# Efficiency of an Estimator

Among a number of estimators of the same class, the estimator having the least variance is called an efficient estimator. Thus, if we have two estimators $\widehat {{\alpha _1}}$ and $\widehat {{\alpha _2}}$ with variances $Var\left( {\widehat {{\alpha _1}}} \right)$ and  $Var\left( {\widehat {{\alpha _2}}} \right)$ respectively, and if $Var\left( {\widehat {{\alpha _1}}} \right) < Var\left( {\widehat {{\alpha _2}}} \right)$, then $\widehat {{\alpha _1}}$ will be an efficient estimator. The ratio of the variances of two estimators denoted by $e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _2}}} \right)$ is known as the efficiency of  $\widehat {{\alpha _1}}$ and $\widehat {{\alpha _2}}$ is defined as follows:

If the value of this ratio is more than 1 then $\widehat {{\alpha _1}}$ will be more efficient, if it is equal to 1 then both $\widehat {{\alpha _1}}$ and $\widehat {{\alpha _2}}$ are equally efficient, and if it is less than 1 then $\widehat {{\alpha _1}}$ will be less efficient. Let us consider the following working example.

Example:

The variances of the sample mean and median are
$\frac{{{\sigma ^2}}}{n}$ and $\frac{\pi }{2}\,\,\,\,\frac{{{\sigma ^2}}}{n}$

Find the efficiency of

1. The median against mean
2. The mean against median

Solution:

• Using the formula  $e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _1}}} \right) = \frac{{Var\left( {\widehat {{\alpha _2}}} \right)}}{{Var\left( {\widehat {{\alpha _1}}} \right)}}$, we have

e (median, mean) $= \frac{{Var\left( {\overline X } \right)}}{{Var\left( {med} \right)}}$
$= \frac{{\frac{{{\sigma ^2}}}{n}}}{{\frac{\pi }{2}\,\,\,\frac{{{\sigma ^2}}}{n}}} = \frac{2}{\pi } = 2 \times \frac{7}{{22}} = 0.63$

Therefore, the efficiency of the median against the mean is only 0.63. This means that a sample mean obtained from a sample of size 63 will be equally as efficient as a sample median obtained from a sample of size 100.

• Using the formula  $e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _1}}} \right) = \frac{{Var\left( {\widehat {{\alpha _2}}} \right)}}{{Var\left( {\widehat {{\alpha _1}}} \right)}}$, we have

e (mean, median) $= \frac{{Var\left( {med} \right)}}{{Var\left( {\overline X } \right)}}$
$= \frac{{\frac{{{\sigma ^2}\pi }}{{2n}}}}{{\frac{{{\sigma ^2}}}{n}}} = \frac{\pi }{2} = \frac{{22}}{7} \times \frac{1}{2} = 1.5714$

Therefore, the efficiency of the mean against the median is 1.57, or in other words the mean is about 57% more efficient than the median.