# Mean absolute deviation vs. standard deviation

In the text book "New Comprehensive Mathematics for O Level" by Greer (1983), I see averaged deviation calculated like this:

Sum up absolute differences between single values and the mean. Then get its average. Througout the chapter the term mean deviation is used.

But I've recently seen several references that use the term standard deviation and this is what they do:

Calculate squares of differences between single values and the mean. Then get their average and finally the root of the answer.

I tried both methods on a common set of data and their answers differ. I'm not a statistician. I got confused while trying to teach deviation to my kids.

So in short, are the terms standard deviation and mean deviation the same or is my old text book wrong?

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The two quantities differ. They weight the data differently. The standard deviation will be larger, and it is relatively more affected by larger values. The standard deviation (most particularly, the n-denominator version) can be thought of as a root-mean-square deviation. Standard deviations are more commonly used. – Glen_b Jan 12 '14 at 16:15
Very closely related: stats.stackexchange.com/questions/118/…. – whuber Jan 12 '14 at 16:44
Gary Kader has a fun way of teaching kids to derive the mean absolute deviation. – Iain Elder Mar 5 '14 at 3:46
Incidentally, one reason that people tend to prefer standard deviation is because variances of sums of unrelated random variables add (and related ones also have a simple formula). That doesn't happen with mean deviation. – Glen_b Sep 15 at 4:38
an important point is that the standard deviation derives from a model of squared errors (L2-norm, think about the normal distribution) while the mean of absolute differences corresponds to the L1-norm (think about the symmetrical exponential distribution): it is therefore more adapted (hear: sensitive) to outliers and sparse distirbutions – meduz Sep 18 at 7:47

An observation that is 1 under the mean is equally "far" from the mean as a value that is 1 above the mean. Hence you should neglect the sign of the deviation. This can be done in two ways:

• Calculate the absolute value of the deviations and sum these.

• Square the deviations and sum these squares. Due to the square, you give more weight to high deviations, and hence the sum of these squares will be different from the sum of the means.

After calculating the "sum of absolute deviations" or the "square root of the sum of squared deviations", you average them to get the "mean deviation" and the "standard deviation" respectively.

The mean deviation is rarely used.

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So when one simply says 'deviation' do they mean 'standard deviation'? – itsols Jan 12 '14 at 10:46
I agree that 1 above or below would indicate a meaningful 'change' or 'dispersion' from a common-man's point-of-view. But squaring it would give larger values and that might not be my 'actual change'. Maybe I'm wrong but that's how I see it :/ – itsols Jan 12 '14 at 10:48
Most of the times the term standard deviation (square root of variance) is used. Calculating the squares is typically done, as it facilitates lots of other calculations. – Kasper Jan 12 '14 at 11:01
@itsols Technically, you should always specify which type of deviation statistic you are calculating for the data set -- the word deviation on its own should refer to the deviation of a single datapoint from the mean (in the way Kasper uses it in the answer). – AmeliaBR Jan 12 '14 at 16:51
@itsols, +1 to Amelia. Indeed, nobody says of a dataset statistic as just "deviation". A statistic is "mean absolute deviation" or "root of mean squared deviation" or such. – ttnphns Jan 12 '14 at 17:19

They are similar measures that try to quantify the same notion. Typically you use st. deviation since it has nice properties, if you make some assumption about the underlying distribution.

On the other hand the absolute value in mean deviation causes some issues from a mathematical perspective since you can't differentiate it and you can't analyse it easily. Some discussion here.

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@itsols, I'll add to Kasper's important notion that The mean deviation is rarely used. Why is standard deviation considered generally a better measure of variability than mean absolute deviation? Because arithmetic mean is the locus of minimal sum of squared (and not sum of absolute) deviations from it.

Suppose you want to assess the degree of altruism. Then you probably won't ask a person about how much he is ready to give money in "general situation" of life. Rather, you'll choose to ask how much he is ready to do it in the constained situation, where he has minimal possible resourses for his own living. I.e. what is the amount of individual altruism in the situation when that amount is individual's minimal?

Likewise, what is the degree of variability of these data? Intuitively, the best measuring index for it is the one which is minimized (or maximized) down to the limit in this context. The context is "around the arithmetic mean". Then st. deviation is the best choice in this sense. If the context were "around the median" then mean |deviation| would be the best choice, because median is the locus of minimal sum of absolute deviations from it.

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Your justification for SD based on Locus is circular. You are justifying SD by putting special importance on arithmetic mean - all this shows is that they have a relationship, not that SD is special. One can similarly place importance on median, which is the locus of minimal sum of absalute loss. The real reasons why SD is used more often is because the maths is easier to work with ... furthermore, it's easier computationally (both because medians require "sorting", and because squares are faster to compute than branch statements). Philosophically absalute deviation has greater value. – samthebest Sep 24 at 10:49

They both measure the same concept, but are not equal.

You are comparing $\frac{1}{n} \sum |x_i-\bar{x}|$ with $\sqrt{\frac{1}{n} \sum (x_i-\bar{x})^2}$. They aren't equal for two reasons:

Firstly the square-root operator is not linear, or $\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$. Therefore the sum of absolute deviations is not equal to the square root of the sum of squared deviations, even though the absolute function can be represented as the square function followed by a square root:
$\sum|x_i-\bar{x}| = \sum \sqrt{(x_i-\bar{x})^2} \neq \sqrt{\sum(x_i-\bar{x})^2}$
as the square root is taken after the sum has been calculated.

Secondly, $n$ is now also under the square root in the standard deviation calculation.

