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Given an unbiased estimator $\hat \theta_n$ of a parameter $\theta$, if the estimator has small variance (approaching $0$ as $n\to\infty$), it seems reasonable to expect that the estimator is consistent (i.e. that $\hat \theta_n$ converges in probability to the constant $\theta$).

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    $\begingroup$ yes. it follows from Markov's inequality $\endgroup$
    – Anthony Quas
    Commented Feb 6, 2012 at 23:37

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Markov's inequality says that for a non-negative random variable $X$ with expected value $\mu$, $$ \Pr(X>a) \le \frac \mu a. $$ (E.g. no more than $1/15$ of the population can have more than $15$ times the average income (assuming all incomes are non-negative), etc.)

So $$ \Pr(|\hat \theta_n - \theta| > \varepsilon) = \Pr((\hat\theta_n-\theta)^2>\varepsilon^2) \le \frac{\sigma_n^2}{\varepsilon^2} \to 0\text{ as }n\to\infty $$ where $\sigma_n^2=\operatorname{var}(\hat\theta_n)$.

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