I'm attending a course in mathematical statistics and it seems the lecturer tacitly assumes that given estimators $T_1,T_2 : \Omega \to \Lambda$ of a parameter $g : \Theta \to \Lambda$, a loss function $L : \Lambda^2 \to [0\,..\infty)$ and its associated risk $R$ if $T_1$ are equivalent $T_2$ in terms of risk, then they are equal $P_\theta$ a.s. for all $\theta \in \Theta$.

By "equivalent" I mean $T_1 \cong T_2$ where:

$$T_1 \cong T_2 \Leftrightarrow R(T_1,\_) = R(T_2,\_) \Leftrightarrow \forall \theta \in \Theta : R(T_1,\theta) =R(T_2,\theta)$$

This does not seem to be true generally though. E.g. if we consider the square loss $L(x,y) = (x-y)^2$ and we let $\Theta = \{0\}$, $\Lambda = \mathbb R$, $P:=P_0$, $g(x) = x$, then surely we can find $\Omega$ with appropriate $\sigma$-algebra and $T_1,T_2$ such that $T_1$ and $T_2$ are not equal $P$-a.s. but that have the same $P$-distribution. Then: $$R(T_1,\theta)= \int f_\theta\,d(P\circ T^{-1}_1) = \int f_\theta\,d(P\circ T^{-1}_2) =R(T_2,\theta)$$ for all $\theta \in \Theta$, where $f_\theta(x) = (x-\theta)^2$.

My question is:

How contrived is this example? Are there usually good reasons why $T_1 \cong T_2$ iff $T_1$ and $T_2$ are equal $P_\theta$-a.s. $\theta\in \Theta$ or is this a phenomenon that usually persists?

If yes I also wonder about properties shared by estimators $T_1,T_2$ s.t. $T_1\cong T_2$.


1 Answer 1


The statement as reported is wrong: A most standard example is provided by the James-Stein estimator: given $X\sim\mathcal{N}_p(\theta,I_p)$ $(p>2)$, assuming $\theta$ is estimated under the square error loss,$$L(\theta,\delta)=||\theta-\delta||^2$$the estimators$$\delta_0(x)=x\qquad\text{and}\qquad \delta_{2(p-2)}(x)=\left(1-\frac{2(p-2)}{||x||^2}\right) x$$have exactly the same risk: $$\mathbb{E}_\theta[||\theta-\delta_0(X)||^2]=\mathbb{E}_\theta[||\theta-\delta_{2(p-2)}(X)||^2]=p$$(The proof goes by Stein's technique of the unbiased estimator of the risk and can be found [e.g.] in my Bayesian Choice book.) Actually, when two estimators share the same risk functions, they are inadmissible under strictly convex losses in that the average of these two estimators is improving the risk: $$R(\theta,\{\delta_1+\delta_2\}/2) < R(\theta,\delta_1)=R(\theta,\delta_2)$$which is what is happening with the James-Stein estimators: $$R(\theta,\{\delta_0+\delta_{2(p-2)}\}/2) < R(\theta,\delta_0)=R(\theta,\delta_{2(p-2)})=p$$with$$\frac{\delta_0+\delta_{2(p-2)}}{2}=\delta_{(p-2)}$$

I presume the confusion stemmed from the concept of completeness, where having a function with constant expectation under all values of the parameter implies that this function is constant. But, for loss functions, the concept does not apply since the loss depends both on the observation (that is complete in a Normal model) and on the parameter.


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