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.