an analyst calculated the following performace statistics for mutual fund over a 15 years period. Geometric mean return 12.5 arithmetic mean return 9.6 standard deviation 19.5 mean absolute deviation 21.9 The analyst most likely made one or more errors in calculationg the measure that are related to A both return and dispersion b neither return nor dispersion c return but not in calculating the measures related to dispersion d dispersion but not in calculation the measures related to return
c - return figures are wrong. Is there any relationship b/w SD and MAD? S
Mad is lower then SD and Geometric is less than Arithmetic And thats all I know folks! Im guessing A Because from seeing so many test questions… One is related to MAD and SD (Dispertion) and other is related to Geo and Arith (returns) I am totally guessing.
C GM<=AM --> but our analyst gets GM>AM, therefore return calculation is wrong MAD>=SD —> dispersion looks correct MAD takes only absolute value (positive) and has to be higher than or equal to SD
Sconnery and thunder, can we get your consensus on MAD and SD please? I am not sure about this -MAD takes only absolute value (positive) and has to be higher than or equal to SD? Considering SD is squared and then taken root? How about the denominator in each of them- SD uses n-1 or n? MAD uses n-1 or n? Thanks for illuminating us. S
now, a^2+b^2+c^2 <= (a+b+c)^2 ==> MAD>=SD
The average absolute deviation from the mean is less than or equal to the standard deviation; one way of proving this relies on Jensen’s inequality. wikipedia? Oh I don’t think you square the MAD thunder.
now, a^2+b^2+c^2 <= (a+b+c)^2 ==> MAD>=SD Lets say mean is X, (and denominator should be same) MAD is calculated as – (a-X)+(b-X)+(c-X)+… = A+B+C+… SD will be Sqrt(A^2+B^2+C^2+…) clearly, A^2+B^2±-- <= (A+B+C+…)^2 Shouldn’t denominators be same, as n-1 is done only as it gives you unbiased estimate of population SD and so does it apply to MAD too. Need experts to confirm this !!!
SConnery Wrote: ------------------------------------------------------- > The average absolute deviation from the mean is > less than or equal to the standard deviation; one > way of proving this relies on Jensen’s > inequality. > > wikipedia? > > Oh I don’t think you square the MAD thunder. Sconnery - I wasn’t squaring MAD, but bringing sqrt from other side. The inequality I wrote will always hold good. But I just searched and you are correct. SD>=MAD. http://en.wikipedia.org/wiki/Median_absolute_deviation thanks