An estimator \(U\) of a parameter \(\theta\) is unbiased when \(E(U)=\theta\). Unbiasedness belongs to the estimator, not to one realised estimate.
Population mean
The sample mean is unbiased: \(E(\overline X)=\mu\). For observed values, the corresponding estimate is \(\widehat\mu=\overline x\).
Population variance
The divisor-\(n\) sample variance is biased because \(E(S_n^2)=\dfrac{n-1}{n}\sigma^2\). Correcting its divisor gives the unbiased estimator.
\[S^2=\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline X)^2,\qquad \widehat{\sigma^2}=s^2=\frac{1}{n-1}\left(\sum x_i^2-\frac{(\sum x_i)^2}{n}\right)\]
Translated data
Suppose the recorded values are \(y_i=x_i+a\), where \(a\) is constant. Translation changes the mean but not the variance.
| Required estimate | Using the transformed values \(y_i=x_i+a\) |
|---|---|
| Population mean | \(\displaystyle\widehat\mu=\overline y-a=\frac{\sum y_i}{n}-a\) |
| Population variance | \(\displaystyle\widehat{\sigma^2}=\frac{1}{n-1}\left(\sum y_i^2-\frac{(\sum y_i)^2}{n}\right)\) |
State which divisor is being used. Dividing by \(n\) describes the sample data; dividing by \(n-1\) gives the usual unbiased estimator of the population variance.
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