Is it enough to show that MSE = 0 as $n\rightarrow\infty$? I also read in my notes something about plim. How do I find plim and use it to show that the estimator is consistent?

**Answer**

**EDIT:** Fixed minor mistakes.

Here’s one way to do it:

An estimator of $\theta$ (let’s call it $T_n$) is consistent if it converges in probability to $\theta$. Using your notation

$\mathrm{plim}_{n\rightarrow\infty}T_n = \theta $.

Convergence in probability, mathematically, means

$\lim\limits_{n\rightarrow\infty} P(|T_n – \theta|\geq \epsilon)= 0$ for all $\epsilon>0$.

The easiest way to show convergence in probability/consistency is to invoke Chebyshev’s Inequality, which states:

$P((T_n – \theta)^2\geq \epsilon^2)\leq \frac{E(T_n – \theta)^2}{\epsilon^2}$.

Thus,

$P(|T_n – \theta|\geq \epsilon)=P((T_n – \theta)^2\geq \epsilon^2)\leq \frac{E(T_n – \theta)^2}{\epsilon^2}$.

And so you need to show that $E(T_n – \theta)^2$ goes to 0 as $n\rightarrow\infty$.

**EDIT 2**: The above requires that the estimator is at least asymptotically unbiased. As G. Jay Kerns points out, consider the estimator $T_n = \bar{X}_n+3$ (for estimating the mean $\mu$). $T_n$ is biased both for finite $n$ and asymptotically, and $\mathrm{Var}(T_n)=\mathrm{Var}(\bar{X}_n)\rightarrow 0$ as $n\rightarrow \infty$. However, $T_n$ is not a consistent estimator of $\mu$.

**EDIT 3**: See cardinal’s points in the comments below.

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