Try calculating $\frac{1}{n}\sum \sqrt{(x_i-\bar{x})^2}$ - it should yield the same answer as the mean deviation and help you to understand.

The reason why the standard deviation is preferred is because it is mathematically easier to work with later on, when calculations become more complicated.

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The absolute value of a sum is not, in general, the same as the sum of absolute values! Neither square, square root, nor absolute functions are linear, which is why the sum after applying the function is different from applying the function after taking the sum. – AmeliaBR Jan 12 '14 at 16:28
@AmeliaBR you are of course perfectly correct! – ltronneberg Jan 12 '14 at 16:47
The rest of the argument was good, though, which is why I decided to edit out the problematic statement. – AmeliaBR Jan 12 '14 at 16:52

One thing worth adding is that the most likely reason your 30-year-old textbook used the absolute mean deviation as opposed to standard deviation is that it is easier to calculate by hand (no squaring / square roots). Now that calculators are readily accessible to high school students, there is no reason not to ask them to calculate standard deviation.

There are still some situations where absolute deviations are used instead of standard deviations in complex model fitting. Absolute deviations are less sensitive to extreme outliers (values far from the mean/trendline) compared to standard deviations because they don't square that distance before adding it to the values from other data points. Since model fitting methods aim to reduce the total deviation from the trendline (according to whichever method deviation is calculation), methods that use standard deviation can end up creating a trendline that diverges away from the majority of points in order to be closer to an outlier. Using absolute deviations reduces this distortion, but at the cost of making calculation of the trendline more complicated.

That's because, as others have noted, the standard deviation has mathematical properties and relationships which generally make it more useful in statistics. But "useful" should never be confused with perfect.

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Today, statistical values are predominantly calculated by computer programs (Excel, ...), not by hand-held calculators anymore . Hence, I would posit that calculating "mean deviation" is no more cumbersome than calculating "standard deviation". Although standard deviation may have "... mathematical properties that make it more useful in statistics", it is, in fact, a distortion of the concept of variance from a mean, since it gives extra weighting to data points far from the mean. It may take some time, but I, for one, hope statisticians evolve back to using "mean deviation" more often when discussing the distribution among data points -- it more accurately represents how we actually think of the distribution.

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You make an extraordinary claim about how (statistically qualified) people "actually think." What is the source of your information about that? – whuber Dec 16 '14 at 18:26
The source is simply people whom I have questioned on this topic, as well as myself. When asked: How do you picture the variation in this data set? the response was always expressed in terms of the linear distance from the mean -- the response never included squares or square roots. Granted, I am an engineer, not a "statistician", but I would ask anyone else to challenge themselves on this topic. Yes, we love the math of standard deviation -- it's fun, but is this really how you picture the deviation from the mean? – andyl Dec 16 '14 at 22:03
It depends on the purpose. For data exploration, I tend to use robust, rank-based estimates of dispersion, such as median deviations from medians, which in spirit are close to your proposal. But for much other work, especially when assessing (even mentally) the potential for statistical significance, estimating appropriate sample sizes, figuring out the value of information, and deciding among competing statistical procedures, thinking in terms of variances (and therefore standard deviations) is essential. Mean deviations are no substitute, as the mathematics clearly shows. – whuber Dec 16 '14 at 23:07

Both measure the dispersion of your data by computing the distance of the data to its mean.

1. the mean absolute deviation is using norm L1 (it is also called Manhattan distance or rectilinear distance)
2. the standard deviation is using norm L2 (also called Euclidean distance)

The difference between the two norms is that the standard deviation is calculating the square of the difference whereas the mean absolute deviation is only looking at the absolute difference. Hence large outliers will create a higher dispersion when using the standard deviation instead of the other method. The Euclidean distance is indeed also more often used. The main reason is that the standard deviation have nice properties when the data is normally distributed. So under this assumption, it is recommended to use it. However people often do this assumption for data which is actually not normally distributed which creates issues. If your data is not normally distributed, you can still use the standard deviation, but you should be careful with the interpretation of the results.

Finally you should know that both measures of dispersion are particular cases of the Minkowski distance, for p=1 and p=2. You can increase p to get other measures of the dispersion of your data.

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There is also a post on math.stackexchange on this topic: math.stackexchange.com/questions/384003/l1-norm-and-l2-norm – RockScience Mar 5 '14 at 2:52

No. You are wrong. Just kidding. There are, however, many viable reasons why one would want to compute mean deviation rather than formal std, and in this way I am in agreement with the viewpoint of my engineering Brethren. Certainly if I am computing statistics to compare with a body of existing work which is expressing qualitative as well as quantitative conclusions, I woud stick with std. But, for example, assume I am trying to run some fast anomaly-detection algorithms on binary, machine-generated data. I'm not after academic comparisons as my final goal. But I am interested in the fundamental inference about the "spread" of a particular flow of data about its mean. I'm also interested in computing this iteratively, and as efficiently as possible. In digital electronic hardware, we play dirty tricks all the time -- we distill multiplications and divisions into left and right shifts, respectively, and for "computing" absolute values, we simply drop the sign bit (and compute one's or two's complement if necessary, both easy transforms). So, my choice is to compute it in the most knuckle-dragging way I can, and apply linear thresholds to my computations for fast anomaly detection over desired time windows.

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The standard deviation can be computed efficiently and simply with an online algorithm, just as any moment can be (including the mean absolute deviation). Thus, the requirement for fast or simple calculation would not rule this out (nor would it rule out any moment-based estimators of spread). – whuber Sep 14 at 17:52