on the first n observations, one can use the sample mean: Tn = (X1 + ... + Xn)/n. This defines a sequence of estimators, indexed by the sample size n. From the properties of the normal distribution, we know the sampling distribution of this statistic: Tn is itself normally distributed, with mean μ and variance σ2/n. Equivalently, has a standard normal distribution: as n tends to infinity, for any fixed . Therefore, the sequence Tn of sample means is consistent for the population mean μ (recalling that is the cumulative distribution of the normal distribution). 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 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 Chebyshev'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): Slutsky’s theorem can be used to combine several different estimators, or an estimator with a non-random convergent sequence. If Tn →dα, and Sn →pβ, then 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. Bias versus consistency Unbiased but not consistent An estimator can be unbiased but not consistent. For example, for an iid sample {x,..., x} one can use T(X) = x as the estimator of the mean E[X]. Note that here the sampling distribution of T is the same as the underlying distribution (for any n, as it ignores all points but the last), so E[T(X)] = E[X] and it is unbiased, but it does not converge to any value. However, if a sequence of estimators is unbiased and converges to a value, then it