MAD vs standard deviation

CFA Level I Quant
#1

Hi All,

while doing assessments found the below statement about MAD and standard deviation which I find confusing. Can you please help to explain the relationship between MAD and stanrdard deviation.

Much appreciated, cheers!

The least accurate statement about measures of dispersion for a distribution is that the:

range provides no information about the shape of the data distribution.

mean absolute deviation will be either less than or equal to the standard deviation.

arithmetic average of the deviations around the mean will be equal to one.

#2

Mean absolute deviation is: (1/n)*sum[abs(x-xbar)]

Standard deviation is : (1/n)*sum[(x-xbar)^2]

Because there is a square in the term for standard deviation, it will always be at least as large as the mean absolute deviation.

Think of it this way, for MAD, you are summing all of the deviations from the mean (irrespective of negative/positive).

For standard deviation, you are squaring each of those distances first, and then adding them.

#3

You should also know that the sum of the deviations from the mean is zero, because of the way in which the mean is calculated (and some other properties would help you prove this). This would imply that the sum of the deviations from the mean is zero, and therefore, the arithmetic average is zero for these deviations (so the third statement is least accurate). This could help you with process of elimination also.

#4

Not correct - what you’ve put as the formula for the SD is actually the formula for variance. If you took the square root, then you have SD. So, since you’re squaring terms then taking the square root, the second statement does not hold.

And even if you were simply looking at squared deviations vs. absolute deviations, if the deviation is less than one, the squared deviation would be smaller than the absolute deviation.

But the third choice is clearly least accurate - the sum of the raw deviations will be zero, not one.

#5

^ Yep, you are right. My bad.

#6

But I guess since you are taking the sqrt(n) for SD, the SD will still be at least as large as MAD.

#7

Not necessarily - you’re squaring, taking the average square, and then taking the root.

I can’t recall the derivation to show the relationship of the square root of the average squared deviation to the average absolute deviation. Any math jocks want to take a stab at it? I believe this would be the relationship of the L1 and L2 norms, if I recall my grad school probability theory.

#8

you’re right, im losin it

#9

Yeah . . . necessarily.

I’d have to fiddle a little to come up with the proof; suffice it to say that the standard deviation is always at least as large as the MAD.

#10

^not if the deviations are less than 1. Then MAD will be greater than stdev.

#11

You need to work harder. If x1 = 0, x2 = ½, and x3 = 1, then:

  • MAD = ⅓ = 0.333…
  • σ = √(1/6) = 0.408248…
#12

I’ll take magician’s word on it, given his math chops.

#13

Hmmm…you’re right. I was working under the impression that a deviation of .2 = MAD of .2, and stdev of .04 (because the .2 is squared, and is therefore smaller than MAD).

But I forgot that you have to take the sqrt of the .04 again at the end. So I guess stdev is always greater than MAD.

#14

I landed on this while preparing my level 1, so I think I should give a more precise answer.

MAD will always be smaller than then standart deviation. Why? Jensen’s inequality, and the square root function is concave. More precisely: It holds that |y| = sqrt(y^2). Because the sqrt is a concave function, by Jensen’s inequality (look it up on wikipedia) the mean of square root is less than the square root of the mean,

where the inequality follows from Jensen’s. I should add that it can very well happen that MAD is greater than the variance, in particular when all the distances to the mean are less than one.

I hope this helps other people that land here looking for answers.