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So there's Standard Deviation, Variance, and Covariance, but is there a co standard deviation?

If not why not? Is there a fundamental mathematical reason or is it just convention?

If so why is it not used more, or at least really hard to find using Google searches?

I don't mean this to be a flippant question, I'm trying to really question statistics rather than just memorize a bunch of formulas.

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    $\begingroup$ Could you clarify what you think a "co standard deviation" would represent? Is there some underlying motivation, or are you just asking (in a meta sense) whether there might be some universal meaning to prepending "co" to the name of any statistic? $\endgroup$ – whuber Oct 16 '17 at 15:07
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    $\begingroup$ I'm assuming the OP is generalizing from variance:covariance :: standard deviation: "costandard deviation", but it wouldn't hurt for the question to be more explicit (assuming they really do mean $\sqrt{\sigma_{XY}}$). $\endgroup$ – Ben Bolker Oct 16 '17 at 15:14
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One useful property of the standard deviation is that it has the same units as the mean, so the magnitudes of $\sigma_X$ and $\bar X$ are directly comparable. I've never seen anyone compute the co-standard deviation (by which I assume you mean the square root of the covariance); if the units of $X$ and $Y$ are denoted as $[X]$ and $[Y]$, then the units of the covariance are $[X][Y]$ and the units of the co-standard deviation would be $\sqrt{[X][Y]}$, which isn't particularly useful. On the other hand, the correlation $\sigma_{XY}/(\sigma_X \sigma_Y)$ is unitless, and is a very common scale for reporting associations.

The variance (in contrast to the standard deviation) is useful because it generally has nicer mathematical properties; in particular

$$ \sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y + 2 \sigma_{XY}, $$ which simplifies nicely when $X$ and $Y$ are independent (hence $\sigma_{XY}=0$).

While you're thinking about ways of scaling variances you could also consider the coefficient of variation $\sigma_X/\bar X$ (which is unitless), or the variance-to-mean ratio $\sigma^2_X/\bar X$ (which has weird units but is meaningful in the context of a count distribution such as the Poisson, which is also unitless).

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    $\begingroup$ Good points, but it does not seem to answer why taking square root of covariance doesn't make sense. $\endgroup$ – Tim Oct 16 '17 at 15:05
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    $\begingroup$ Here's one way to exploit your formula: use it to observe that the covariance can be defined as $$\sigma_{XY} = (\sigma^2_{X+Y}-\sigma_X^2-\sigma_Y^2)/2.$$ So why not then simply define a "co-SD"--let's call it $\tau$, say--as $$\tau_{XY}=(\sigma_{X+Y}-\sigma_X-\sigma_Y)/2?$$ This hints at the difficulty of answering the original question without knowing what the "co" of anything might possibly mean: you can't demonstrate much just by showing that one particular generalization is nonsense or useless; you have to consider all possible ways to generalize a concept! $\endgroup$ – whuber Oct 16 '17 at 21:25
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The question seems back-to-front. In mathematics we don't invent names for quantities "just because we can", but because the named quantity is useful for something.

The OP's question doesn't give and reasons why he/she thinks there is a useful quantity that might be named "coStandard Deviation" and the answers are guessing at things that might be useful.

To generalize the concept to multi-variable linear regression with $n$ variables, the "covariance" becomes an $n \times n$ symmetric matrix. You can certainly make a sensible definition of the "square root of a symmetric matrix" so long as it is positive definite or semi-definite, but it's hard to think of a use for it in this context - and it isn't the same as taking the square root of each term of the matrix separately!

Of course the square root of a diagonal matrix (e.g. the variance matrix) is just the square root of the individual terms, so the concept of "standard deviation" does generalize in an obvious and useful way - but "coStandard Deviation" doesn't, IMO. And in general, the "square root of a matrix" isn't even uniquely defined, so which particular square root do you want to choose as the coStandard Deviation?

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Covariance can be both positive and negative.

So the square root of the covariance could be real or imaginary.

You can compare a real number with an imaginary number for size. The units for "standard co-deviation" would be inconvenient. There's no benefit in taking the square root.

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