Geometric mean and arithmetic mean

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Xander086

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We know that the arithmetic mean is always greater or at least equal to the geometric mean. But anyone knows what does it mean by “the difference between the geometric mean and the arithmetic mean increases with an icrease in variability between period-to-period observations” and why is that so? Cheers
 
It means that the difference between arithmetic mean and geometric mean is bigger for [1.1, 0.9] than for [1.05, 0.95]. In a geometric mean, the lower values have a stronger affect than they would in an arithmetic mean. More dispersed values make this effect more pronounced.
 
Xander086 wrote:
We know that the arithmetic mean is always greater or at least equal to the geometric mean. But anyone knows what does it mean by “the difference between the geometric mean and the arithmetic mean increases with an icrease in variability between period-to-period observations” and why is that so? Cheers
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I am not sure what is the best way to explain.
Here is my simple explanation. Think of a two period model with one positive return +a and one negative return -a. The arithmetic return is (a+(-a))/2 = 0. The geometric mean return would be sqrt([1+a]*[1-a])-1=sqrt(1-a^2)-1. Clearly, as a grows (the higher variability, the higher is a), the geometric mean return would decline.
Another way to look at that would be to think about reinvestment problem. If you lose a lot of money due to high variability in returns, it’s really hard to make it all back because the base is much lower. If you lose money at a higher base and make money with a lower base, that lowers your return.
Another way is to check out lognormal distribution: http://en.wikipedia.org/wiki/Log-normal_distribution You can see that its expectation grows with variability for given mu.
Does that help at all?
 
GM is approx = AM - 1/2 * (Std Dev).
This is the same thing that Maratikus is pointing out above. This is the “mathematical” manifestation of the same.
 
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