1
$\begingroup$

It is well accepted that $$S^2 = \frac{1}{n}\sum_{i = 1}^n (W_i - \bar{W})^2$$ is an estimator of the variance, $\sigma^2$, of a bounded random variable $W$.

However, because the estimator is biased (i.e. it's expectation is $\frac{(n-1)}{n}\sigma^2$).

We also have the unbiased estimator $$\hat{\sigma}^2 = \frac{1}{n-1}\sum_{i = 1}^n (W_i - \bar{W})^2$$

How would I go about showing that $S^2$ is in fact a consistent estimator of $\sigma^2$?

I know that by Chebychev's inequality for any $\varepsilon>0$, we get the following: \begin{equation} \mathbb{P}\left(|S^2 - \frac{n-1}{n}\sigma^2|\geq \varepsilon\right) \leq \frac{\left(\frac{n-1}{n}\right)^2\text{Var}(\hat{\sigma}^2)}{\varepsilon^2}. \end{equation}

However, taking the limit only allows me to show that $S^2$ is a consistent estimator of $\frac{n-1}{n}\sigma^2$, not $\sigma^2$ itself.

It seems that this should be pretty straight forward, but somehow I am getting hung up.

$\endgroup$
  • 1
    $\begingroup$ All you need to do is show that the expectation converges to the actual value as $n \to \infty$, which can be done without Chebychev; what does $[(n-1)/n]\sigma^2$ converge to as $n \to \infty$? $\endgroup$ – jbowman Mar 24 '18 at 19:29
  • 1
    $\begingroup$ If some estimator sequence $X_n$ is consistent for some parameter $\theta$, and $a_n$ is a deterministic sequence converging to 1, then $a_n X_n$ is also consistent for $\theta$. Can you prove that? $\endgroup$ – kjetil b halvorsen Mar 24 '18 at 19:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.