I came across the following statement (marked as true) in multiple-choice section of an old exam:

The variance of a consistent estimator goes to zero with the growing sample size.

As far as I can tell, it can be translated as

Convergence in probability to a constant implies convergence in $L^2$.

Which is clearly false.

Is there a way to repair the statement? I mean maybe the professor forgot to mention some additional assumption typical for the context. (E.g. how convergence in probability and uniform integrability together imply $L^1$ convergence, but it seems to be irrelevant for the above statement.)


There are different concepts of consistency, knitted to different convergence concepts. You have almost-sure consistency, consistency in probability (defined by convergence in probability), $L^2$-consistency, and so on.


The claim is not true, using the definition by convergence in probability. Maybe it is true by some stronger definition. I will give a (somewhat artificial) counterexample: Let $$ \hat{\theta}_n = \begin{cases} \theta ~~\text{with probability $\frac{n-1}{n}$}\\ \theta+n ~~ \text{with probability $\frac1n$} \end{cases} $$ then we have convergence in probability, but also $E(\hat{\theta}_n-\theta)^2 = 0\cdot \frac{n-1}{n} + n^2\cdot \frac1n=n \rightarrow \infty$ disproving the claim.

  • $\begingroup$ But convergence in probability is exactly what I had in mind. Could you elaborate why is the cited claim true? $\endgroup$ – Leo Apr 5 '15 at 19:11
  • $\begingroup$ I corrected the answer $\endgroup$ – kjetil b halvorsen Apr 5 '15 at 19:26
  • $\begingroup$ A slighly less artificial example would be an estimator with a Cauchy density, centered on the mean and which width (scale) tends to zero as $n \to \infty$. This would be a consistent estimator, however its variance is infinite for all $n$. What it's true is that almost all estimators in practice converge also in $L^2$, so many people (including some teachers) believe that vanishing variance is necessary for consistency, but that's wrong, it's just sufficient (together with asymptotic unbiasedness). $\endgroup$ – leonbloy Apr 5 '15 at 19:46

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