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?


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:

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


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

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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    $\begingroup$ @G.JayKerns Unbiasedness is unnecessary for this. Consider $S_n = \sqrt{\frac{1}{n-1}\sum_{i=1}^n (X_i - \bar{X_n})^2}$. $S_n$ is a biased estimator of the standard deviation yet you can use the above argument to show that it's consistent. $\endgroup$ – user5594 Oct 29 '11 at 15:54
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    $\begingroup$ Looks good (+1); and I'll delete my earlier comments. $\endgroup$ – user1108 Oct 29 '11 at 17:19
  • $\begingroup$ @G.JayKerns Your comments were a necessary addition. We must always make sure we're aware of the assumptions we're working under. $\endgroup$ – user5594 Oct 29 '11 at 17:20
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    $\begingroup$ @MikeWierzbicki: I think we need to be very careful, in particular with what we mean by asymptotically unbiased. There are at least two different concepts that often receive this name and it's important to distinguish them. Note that it is not true in general that a consistent estimator is asymptotically unbiased in the sense that $\mathbb E T_n \to \theta$ even when the mean $\theta_n = \mathbb E T_n$ exists for all $n$. Many people call the convergence $\mathbb E T_n \to \theta$ unbiasedness in the limit or approximate unbiasedness...(cont.) $\endgroup$ – cardinal Oct 29 '11 at 18:56
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    $\begingroup$ Obviously, in order for a consistent estimator to be biased in the limit, the convergence in $L_2$ must fail since $\mathbb E(T_n - \theta)^2 = \mathrm{Var}(T_n) + (\theta_n - \theta)^2$ where $\theta_n = \mathbb E T_n$. $\endgroup$ – cardinal Oct 29 '11 at 18:57

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