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I'm trying to understand the concept of consistency in point estimation. Could somebody give me an illustrative example of a (simply) consistent estimator that is not MSE consistent?

Just a recall of definitions and theorems (I hope I have them right):

  • $\hat{\theta}_m$ is simply consistent estimator of parameter $\theta$ if $\forall \epsilon > 0$: $$\lim_{m \to \infty} \mathbb{P}[| \hat{\theta}_m - \theta | > \epsilon] = 0$$

  • $\hat{\theta}_m$ is MSE consistent estimator of parameter $\theta$ if: $$\lim_{m \to \infty} \mathbb{E}[(\hat{\theta}_m - \theta)^2] = 0$$

  • MSE can be decomposed into variance and bias: $$ \mathbb{E}[(\hat{\theta} - \theta)^2] = \underbrace{\mathbb{E}[(\hat{\theta} - \mathbb{E}[\hat{\theta}])^2]}_{\text{Var}[\hat{\theta}]} + \underbrace{(\mathbb{E}[\hat{\theta}] - \theta)^2}_{\text{Bias}_\theta[\hat{\theta}]^2} $$

  • MSE consistency implies simple consistency from Chebyshev's Inequality: $$\mathbb{P}[|\hat{\theta} - \theta| \geq \epsilon] \leq \frac{\mathbb{E}[(\hat{\theta} - \theta)^2]}{\epsilon^2}$$

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    $\begingroup$ The question is really about the difference between convergence in probability and in $L^2$. One can construct a sequence of r.v.'s (estimators are r.v.'s) that converge in probability but not $L^1$. $\endgroup$ – Michael Dec 15 '17 at 21:54
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I think you are trying to say that convergence in probability does not imply $L^p$ convergence. see here

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  • $\begingroup$ Agree, but an example of this in the context of an estimation problem would be interesting. $\endgroup$ – alexpghayes Aug 17 at 20:30

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