Let $X_1, \dots, X_n \in \mathbb{R}^p$ be i.i.d. samples from the $p$-normal distribution $N(\theta, \tau^2 I)$. Suppose we are interested in estimating $\theta$ with known variance $\tau^2$. Take the loss to be squared error.

Define the estimator $\delta(X_1, \dots, X_n)$. Is there someway to prove that a suitably constructed estimator $\tilde{\delta}(\bar{X})$ will never be dominated by $\delta(X_1, \dots, X_n)$, for the sample mean $\bar{X} = \sum_{i=1}^n X_i / n$? Some facts that might be helpful:

  • The sample mean $\bar{X}$ is sufficient for $\theta$
  • The loss is convex

I ask because often texts will only focus on the $n=1$ case when talking about the e.g. James-Stein estimator, suggesting that some averaging has occurred beforehand.


1 Answer 1


By the Rao-Blackwell theorem, any estimator $\delta(X_1,\ldots,X_n)$ can be improved (under any strictly convex loss) by its conditional expectation $$\tilde\delta(\bar X_n) = \mathbb E[\delta(X_1,\ldots,X_n)|\bar X_n]$$ since it keeps the same expectation but enjoys a smaller risk. To quote from the historical paper by Girshick and Savage (Proceedings of the Second Berkeley Symposium on Mathematical Statistics and Probability, 1951):

In particular it is known by a theorem of Blackwell (1951) that if the loss is quadratic in any function of $\tau$, only functions of [a sufficient statistic] $u$ can be admissible, and if any estimate is minimax there is a minimax function of $u$.

  • $\begingroup$ You're the best! $\endgroup$
    – user551504
    Jan 30, 2022 at 16:50

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