^{\nu }) \ge f(\theta )\).

Fix \(\epsilon >0\). Since f is upper semi-continuous at \(\theta \), there is a \(\delta >0\) such that \(\left| \left| z-\theta \right| \right| < 2 \delta \) gives \(f(z) - f(\theta ) < \epsilon \).

Consider any sequence \(\theta ^{\nu } \rightarrow \theta \). We have that

$$\begin{aligned} f^{\nu }(\theta ^{\nu })- f(\theta )= & {} s_{n} \cdot \nu ^{n} \cdot \int _{\left| \left| \theta ^{\nu }-z\right| \right| > \frac{1}{\nu }} (f(z) - f(\theta )) dz \\= & {} s_{n} \cdot \nu ^{n} \cdot \int _{\left| \left| z\right| \right| < \frac{1}{\nu }} (f(z+\theta ^{\nu })-f(\theta )) \, dz. \end{aligned}$$

Choose \(\nu _{0} \in \mathbb {N}\) so that \(\left| \left| \theta -\theta ^{\nu }\right| \right| <\delta \) and \(\frac{1}{\nu } < \delta \) for each \(\nu > \nu _{0}\). Then for any \(\nu > \nu _{0}\),

$$\begin{aligned} s_{n} \cdot \nu ^n \cdot \int _{\left| \left| z\right| \right|< \frac{1}{\nu }} \left( f(z+\theta ^{\nu })-f(\theta ) \right) \, dz \le s_{n} \cdot \nu ^n \cdot \epsilon \cdot \int _{\left| \left| z\right| \right| < \frac{1}{\nu }} \, dz = \epsilon . \end{aligned}$$

Thus \(\limsup _{\nu } f^{\nu }(\theta ^{\nu }) \le f(\theta )\).

To establish the second part of the hypo-convergence definition, we focus our attention on constructing a sequence that satisfies the required inequality.

Consider any \(\eta \in \mathbb {N}\). Recall that \(f^{\nu }\) is an upper semi-continuous density. Let C be the set where f is continuous. Because C is dense, for each \(\nu \in \mathbb {N}\), there is a \(y^{\nu } \in C\) such that \(\left| \left| y^{\nu }-x\right| \right| < \frac{1}{\nu }\). Furthermore, \(y^{\nu } \in C\) means that there is a \(\delta (y^{\nu },\eta ) >0\) such that any \(z \in {\varTheta }\) which satisfies \(\left| \left| y^{\nu } - z\right| \right| < \delta (y^{\nu },\eta )\) also has

$$\begin{aligned} \left| f(y^{\nu }) - f(z) \right| < \frac{1}{\eta }. \end{aligned}$$

Here we use function notation for \(\delta \) to emphasize that \(\delta \) depends on both \(y^{\nu }\) and \(\eta \).

For each \(\eta \), define a sequence such that

$$\begin{aligned} z^{\nu , \eta } = {\left\{ \begin{array}{ll} 0 &{}\quad \text { when } \frac{1}{\nu } >\delta (y^{1}, \eta ) \\ y^{1} &{}\quad \text { when } \delta (y^{2}, \eta ) \le \frac{1}{\nu }< \delta (y^{1},\eta ) \\ y^{2} &{}\quad \text { when } \delta (y^{3}, \eta ) \le \frac{1}{\nu }< \min \{\delta (y^{2}, \eta ), \delta (y^{1},\eta ) \} \\ y^{3} &{}\quad \text { when } \delta (y^{4}, \eta ) \le \frac{1}{\nu } < \min _{i \le 4}{\delta (y^{i}, \eta )} \\ \vdots &{}\quad \vdots \end{array}\right. } \end{aligned}$$

Extracting a diagonal subsequence from the sequences generated according to this procedure gives a sequence \(\theta ^{\nu }\) such that \(\theta ^{\nu } \rightarrow \theta \) and \(\frac{1}{\nu } < \delta (\theta ^{\nu }, \nu )\). In particular, \(\left| f(\theta ^{\nu }) - f(z) \right| < \frac{1}{\nu }\) for z with \(\left| \left| \theta ^{\nu }-z\right| \right| < \frac{1}{\nu } \).

Hence, for any \(\epsilon > 0\), choosing \(\nu > \frac{2}{\epsilon }\) gives

$$\begin{aligned} \left| f^{\nu }(\theta ^{\nu }) - f(\theta ) \right|&\le \left| f^{\nu }(\theta ^{\nu }) - f(\theta ^{\nu }) \right| + \left| f(\theta ^{\nu }) -f(\theta ) \right| \\&\le \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon \end{aligned}$$

We conclude that \(\lim _{\nu } f^{\nu }(\theta ^{\nu }) = f(\theta )\), so the result is proven. \(\square \)

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Bassett, R., Deride, J. Maximum a posteriori estimators as a limit of Bayes estimators. Math. Program. 174, 129–144 (2019).

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Mathematics Subject Classification

  • 62C10
  • 62F10
  • 62F15
  • 65K10