Why should I prefer the standard deviation over other measures of variance? [duplicate]

The most common kind of deviation is the standard deviation.

$$\text{Sd}(x) = \sqrt{\text{Mean}((x - \text{Mean}(x))^2)}$$

The standard deviation is very similar to the mean absolute deviance or

$$\text{MAD}(x) = \text{Mean}(|x - \text{Mean}(x)|)$$

but is often simpler to calculate or obeys nice algebraic properties.

But there are a lot of other measures of variance. For example, there is the most common absolute deviance from the mean value: $\text{Mode}(|x - \text{Mean}(x)|)$.

It is not clear to me why I should prefer the standard deviation over other kinds of measures of dispersion. I suppose the simplest answer is that this measure of dispersion is the most highly studied and well known and using other methods of dispersion will confuse people you try to communicate with.

I guess I would have to use these kind of alternative measures for pathological distributions like the Cauchy distribution though.

marked as duplicate by Aksakal, COOLSerdash, usεr11852, Silverfish, gung♦Aug 9 '16 at 23:46

• Try to look for related threads. I think this question has been answered earlier, just cannot find the duplicate right now. – Richard Hardy Aug 9 '16 at 20:03
• One reason variance is popular (and hence standard deviation) is that the variance of a sum of random variables has a very simple form (doubly so if they're independent). None of the main competitors does. – Glen_b Aug 9 '16 at 20:06
• @Aksakal This is not a duplicate. This is not about squaring or square rooting but about the mean versus the mode and similar. – Steven Stewart-Gallus Aug 9 '16 at 20:10
• This question is being asked in different forms over and over, that's my point. It's impossible to compare variance to every measure of dispersion. – Aksakal Aug 9 '16 at 20:12
• I'd agree that the questions were not phrased identically, eg the other question talked about squaring vs square rooting, but the actual answers received there were far more general than the title suggests. – Silverfish Aug 9 '16 at 23:12

You may come up with infinite number of dispersion measures. It's a lost cause to compare the variance to each and everyone of them.

There are two features of variance that are attractive to me. First, it's a smooth function. For instance, the mean absolute deviation is not.

Second, it's one of the central moments: $$\mu_k=\sum_ip_i(x_i-\mu_1)^k$$ Here $\mu_2$ is a variance.

Being a moment is important, for it defines the distribution when combined with all other moments. Other measures of dispersion as stand-alone metrics.

• Well you could take the moment around the mean but couldn't you just as easily take the moment around the median? The nth moment is $\int^{\infty}_{-\infty} (x-c)^n f(x) \, \mathrm{d} x$ but c does not need to be the mean at all. – Steven Stewart-Gallus Aug 9 '16 at 20:23
• @StevenStewart-Gallus, that wouldn't be a central moment. In fact non-central moment is defined as $\gamma_k=\sum_ip_ix_i^k$. So, the mean is not important here. The important part is squaring. – Aksakal Aug 9 '16 at 20:26
• I don't follow. Why should I care for the central moment and not some other moment? – Steven Stewart-Gallus Aug 9 '16 at 20:41
• @StevenStewart-Gallus, a full set of moments defines the distribution, central or non-central. The central second moment has a nice interpretation as a dispersion measure. All I'm saying is that this measure of dispersion is not a stand alone, it's like a piece which completes the puzzle, unlike other measures of dispersion – Aksakal Aug 9 '16 at 20:45
• but you just said that other moments work too? The moment around the median seems to have just as nice an interpretation. – Steven Stewart-Gallus Aug 9 '16 at 21:15

I'm surprised no one has mentioned the really essential property that Variance is additive for independent random variables:

$$\mbox{Var}(a_1X_1+\cdots+a_nX_n)=\sum_{i=1}^na_i^2\mbox{Var}(X_i),$$

and the equally nice linearity properties that covariance shares. This becomes completely intractable without the square inside the expectation in the definition of variance.

As well for Central limit theorem it is variance, and not L1 that gives rise to the CLT. Specifically L1 gives rise to the strong law of large numbers, but not the fluctuations therein.

In some sense, a related and deeper question is why do people tend to use the $L_2$ norm instead of the $L_1$ norm or indeed other norms? In the two-dimensional Euclidean vector space, why do people tend to use $\sqrt{x^2+y^2}$ (i.e. $L_2$ norm) as a measure of distance instead of $|x| + |y|$ (i.e. $L_1$ norm)?

How the $L_2$ and $L_1$ norm are related to standard deviation and mean absolute deviation respectively:

Let $x$ be a mean zero random variable and $P$ be a probability measure.

The standard deviation is simply the $L_2$ norm:

$$\left( \int |x|^2 \;dP \right) ^\frac{1}{2}$$

And the mean absolute deviation is simply the $L_1$ norm. $$\left( \int |x| \; dP \right)^ \$$

• This is not what I'm asking about. I'm asking about the mean versus the mode and similar and not square versus absolute value. – Steven Stewart-Gallus Aug 11 '16 at 1:58
• $L_1$ is not rotation invariant, while $L_2$ is. – Aksakal Aug 11 '16 at 14:07