I have the following estimator: $\hat{\sigma}^2_N = \frac{1}{h^2}\sum\limits_{i=1}^{N}x^2_i$, where $x_i \sim i.i.d. \; \mathcal{N}(\mu\frac{h}{N}, \sigma^2\frac{h}{N})$. We can show that $E[\hat{\sigma}^2_N] = \sigma^2 + \mu^2 \frac{h}{N}$ and that $var[\hat{\sigma}^2_N] = 2\sigma^4 \frac{1}{N} + 4\mu^2 \sigma^2 \frac{h}{N^2}$. I know that it is possible to prove that $\hat{\sigma}^2_N$ is a consistent estimator of $\sigma$, but I don't know how to proceed.
My attempt is: since $E[\hat{\sigma}^2_N] \to \sigma^2$ as $N \to \infty$, $\hat{\sigma}^2_N$ is asymptotically unbiased. Moreover, we have that $var[\hat{\sigma}^2_N] \to 0$ as $N \to \infty$, hence the estimator is consistent.
I know it is not very rigorous, but is this reasoning valid? If not, how is it possible to prove that the estimator is consistent?

  • $\begingroup$ Do the techniques in this post help you? stats.stackexchange.com/q/414265/354041 $\endgroup$
    – Ben
    Jan 27 at 14:35
  • $\begingroup$ Unfortunately, my estimator is biased, hence it is not possible to use the procedure mentioned in the post. Thanks anyway for the comment $\endgroup$ Jan 27 at 14:56
  • $\begingroup$ Hi, the machinery in the post is still useful despite your estimator being biased. Please consider how to use their approach. $\endgroup$
    – Ben
    Jan 27 at 15:06
  • $\begingroup$ From what I understand the only possible solution with the method mentioned in the post would be to prove that $E[(\hat{\sigma}^2_N - \sigma^2)^2]$ goes to zero as $N \to \infty$. However this leads to quite involved calculation. It would be nice to know if the simple argument in my question is valid. $\endgroup$ Jan 27 at 15:11
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    $\begingroup$ That looks great. That rigorously establishes the consistency and it shows why the bias and the variance approaching 0 (which you show) is enough. P.S. the expected squared error above is just the squared bias plus variance, which slightly simplifies the calculation. $\endgroup$
    – Ben
    Jan 27 at 16:15

1 Answer 1


The estimator $\hat{\sigma}^2_N$ is consistent if it converges in probability to $\sigma^2$. To prove consistency it is sufficient to show that $E[(\hat{\sigma}^2_N - \sigma^2)^2]$ goes to $0$ as $N \to \infty$. In order to show this, we exploit the fact that $E[(\hat{\sigma}^2_N - \sigma^2)^2] = E[(\hat{\sigma}^2_N)^2] -2\sigma^2E[\hat{\sigma}^2_N] + \sigma^4 = 2\sigma^4 \frac{1}{N} + 4\mu^2 \sigma^2 \frac{h}{N} + \mu^4 \frac{h^2}{N^2}$, which is equivalent to the sum of the variance and the bias. It is immediate to see that this quantity goes to $0$ as $N \to \infty$. Therefore we proved that the estimator is consistent.


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