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A Bayes estimator is one which minimizes the Bayes risk. Specifically, if and only if

$$\delta_{\Lambda} = \arg\min \operatorname{BR}(\Lambda,\delta) := \int R(\theta, \delta) d \Lambda(\theta) = \int \left( \int L(\theta, \delta(x))dx \right) d \Lambda(\theta) $$ where $L(\theta, \delta(X))$ is a given loss function, $R(\theta, \delta)$ is the corresponding risk function, and $\operatorname{BR}(\Lambda, \delta)$ is defined to be the Bayes risk, is $\delta_{\Lambda}$ a Bayes estimator.

Theorem 4.1.1 on p. 228 of Casella, Lehmann, Theory of Point Estimation, as well as Theorem 7.1 on p. 116 of Keener, Theoretical Statistics: Topics for a Core Course, state the following sufficient condition for $\delta_{\Lambda}$ to be a Bayes estimator:

$$\forall x, \quad \delta_{\Lambda} = \arg\min \mathbb{E}\left[ L(\Theta, \delta(X))| X = x \right] $$

It is obvious why this is a sufficient condition: integrating first over $x$, we get by monotonicity of integrals an $\arg\min$ for $\mathbb{E}[L(\Theta, \delta(X))] = \int L(\Theta, \delta(x)) dx = R(\Theta, \delta)$. Then, integrating over $\theta$, we get an $\arg\min$ for the Bayes risk, again by monotonicity of integrals.

Question: Is the above condition necessary for $\delta_{\Lambda}$ to be a Bayes estimator?

Intuitively, I don't see any reason why it is necessary unless we have additional conditions guaranteeing uniqueness ($\mathcal{P}$-a.s.) of the Bayes estimator. Also, the proofs in both of the books I mentioned above only seem to show sufficiency, not necessity.

However, Wikipedia says that: "An estimator... is said to be a Bayes estimator if it minimizes the Bayes risk among all estimators. Equivalently, the estimator which minimizes the posterior expected loss ... for each x ." I.e. it seems to be implying that the two conditions are equivalent, i.e. that the latter condition is not only sufficient, but necessary. Is this actually true in general?

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First, if the condition$$\delta_{\Lambda} = \arg\min \mathbb{E}\left[ L(\Theta, \delta(X))| X = x \right]$$ holds almost surely in $x$, the same argument applies. Hence, the Bayes estimator is defined almost surely and hence can arbitrarily vary on an arbitrary set of measure zero.

Second, there are situations where there are multiple Bayes estimators. For instance, here is an exercise from my book:

2.40 Consider $\pi(\theta) = (1/3)(\mathscr{U}_{[0,1]}(\theta)+\mathscr{U}_{[2,3]}(\theta)+\mathscr{U}_{[4,5]}(\theta))$ and $f(x|\theta) = \theta e^{-\theta x}$. Show that, under the loss (2.5.4), for every $x$, there exist values of $k_1$ and $k_2$ such that the Bayes estimator is not unique.

where the loss (2.5.4) is \begin{eqnarray} \mathrm{L}_{k_1,k_2} (\theta ,d) = \begin{cases} k_2(\theta -d) & \text{if }\theta > d, \cr k_1(d-\theta ) &\text{otherwise.} \cr\end{cases} \end{eqnarray}

Third, when the Bayes risk $\min \operatorname{BR}(\Lambda,\delta)$ is infinite, any estimator is a Bayes estimator.

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  • $\begingroup$ Never mind, after re-reading my original question, I see how this addresses a remark made in the question. This doesn't seem to answer the main question though, although that may not have been your intent anyway. But it is helpful in that it clarifies a related point which I raised implicitly. $\endgroup$ – Chill2Macht Dec 27 '17 at 15:40

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