How to show that an estimator is consistent? 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?
 A: 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.
