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:
\[e\left( {\widehat {{\alpha _1}},\widehat {{\alpha _1}}} \right) = \frac{{Var\left( {\widehat {{\alpha _2}}} \right)}}{{Var\left( {\widehat {{\alpha _1}}} \right)}}\]
If the value of this ratio is more than 1 then $$\widehat {{\alpha _1}}$$ will be more efficient, if it is equal to 1 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
 The median against mean
 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.