Unbiasedness of an Estimator

This is probably the most important property that a good estimator should possess. According to this property if the statistic \widehat \alpha is an estimator of \alpha ,\widehat \alpha will be an unbiased estimator if the expected value of  \widehat \alpha equals the true value of the parameter \alpha

i.e.

E\left( {\widehat \alpha } \right) = \alpha

Consider the following worked example

Example:

Show that the sample mean \overline X is an unbiased estimator of the population mean\mu .

Solution:

In order to show that \overline X is an unbiased estimator, we need to prove that

E\left( {\overline X } \right) = \mu

We have

\begin{gathered} \overline X = \frac{{\sum X}}{n} = \frac{{{X_1} + {X_2} + {X_3} + \cdots + {X_n}}}{n} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{X_1}}}{n} + \frac{{{X_2}}}{n} + \frac{{{X_3}}}{n} + \cdots + \frac{{{X_n}}}{n} \\ \end{gathered}

Therefore,

E\left( {\overline X } \right) = E\left( {\frac{{{X_1}}}{n} + \frac{{{X_2}}}{n} + \frac{{{X_3}}}{n} + \cdots + \frac{{{X_n}}}{n}} \right)

From the rule of expectation, the expected value of a linear combination is equal to the linear combination of their expectations, we have

E\left( {\overline X } \right) = \frac{1}{n}E\left( {{X_1}} \right) + \frac{1}{n}E\left( {{X_2}} \right) + \frac{1}{n}E\left( {{X_3}} \right) + \cdots + \frac{1}{n}E\left( {{X_n}} \right)

Since {X_1},{X_2},{X_3}, \ldots ,{X_n} are each random variable, their expected value will be equal to the probability mean\mu ,

\begin{gathered} E\left( {\overline X } \right) = \frac{1}{n}\mu + \frac{1}{n}\mu + \frac{1}{n}\mu + \cdots + \frac{1}{n}\mu \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n\mu }}{n} = \mu \\ \end{gathered}

Therefore, E\left( {\overline X } \right) = \mu

Hence \overline X is an unbiased estimator of the population mean\mu .