# Proving estimator consistency

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?

• Do the techniques in this post help you? stats.stackexchange.com/q/414265/354041
– Ben
Jan 27 at 14:35
• Unfortunately, my estimator is biased, hence it is not possible to use the procedure mentioned in the post. Thanks anyway for the comment Jan 27 at 14:56
• Hi, the machinery in the post is still useful despite your estimator being biased. Please consider how to use their approach.
– Ben
Jan 27 at 15:06
• 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. Jan 27 at 15:11
• 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.
– Ben
Jan 27 at 16:15

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.