Consistent estimator · Lablypedia

In practice one constructs an estimator as a function of an available sample of size n, and then imagines being able to keep collecting data and expanding the sample ad infinitum. In this way one would obtain a sequence of estimates indexed by n, and consistency is a property of what occurs as the sample size “grows to infinity”. If the sequence of estimates can be mathematically shown to converge in probability to the true value θ₀, it is called a consistent estimator; otherwise the estimator is said to be inconsistent.

Consistency as defined here is sometimes referred to as weak consistency. When we replace convergence in probability with almost sure convergence, then the estimator is said to be strongly consistent. Consistency is related to bias; see bias versus consistency.

Definition

Formally speaking, an estimator T_n of parameter θ is said to be weakly consistent, if it converges in probability to the true value of the parameter:

\underset{n \to \infty}{plim} T_{n} = θ .

i.e. if, for all ε > 0

lim_{n \to \infty} Pr (| T_{n} - θ | > ε) = 0.

An estimator T_n of parameter θ is said to be strongly consistent, if it converges almost surely to the true value of the parameter:

Pr (lim_{n \to \infty} T_{n} = θ) = 1.

A more rigorous definition takes into account the fact that θ is actually unknown, and thus, the convergence in probability must take place for every possible value of this parameter. Suppose {p_θ: θ ∈ Θ} is a family of distributions (the parametric model), and X^θ = {X₁, X₂, … : X_i ~ p_θ} is an infinite sample from the distribution p_θ. Let { T_n(X^θ) } be a sequence of estimators for some parameter g(θ). Usually, T_n will be based on the first n observations of a sample. Then this sequence {T_n} is said to be (weakly) consistent if

\underset{n \to \infty}{plim} T_{n} (X^{θ}) = g (θ), for all θ \in Θ .

This definition uses g(θ) instead of simply θ, because often one is interested in estimating a certain function or a sub-vector of the underlying parameter. In the next example, we estimate the location parameter of the model, but not the scale:

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Examples

Sample mean of a normal random variable

Suppose one has a sequence of statistically independent observations {X₁, X₂, ...} from a normal N(μ, σ²) distribution. To estimate μ based on the first n observations, one can use the sample mean: T_n = (X₁ + ... + X_n)/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: T_n is itself normally distributed, with mean μ and variance σ²/n. Equivalently, $(T_{n} - μ) / (σ / \sqrt{n})$ has a standard normal distribution:

Pr [| T_{n} - μ | \geq ε] = Pr [\frac{\sqrt{n} | T_{n} - μ |}{σ} \geq \sqrt{n} ε / σ] = 2 (1 - Φ (\frac{\sqrt{n} ε}{σ})) \to 0

as n tends to infinity, for any fixed ε > 0. Therefore, the sequence T_n of sample means is consistent for the population mean μ (recalling that $Φ$ is the cumulative distribution of the standard 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

Pr [h (T_{n} - θ) \geq ε] \leq \frac{E [h (T_{n} - θ)]}{h (ε)},

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 T_n is consistent for θ and g(·) is a real-valued function continuous at the point θ, then g(T_n) will be consistent for g(θ):

T_{n} \overset{p}{\to} θ \Rightarrow g (T_{n}) \overset{p}{\to} g (θ)

Slutsky's theorem can be used to combine several different estimators, or an estimator with a non-random convergent sequence. If T_n →^dα, and S_n →^pβ, then

\begin{matrix} T_{n} + S_{n} \overset{d}{\to} α + β, \\ T_{n} S_{n} \overset{d}{\to} α β, \\ T_{n} / S_{n} \overset{d}{\to} α / β, provided that β \neq 0 \end{matrix}

If estimator T_n 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 {X_n} of random variables and under suitable conditions,

\frac{1}{n} \sum_{i = 1}^{n} g (X_{i}) \overset{p}{\to} E [g (X)]

If estimator T_n 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
₁,..., x
_n} one can use T
_n(X) = x
_n as the estimator of the mean E[X]. Note that here the sampling distribution of T
_n is the same as the underlying distribution (for any n, as it ignores all points but the last). So E[T
_n(X)] = E[X] for any n, hence 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 is consistent, as it must converge to the correct value.

Biased but consistent

Alternatively, an estimator can be biased but consistent. For example, if the mean is estimated by $\frac{1}{n} \sum x_{i} + \frac{1}{n}$ it is biased, but as $n \to \infty$ , it approaches the correct value, and so it is consistent.

Important examples include the sample variance and sample standard deviation. Without Bessel's correction (that is, when using the sample size $n$ instead of the degrees of freedom $n - 1$ ), these are both negatively biased but consistent estimators. With the correction, the corrected sample variance is unbiased, while the corrected sample standard deviation is still biased, but less so, and both are still consistent: the correction factor converges to 1 as sample size grows.

Here is another example. Let $T_{n}$ be a sequence of estimators for $θ$ .

Pr (T_{n}) = {\begin{cases} 1 - 1 / n, & if T_{n} = θ \\ 1 / n, & if T_{n} = n δ + θ \end{cases}

We can see that $T_{n} \overset{p}{\to} θ$ , $E [T_{n}] = θ + δ$ , and the bias does not converge to zero.