# Is CoStandard Deviation a thing?

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.

• 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? – whuber Oct 16 '17 at 15:07
• 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 $\sqrt{\sigma_{XY}}$). – Ben Bolker Oct 16 '17 at 15:14

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).

• Good points, but it does not seem to answer why taking square root of covariance doesn't make sense. – Tim Oct 16 '17 at 15:05
• 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! – whuber Oct 16 '17 at 21:25

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?

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.