# Are the MSEs of all estimators that have an MSE, equivalent asymptotically?

Are the MSEs of all estimators that have an MSE, equivalent asymptotically or are some estimators terribly bad even when sample size approaches infinity?

I was thinking about estimators with MSEs dependent on sample size and those have MSE of zero as limit as n approaches infinity.

There is no restriction on the term estimator, I can define an estimator for an unknown parameter $$\theta$$, as $$\hat\theta=1$$ and call it my estimator. It's a bad estimator that uses no data, and doesn't make MSE shrink towards zero. The MSE of an estimator is actually $$MSE(\hat\theta)=E[(\hat\theta-\theta)^2]$$, where it has variance and bias terms. These terms not necessarily approach zero.

Counterexample time!

Let's take a random sample $$X_1,\dots,X_n \overset{iid}{\sim} N(\mu,1)$$. As our first estimate of $$\mu$$, let's use $$\hat{\mu}_1 = \bar{X}$$. As an alternative estimator of $$\mu$$, lets use $$\hat{\mu}_2 = 2$$. Despite this looking silly, this is an admissible estimator in the sense that it has lower $$MSE$$ than $$\hat{\mu}_1$$ when the truth is that $$\mu=2$$.

We can decompose $$MSE$$ into the bias and variance: $$MSE = bias^2 + variance$$. Let's look at the $$MSE$$ of each estimator.

$$MSE_1 = 0^2 + \frac{1}{n}$$. As $$n \rightarrow 0$$, $$MSE_1 \rightarrow 0$$.

$$MSE_2 = (\mu-2)^2 + 0$$. As $$n \rightarrow 0$$, $$MSE_2 \rightarrow \mu^2 - 4\mu + 4$$.

No, these are not asymptotically equivalent.