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, also 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

e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _1}}} \right) = \frac{{Var\left( {\widehat {{\alpha _2}}} \right)}}{{Var\left( {\widehat {{\alpha _1}}} \right)}}

If the value of this ratio is more than 1, then \widehat {{\alpha _1}} will be more efficient, if it is equal to 1, 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 worked example.

Example:

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

Find the efficiency of

  1. median against mean
  2. 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 median against the mean is only 0.63. It means that a sample mean obtained from a sample of size 63 will be equally efficient to a sample median obtained from a sample of size 100.

  • Also, 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 mean against median is 1.57 or in other words mean is about 57% more efficient than the median.