# 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 $\alpha ,\widehat \alpha$, it 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 working 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. So 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 variables, their expected values 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 $\mu$.