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I was practicing an exam, and came along this question:

A consistent estimator converges in probability to the true parameter value. Therefore, the variance of such an estimator converges to zero with increasing sample size. This means that the asymptotic variance of a consistent estimator is zero.

I think I disagree with "the variance of such an estimator converges to zero with increasing sample size". The variance does not have to exist (i.e. infinity) in order for the estimator to converge in probability to the true value, Kolmogorov's SLLN tells us. Could anyone tell me if this reasoning is correct?

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Your reasoning is correct. Example: If $X_1, X_2, \ldots, X_n$ are iid from the density $$ f_\theta(x) = {c\over 1 + |x-\theta|^3},$$ where $c$ is a normalizing constant, then by the SLLN the sample mean converges almost surely to $E_\theta(X)=\theta$, hence $\bar X$ is a consistent estimator for $\theta$. However, the distribution lacks a second moment so $\operatorname{Var}(\bar X)=\infty$ for every $n$.

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