consistent estimator

Given a set of samples $X_1,\mathrm{\dots },X_n$ from a given probability distribution $f$ with an unknown parameter $\theta \in \mathrm{\Theta }$, where $\mathrm{\Theta }$ is the parameter space that is a subset of ${ℝ}^m$. Let $U(=U(X_1,\mathrm{\dots },X_n))$ be an estimator of $\theta$. Allowing the sample size $n$ to vary, we get a sequence of estimators for $\theta$:

$U_1$ $=$ $U(X_1),$

$\mathrm{⋮}$

$U_n$ $=$ $U(X_1,\mathrm{\dots },X_n),$

$\mathrm{⋮}$

We say that the sequence of estimators $\{U_n\}$ consistent (or that $U$ is a consistent estimator of $\theta$), if $U_i$ converges in probability to $\theta$ for every $\theta \in \mathrm{\Theta }$. That is, for every $\epsilon >0$,

$$\underset{n\to \mathrm{\infty }}{lim}P(|h_n-\theta |\ge \epsilon )=0$$

for all $\theta \in \mathrm{\Theta }$.

Remark. Suppose $U$ is an estimator of $\theta$ such that the sequence $\{U_n\}$ is consistent. If ${\alpha }_n\to \alpha \in ℝ$ and ${\beta }_n\to \beta \in {ℝ}^m$ are two convergent sequences of constants with and , then the sequence $\{V_n\}$, defined by $V_n:={\alpha }_n{U}_n+{\beta }_n$, is consistent, $V$ is an estimator of $\alpha \theta +\beta$.

Proof.

First, observe that

$|V_n-(\alpha \theta +\beta )|$ $=$ $|{\alpha }_n{U}_n+{\beta }_n-\alpha \theta -\beta |$

$\le$ $|{\alpha }_n{U}_n-\alpha \theta |+|{\beta }_n-\beta |$

$=$ $|{\alpha }_n{U}_n-{\alpha }_n\theta +{\alpha }_n\theta -\alpha \theta |+|{\beta }_n-\beta |$

$\le$ $|{\alpha }_n{U}_n-{\alpha }_n\theta |+|{\alpha }_n\theta -\alpha \theta |+|{\beta }_n-\beta |$

$=$ $|{\alpha }_n||U_n-\theta |+|{\alpha }_n-\alpha ||\theta |+|{\beta }_n-\beta |.$

This implies

$P(|V_n-(\alpha \theta +\beta )|\ge \epsilon )$

$\le$ $P(|{\alpha }_n||U_n-\theta |+|{\alpha }_n-\alpha ||\theta |+|{\beta }_n-\beta |\ge \epsilon )$

$=$ $P(|U_n-\theta |\ge {\displaystyle \frac{\epsilon -|{\beta }_n-\beta |-|{\alpha }_n-\alpha ||\theta |}{|{\alpha }_n|}}).$

As $n\to \mathrm{\infty }$, $|{\beta }_n-\beta |\to 0$, $|{\alpha }_n-\alpha ||\theta |\to 0$, and $|{\alpha }_n|\to |\alpha |\ne 0$. So the last expression goes to $0$ as $n\to \mathrm{\infty }$. Therefore,

$$\underset{n\to \mathrm{\infty }}{lim}P(|V_n-(\alpha \theta +\beta )|\ge \epsilon )=0,$$

and thus $\{V_n\}$ is a consistent sequence of estimators of $\alpha \theta +\beta$. ∎

Title	consistent estimator
Canonical name	ConsistentEstimator
Date of creation	2013-03-22 15:26:34
Last modified on	2013-03-22 15:26:34
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	5
Author	CWoo (3771)
Entry type	Definition
Classification	msc 62F12
Defines	consistent sequence of estimators