Bias of an estimator

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This article is about bias of statistical estimators. For other uses, see Bias (disambiguation).

In statistics, the difference between an estimator's expected value and the true value of the parameter being estimated is called the bias. An estimator or decision rule having nonzero bias is said to be biased.

Although the term bias sounds pejorative, it is not necessarily used in that way in statistics. Biased estimators may have desirable properties. Not only do they sometimes have a smaller mean squared error than any unbiased estimator, but in some cases the only unbiased estimators are not even within the convex hull of the parameter space, so their use is absurd.

1 Definition
2 Examples
3 See also
4 External link

[edit] Definition

Suppose we are trying to estimate the parameter $θ$ using an estimator $\widehat{\theta}$ (that is, some function of the observed data). Then the bias of $\widehat{\theta}$ is defined to be

$\operatorname{E}(\widehat{\theta})-\theta.\,$

In words, this would be "the expected value of the estimator $\widehat{\theta}$ minus the true value $θ$ ." This may be rewritten as

$\operatorname{E}(\widehat{\theta}-\theta).\,$

which would read "the expected value of the difference between the estimator and the true value" (the expected value of $θ$ is precisely $θ$ ).

[edit] Examples

[edit] Estimating variance

Suppose X₁, ..., X_n are independent and identically distributed random variables with expectation μ and variance σ². Let

$\overline{X}=(X_1+\cdots+X_n)/n$

be the "sample average", and let

$S^2=\frac{1}{n}\sum_{i=1}^n(X_i-\overline{X}\,)^2$

be a "sample variance". Then S² is a "biased estimator" of σ² because

$\operatorname{E}(S^2)=\frac{n-1}{n}\sigma^2\neq\sigma^2.$

Note that when a transformation is applied to an unbiased estimator, the result is not necessarily itself an unbiased estimate of its corresponding population statistic. That is, for a non-linear function f and an unbiased estimator U of a parameter p, f(U) is usually not an unbiased estimator of f(p). For example the square root of the unbiased estimator of the population variance is not an unbiased estimator of the population standard deviation.

Bias is not the only consideration when choosing a statistic, however. Bias refers to the central tendency of the sampling distribution of a statistic, but the variance of the sampling distribution can also be an important consideration. Specifically, statistics with smaller sampling variances will yield greater statistical power. For example, while S² above is more biased than the traditional sample calculation

$S_\mathrm{sample}^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X}\,)^2,$

S² has a lower estimation variability than S²_sample because the denominator dividing the sum of squares is larger in the calculation of S², resulting in a smaller scale of final values, and therefore lower estimation variability, than that of S²_sample. Practically, this demonstrates that for some applications (where the amount of bias can be equated between groups/conditions) it is possible that a biased estimator can prove to be a more powerful, and therefore useful, statistic.

[edit] Estimating a Poisson probability

A far more extreme case of a biased estimator being better than any unbiased estimator is well-known: Suppose X has a Poisson distribution with expectation λ. It is desired to estimate

$\operatorname{P}(X=0)^2=e^{-2\lambda}.\quad$

(For example, when incoming calls at a telephone switchboard are modeled as a Poisson process, and λ is the average number of calls per minute, then e^−2λ is the probability that no calls arrive in the next two minutes.)

The only function of the data constituting an unbiased estimator is

$\delta(X)=(-1)^X.\quad$

If the observed value of X is 100, then the estimate is 1, although the true value of the quantity being estimated is obviously very likely to be near 0, which is the opposite extreme. And if X is observed to be 101, then the estimate is even more absurd: it is −1, although the quantity being estimated obviously must be positive.

The (biased) maximum-likelihood estimator

$e^{-2X}\quad$

is far better than this unbiased estimator, and not only in the mean-squared error sense.

in the sense that the mean squared error

$e^{-4\lambda}-2e^{\lambda(1/e^2-3)}+e^{\lambda(1/e^4-1)}$

is smaller. Compare the unbiased estimator's MSE of

1 - e - 4λ

The MSE is a function of the true value λ. The bias of the maximum-likelihood estimator is:

$e^{-2\lambda}-e^{\lambda(1/e^2-1)}$ .

[edit] Maximum of a discrete uniform distribution

The bias of maximum-likelihood estimators can be substantial. Consider a case where n tickets numbered from 1 through to n are placed in a box and one is selected at random, giving a value X. If n is unknown, then the maximum-likelihood estimator of n is X, even though the expectation of X is only (n + 1)/2; we can only be certain that n is at least X and is probably more. In this case, the natural unbiased estimator is 2X − 1.