Consistent Estimator - Establishing Consistency

Establishing Consistency

The notion of asymptotic consistency is very close, almost synonymous to the notion of convergence in probability. As such, any theorem, lemma, or property which establishes convergence in probability may be used to prove the consistency. Many such tools exist:

  • In order to demonstrate consistency directly from the definition one can use the inequality
\Pr\!\big \leq \frac{\operatorname{E}\big}{\varepsilon},

the most common choice for function h being either the absolute value (in which case it is known as Markov inequality), or the quadratic function (respectively Chebychev's inequality).

  • Another useful result is the continuous mapping theorem: if Tn is consistent for θ and g(·) is a real-valued function continuous at point θ, then g(Tn) will be consistent for g(θ):
T_n\ \xrightarrow{p}\ \theta\ \quad\Rightarrow\quad g(T_n)\ \xrightarrow{p}\ g(\theta)
  • Slutsky’s theorem can be used to combine several different estimators, or an estimator with a non-random covergent sequence. If Tn →pα, and Sn →pβ, then
\begin{align} & T_n + S_n \ \xrightarrow{p}\ \alpha+\beta, \\ & T_n S_n \ \xrightarrow{p}\ \alpha \beta, \\ & T_n / S_n \ \xrightarrow{p}\ \alpha/\beta, \text{ provided that }\beta\neq0 \end{align}
  • If estimator Tn is given by an explicit formula, then most likely the formula will employ sums of random variables, and then the law of large numbers can be used: for a sequence {Xn} of random variables and under suitable conditions,
  • If estimator Tn is defined implicitly, for example as a value that maximizes certain objective function (see extremum estimator), then a more complicated argument involving stochastic equicontinuity has to be used.

Read more about this topic:  Consistent Estimator

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