Hello,The relation between the geometric mean and arithmetic mean of a time series is that geometric mean is always lower than or equal to the arithmetic mean, and equal if and only if the data points in the time series are equal, right? I wonder if there is an equal relation between the geometric mean of a time series and the arithmetic mean of the logarithms of that same time series? Is there a way to establish a relation between those two ways of computing the mean?

Lets say that I have a geomeatric time series, then when I take geomean("time series")-1 (working in Excel) and compare it to the arithmetic mean of the logarithms of that time series the arithmetic mean is always lower... at least in the time series i've been working with.Is there a way to prove that this relation always holds?Thank you in advance!

QuoteOriginally posted by: ACDSomething like this?Not really. Lets say my time series is returns on geometric form, for example with the data points 1.05, 1.02, 1.07 and 0.97. That would give me an return (geometric) of ~2.7023% and if I take the average of the log of these datapoints I get ~2.6665%, that is, the result of the arithmetic mean is lower than the geometric mean. Does this relation always hold and if so, how do I prove it?

Sorry missed your second post, yes the geometric mean will normally always be greater than the arithmetic mean of the log returns. Using the relationship I gave in the previous post we have:And:With the equality only occurring when x=0 which is the only time the arithmetic mean of the logs won't be less than the geometric mean. Hope that solves the issue for you.

Last edited by ACD on August 29th, 2011, 10:00 pm, edited 1 time in total.

QuoteOriginally posted by: ACDSorry missed your second post, yes the geometric mean will normally always be greater than the arithmetic mean of the log returns. Using the relationship I gave in the previous post we have:And:With the equality only occurring when x=0 which is the only time the arithmetic mean of the logs won't be less than the geometric mean. Hope that solves the issue for you.Thanks, but I don't know if I get that last expression, exp(x) - 1 >= x, how do you come to that conclusion?

QuoteOriginally posted by: lampisQuoteOriginally posted by: ACDSorry missed your second post, yes the geometric mean will normally always be greater than the arithmetic mean of the log returns. Using the relationship I gave in the previous post we have:And:With the equality only occurring when x=0 which is the only time the arithmetic mean of the logs won't be less than the geometric mean. Hope that solves the issue for you.Thanks, but I don't know if I get that last expression, exp(x) - 1 >= x, how do you come to that conclusion?Well you could prove that in a number of ways, a simple way I could see is as follows:Let's define:g(x) = exp[x] - 1f(x)= xa)g(0) = f(0)b)f'(x)=1g'(x)=exp(x)c)Due to (b) we have:g'(x)>1 for x=(0,inf)g'(x)<1 for x=(-inf,0)Therefore from (c) g(x) is increasing faster than f(x) on x=(0,inf) meaning g(x)>f(x) on x=(0,inf) due to (a) & (c)And from (c) g(x) is decreasing slower than f(x) on x=(-inf,0) meaning that g(x)>f(x) on x=(-inf,0) again due to (a) & (c) Put it all together and you have the inequality. Although looking at the graphs of f(x) and g(x) it should be obvious.

Last edited by ACD on August 29th, 2011, 10:00 pm, edited 1 time in total.

QuoteOriginally posted by: ACDQuoteOriginally posted by: lampisQuoteOriginally posted by: ACDSorry missed your second post, yes the geometric mean will normally always be greater than the arithmetic mean of the log returns. Using the relationship I gave in the previous post we have:And:With the equality only occurring when x=0 which is the only time the arithmetic mean of the logs won't be less than the geometric mean. Hope that solves the issue for you.Thanks, but I don't know if I get that last expression, exp(x) - 1 >= x, how do you come to that conclusion?Well you could prove that in a number of ways, a simple way I could see is as follows:Let's define:g(x) = exp[x] - 1f(x)= xa)g(0) = f(0)b)f'(x)=1g'(x)=exp(x)c)Due to (b) we have:g'(x)>1 for x=(0,inf)g'(x)<1 for x=(-inf,0)Therefore g(x) is increasing faster than f(x) on x=(0,inf) meaning g(x)>f(x) on x=(0,inf) due to (a) & (c)And g(x) is decreasing slower than f(x) on x=(-inf,0) meaning that g(x)>f(x) on x=(-inf,0) again due to (a) & (c) Put it all together and you have the inequality.Yes and I just found out you can prove it with taylor series of exp(x) also. Thank you!

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