Mean Absolute Deviation: Calculation and Use
In your quest to find a measure of dispersion, you think of taking the average of all these deviations, or the average mean deviation, in the form
$$\displaystyle \dfrac{\sum_{i=1}^{n}(X_{i}-{\bar X})}{n-1}$$.
Will that get you the information you're looking for?
Correct.
Incorrect. This approach would take the average over
$$\displaystyle X_{i}-{\bar X}$$,
which is the same as the average of the $$X_i$$ terms minus the average of $$X_m$$.
This would give you
$$\displaystyle {\bar X}-{\bar X}=0$$,
which does not sound right.
Perhaps the problem lies in the idea of using the deviation from the mean $$X_i-X_m$$ as the primary variable. Instead, you could consider the absolute value of the deviation from the mean,
$$\displaystyle |X_{i}-{\bar X}|$$.
The __mean absolute deviation__ uses the absolute deviation $$|X_{i}-{\bar X}|$$, and its formula is:
$$\displaystyle MAD={{\sum_{i=1}^{n}|X_{i}-{\bar X}|}\over{n}} $$.
Why might this formula work better as a measure of dispersion?
Incorrect. The absolute value switches the sign of negative values but does not change its scale. The real reason that this will work better is that the absolute value of deviations always has the same sign.
Now you can submit your report to your boss with a straight face. MAD saved the day.
It isn't always the statistic of choice for measuring dispersion, though. Why do you think that is?
Incorrect. The simple absolute deviation does a perfect job of capturing dispersion.
Correct.
To summarize:
[[summary]]
Correct.
Suppose that you have a sample of returns from a particular portfolio, and that it contains the ordered 15 observations: 2,2,3,3,3,5,5,5,5,5,7,7,7,8,8. Having calculated the sample mean of 5, along with the range, you go back to your boss.
Your boss politely thanks you for your effort, but asks for more details about the dispersion of the portfolio’s returns. Back at your desk, you realize that you will need to measure the deviation of each observation from the sample mean, namely
$$\displaystyle X_{i}-{\bar X}$$.
If you take the average of the deviations $$X_{i}-{\bar X}$$, the positive and negative deviations would cancel each other, so your calculation would return 0.
In your sample, the mean deviations $$X_{i}-{\bar X}$$ are: -3,-3,-2,-2,-2,0,0,0,0,0,2,2,2,3,3. Their average is 0. This is in spite of the obvious fact that there is a dispersion in the sample. Your boss is not going to like that result.
Using the absolute value turns a negative deviation into a positive value. A negative deviation, say, -3, becomes positive, since $$|-3| = 3$$. Therefore, the mean absolute deviation will have a positive value, and the larger it will be as the absolute value of the deviations gets wider.
With your portfolio sample, the mean absolute deviation is
$$\dfrac{|-3|+|-3|+|-2|+|-2|+|-2|+0+0+0+0+2+2+2+3+3}{15}$$
> $$= \dfrac{12+0+12}{15} = \dfrac{24}{15} = \dfrac{8}{5}$$.
In practical use, the absolute deviation is difficult to work with sometimes, as the math can get complicated. So people have looked for other measures of dispersion that do not rely on absolute deviations.
Yes
No, this fails to take the average of deviations
The absolute value of the deviations is generally larger
The absolute value of deviations always has the same sign
The absolute deviation is missing important aspects of dispersion
Analysts do not like working with the absolute deviation
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