Suppose that $\mu$ is an unknown $k$-vector, which one seeks to guess. You observe $n$ i.i.d. normal variates $x_i \sim \mathcal{N}\left(\mu,\Sigma\right)$, say stacked in the matrix $X$, then produce estimate $\hat{\mu}\left(X\right)$. Cramer-Rao theorem gives a lower bound on $Var\left(\hat{\mu}\left(X\right)\right)$ when $\hat{\mu}$ is unbiased.

However, the variance (and MSE) for a biased $\hat{\mu}$ might be lower than an unbiased one. You might be tempted to choose a biased estimator. My question is whether there is a "free lunch" type argument against a biased estimator. Suppose that one will draw $\mu$ uniformly from all $k$-vectors with unit norm, then generate the $X$, and compute the estimate. Is there a lower bound for $$ E_{\mu}\left[E_{X} \left[\left(\hat{\mu}\left(X\right) - \mu\right)^2\right] \right] $$ that applies to all estimators?

  • 1
    $\begingroup$ I'm not sure to get the question right, but the Cramer-Rao bound is still valide for a biased estimator. For a one dimensionnal estimator, you have $E((\hat\mu(X) - \mu)^2) \geq (1 + bias^2) / J$ where J is the Fisher information such that $E_\mu[E((\hat\mu(X) - \mu)^2)] \geq E_\mu[(1 + bias^2) / J]$. $\endgroup$
    – Pohoua
    Jan 21, 2021 at 8:38

1 Answer 1


If one picks a probability distribution $\pi$ on $\mu$ and minimises the integrated risk, this is equivalent to running a Bayesian decision analysis with this prior distribution. In this case the optimal solution is the posterior mean $$\hat\mu(x)=\mathbb E^\pi[\mu|x]$$ and the estimator is always biased. There is thus gains to be found by reaching outside unbiasedness.

  • $\begingroup$ I'm ashamed to say I am not familiar with Bayesian decision analysis, but this looks to be just what I want. Am reading some references now. Thanks, $\endgroup$
    – shabbychef
    Jan 22, 2021 at 6:38
  • $\begingroup$ Is Berger's Statistical Decision Theory still the right text for an intro to the topic? $\endgroup$
    – shabbychef
    Jan 27, 2021 at 6:58
  • $\begingroup$ Yes, Jim's book is completely appropriate for such an introduction. $\endgroup$
    – Xi'an
    Jan 27, 2021 at 7:18

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