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 $$\mu $$.