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]
because (sqrt n)
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)
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
no idea… are you sure that MAD is ALWAYS smaller than st dev?
is there any reason why we need to know that MAD is always smaller than ST dev? or even worse why?
@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.
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 )
well i guess it’s a problem for Dec. 2nd right now… thanks to all participants so far…
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.
the other answer was that the arithmetic average of deviations around the mean is always 1.
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.