I am reading a proof of the consistency of sample variance estimator.


On page two, it argues that, since $A=\frac{1}{n}\sum_{i=1}^nX_i^2$ is an unbiased estimator of $\Bbb E(X^2)$ and since $Var(A)\to 0$ as $n\to\infty$, the estimator converges in probability to $\Bbb E(X^2)$, or $A\overset{\Bbb P}{\to}\Bbb E(X^2)$.

I am new to these convergence concepts, so how does this follow? Is it true that $$Unbiased + Vanishing\text{ }Variance \implies Convergence\text{ }in\text{ } Probability$$ holds in general?


1 Answer 1


Unbiasedness + vanishing variance implies convergence in mean-square; convergence in mean-square implies convergence in probability.

If $A$ is an unbiased estimator of some parameter $\alpha$ then (by definition) we have $\mathbb{E}(A) = \alpha$. Then, using Chebyshev's inequality we have:

$$\mathbb{P}(|A - \alpha| > \epsilon) = \mathbb{P}(|A - \mathbb{E}(A)| > \epsilon) \leqslant \frac{\mathbb{V}(A)}{\epsilon^2}.$$

If $\lim_{n \rightarrow \infty} \mathbb{V}(A) = 0$ then $\lim_{n \rightarrow \infty} \mathbb{P}(|A - \alpha| > \epsilon) = 0$ for all $\epsilon > 0$, which means $A \xrightarrow{P} \alpha$.

In your particular question, you have $\alpha = \mathbb{E}(X^2)$, so the result you give is correct. It is a specific case of a more general rule that unbiasedness and vanishing variance implies convergence in mean-square, which then implies convergence in probability.

  • 1
    $\begingroup$ Thank you so much for your input, @Ben. This is really helpful. But did you mean to say $\Bbb E(A)$ rather than $\Bbb E(a)$ in the equation? $\endgroup$ Jan 29, 2018 at 4:32
  • $\begingroup$ Yes I did - edited. Thanks for pointing that out. $\endgroup$
    – Ben
    Jan 29, 2018 at 4:39

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