MAD vs. std.dev

CFA Level I
#1

why is mean absolute deviation ALWAYS smaller than std. deviation? i would like a mathematical solution. can someone help? !x-xbar!/n < sqrt[((x-xbar)^2)/n]

#2

because (sqrt n)

#3

long, indeed the denominator is smaller for the std.dev. but what about the nominator. sum of absolute deviations <=> sum of squared deviations from the mean? MAD…Std.dev sum!x-xbar!/n…sqrt(sum(x-xbar)^2)/sqrt(n)

#4

x1=5 x2=10 x3=15 xbar=10 MAD: 10/3=3.33 Std.Dev.: sqrt(50/3)=4.08 so in this example MAD

#5

no idea… are you sure that MAD is ALWAYS smaller than st dev?

#6

is there any reason why we need to know that MAD is always smaller than ST dev? or even worse why?

#7

@long, it was on a cfai sample exam. there was definitely an answer which was “less accurate” but i was just wondering. @bluey. i even do not know if this is always the case. that’s why i tried to question this situation. maybe joey can enlighten this.

#8

we r comparing |xi - mean| and sqrt(sum(xi-mean)^2) maybe when the size of a population approaches infinity we can assume that they become equal? a+b+c+d+… = sqrt(a^2+b^2+c^2+d^2+…) considering 1/n < 1/sqrt(n) - mad< st.dev no, the proof is too weak well, another distraction )

#9

well i guess it’s a problem for Dec. 2nd right now… thanks to all participants so far… :slight_smile:

#10

I remember this question. It is not always but mostly true. If all the value are equal, then both MAD and STD are 0. But the other choise is always wrong. I think in that logic we can choose the other answer.

#11

the other answer was that the arithmetic average of deviations around the mean is always 1.

#12

barthezz Wrote: ------------------------------------------------------- > why is > > mean absolute deviation ALWAYS smaller than std. > deviation? > > i would like a mathematical solution. can someone > help? > > !x-xbar!/n < sqrt[((x-xbar)^2)/n] This is called the Cauchy-Schwarz Inequality or even more generally Holder’s Inequality both of which have much more general statements than this. I could prove those to you but I would just be reciting something from a linear algebra book that is widely available on the web. If you really want a proof, Google it.