While the arithmetic mean is obtained from the sum of observations, the product of observations lead to geometric mean. Geometric mean finds its application in investment portfolios. Like arithmetic mean, geometric mean is also well defined and suitable for algebraic treatment. In contrast to arithmetic mean geometric mean is not much affected by the presence of extreme values in the data set. GM(geometric mean) can also be calculated only for positive valued observations.

Geometric mean is defined to be the nth root of n values.
Thus the geometric mean of n observations x1, x2, ........xn is given by
GM = $\sqrt[n]{x_{1}.x_{2}......x_{n}}$ = $(x_{1}.x_{2}.....x_{n})$$^{\frac{1}{n}}$
This can also be written using 'Pi' notation as GM = $(\prod_{i=1}^{n}x_{n})$$^{\frac{1}{n}}$
If x1, x2, ........xn occur with corresponding frequencies $f_1, f_2, ......f_n$ then the Geometric Mean is given by
GM = $\sqrt[N]{x_{1}^{f_{1}}.x_{2}^{f_{2}}......x_{n}^{f_{n}}}$ = $(x_{1}^{f_{1}}.x_{2}^{f_{2}}......x_{n}^{f_{n}})$$^{\frac{1}{N}}$
where N = f1 + f2 + ....... + fn.
The geometric mean of two numbers a and b is given by $\sqrt{ab}$.

Example:
The geometric mean of 3 and 12 is
= √(3 x 12) = √36 = 6.


Geometric mean of two groups of items


Suppose G1 and G2 are geometric means of two groups of items with corresponding observations n1 and n2, then the combined geometric mean of two groups is given by
G = $(G_{1}^{n_{1}}.G_{2}^{n_{2}})$$^{\frac{1}{n_{1}+n_{2}}}$

Example:
If the geometric means of 6 test scores in each of Math and Science is 92 and 87, then the combined geometric mean of Math and Science in a total of 12 tests

= $(92^{6}87^{6})$$^{\frac{1}{6+6}}$ = 89.47
In a right triangle, the altitude on the hypotenuse divides the altitude into two segments. The geometric mean theorem provides proportions relating the lengths of these segments of hypotenuse to the lengths of legs and the length of the altitude.

Geometric Mean Theorem

In the above diagram, triangle ABC is a right triangle right angled at C. CD is the altitude drawn on hypotenuse AB which divides AB into two segments AD and DB.

In a right triangle the length of the altitude from the right angle to the hypotenuse is the geometric mean of the length of the two segments of the hypotenuse made by the altitude.

With reference to the diagram given above the statement of the above theorem can be written as
CD2 = AD . BD  or CD = $\sqrt{AD.BD}$

If the length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the segment of the hypotenuse adjacent to it.
Using the diagram above the above statement can be explained by the following two equations:
AC2 = AB . AD or AC = $\sqrt{AB.AD}$
and
BC2 = AB . BD or BC = $\sqrt{AB.BD}$


Solved Example

Question: Arthur's investment in mutual funds earned him 25% in the first year. But it came out to be a loss of 10% and 12% in the second and third years correspondingly. Find the geometric mean of annual rate of return.
Solution:
 
To find the geometric mean of annual rate of return we need to find the geometric mean of growth factors of the investment in three yeas.

   Growth factor in year I = 1 + 0.25 = 1.25
                         in year II = 1 - 0.1 = 0.90
                         in year III = 1 - 0.12 = 0.88
   The geometric mean of growth factors = [(1.25) (0.9) (0.88)]1/3 = 0.9967
   This means the investment gives an annual return of 99.67 dollars for every 100 dollars invested.
   This method is helpful in comparing the returns of two or more investments, without actually knowing the actual amounts invested.