Since the geometric mean is less than or equal to the arithmetic mean many fund managers inflate "average returns" by reporting the latter. So wouldn't it be nice to have a proof that the geometric mean is only in a special case as large as the arithmetic mean (for a given number of returns) readily available? In other words: Can we* construct a proof by using a method that is well understood among economists? ¡Claro que sí!
For the sake of simplicity we take the case n = 2 (x1, x2 > 0):
To proof this inequality, we can consider this problem from an optimization viewpoint by restating it as follows: "What is the maximum value that f(x1, x2) = Sqrt(x1x2) can achieve if we must have in addition (x1 + x2)/2 = A for some fixed A?" (A stands for average.) The problem is to solve:
This is a simple Lagrange multiplier problem which involves finding the critical values for the Lagrangean
The critical points of L are the solutions to:
Obviously, x1 = x2. Together with the third equation we have x1 = x2 = A. It is easy to see that this is a maximum, and thus the maximum value of f(x1, x2) is at
f(A,A) = A. Ergo
and both sides are equal only if x1 = x2.
*I somehow got inspired by this nice handout.
For the sake of simplicity we take the case n = 2 (x1, x2 > 0):
To proof this inequality, we can consider this problem from an optimization viewpoint by restating it as follows: "What is the maximum value that f(x1, x2) = Sqrt(x1x2) can achieve if we must have in addition (x1 + x2)/2 = A for some fixed A?" (A stands for average.) The problem is to solve:This is a simple Lagrange multiplier problem which involves finding the critical values for the Lagrangean
The critical points of L are the solutions to:
Obviously, x1 = x2. Together with the third equation we have x1 = x2 = A. It is easy to see that this is a maximum, and thus the maximum value of f(x1, x2) is at
f(A,A) = A. Ergo
and both sides are equal only if x1 = x2.
*I somehow got inspired by this nice handout.
Mahalanobis - am 2005-01-30 01:40 - Rubrik: mathstat
