# A measure of “variance” from the covariance matrix?

If the data is 1d, the variance shows the extent to which the data points are different from each other. If the data is multi-dimensional, we'll get a covariance matrix.

Is there a measure that gives a single number of how the data points are different from each other in general for multi-dimensional data?

I feel that there might be many solutions already, but I'm not sure the correct term to use to search for them.

Maybe I can do something like adding up the eigenvalues of the covariance matrix, does that sound sensible?

• Determinant of the covariance matrix. I will post a more fleshed out answer soon. – user603 Jul 25 '16 at 7:17
• Trace is used very often. E.g. in PCA, fraction of variance explained by each component is a fraction of the "total variance" which is defined as the trace of the covariance matrix. @user603 Looking forward to your answer. – amoeba Jul 25 '16 at 9:41
• adding up the eigenvalues of the covariance matrix is equal to the trace amoeba mentioned right above. – ttnphns Jul 25 '16 at 20:40
• What is/was the measure going to be used for? – HelloGoodbye May 11 '17 at 12:43
• @HelloGoodbye hi actually I have some [noisy] data with labels, and I assume in advance that the [true] data points within the same category should not be very different. I'm looking for a way to measure the degree of differences of the data points within each category, so that I can get an idea of how noisy the data are for each category. – dontloo May 13 '17 at 6:15

(The answer below merely introduces and states the theorem proven in Eq. (0) The beauty in that paper is that most of the arguments are made in terms of basic linear algebra. To answer this question it will be enough to state the main results, but by all means, go check the original source).

In any situation where the multivariate pattern of the data can be described by a $$k$$-variate elliptical distribution, statistical inference will, by definition, reduce it to the problem of fitting (and characterizing) a $$k$$-variate location vector (say $$\boldsymbol\theta$$) and a $$k\times k$$ symmetric semi-positive definite (SPSD) matrix (say $$\boldsymbol\varSigma$$) to the data. For reasons explained below (which are assumed as premises) it will often be more meaningful to decompose $$\boldsymbol\varSigma$$ into its shape component (a SPSD matrix of the same size as $$\boldsymbol\varSigma$$) accounting for the shape of the density contours of your multivariate distribution and a scalar $$\sigma_S$$ expressing the scale of these contours.

In univariate data ($$k=1$$), $$\boldsymbol\varSigma$$, the covariance matrix of your data is a scalar and, as will follow from the discussion below, the shape component of $$\boldsymbol\varSigma$$ is 1 so that $$\boldsymbol\varSigma$$ equals its scale component $$\boldsymbol\varSigma=\sigma_S$$ always and no ambiguity is possible.

In multivariate data, there are many possible choices for scaling functions $$\sigma_S$$. One in particular ($$\sigma_S=|\pmb\varSigma|^{1/k}$$) stands out in having a key desirable propriety, making it the preferred choice of scaling functions in the context of elliptical families.

Many problems in MV-statistics involve estimation of a scatter matrix, defined as a function(al) SPSD matrix in $$\mathbb{R}^{k\times k}$$ ($$\boldsymbol\varSigma$$) satisfying:

$$(0)\quad\boldsymbol\varSigma(\boldsymbol A\boldsymbol X+\boldsymbol b)=\boldsymbol A\boldsymbol\varSigma(\boldsymbol X)\boldsymbol A^\top$$ (for non singular matrices $$\boldsymbol A$$ and vectors $$\boldsymbol b$$). For example the classical estimate of covariance satisfies (0) but it is by no means the only one.

In the presence of elliptical distributed data, where all the density contours are ellipses defined by the same shape matrix, up to multiplication by a scalar, it is natural to consider normalized versions of $$\boldsymbol\varSigma$$ of the form:

$$\boldsymbol V_S = \boldsymbol\varSigma / S(\boldsymbol\varSigma)$$

where $$S$$ is a 1-honogenous function satisfying:

$$(1)\quad S(\lambda \boldsymbol\varSigma)=\lambda S(\boldsymbol\varSigma)$$

for all $$\lambda>0$$. Then, $$\boldsymbol V_S$$ is called the shape component of the scatter matrix (in short shape matrix) and $$\sigma_S=S^{1/2}(\boldsymbol\varSigma)$$ is called the scale component of the scatter matrix. Examples of multivariate estimation problems where the loss function only depends on $$\boldsymbol\varSigma$$ through its shape component $$\boldsymbol V_S$$ include tests of sphericity, PCA and CCA among others.

Of course, there are many possible scaling functions so this still leaves the open the question of which (if any) of several choices of normalization function $$S$$ are in some sense optimal. For example:

• $$S=\text{tr}(\boldsymbol\varSigma)/k$$ (for example the one proposed by @amoeba in his comment below the OP's question as well as @HelloGoodbye's answer below. See also [1], [2], [3])
• $$S=|\boldsymbol\varSigma|^{1/k}$$ ([4], [5], [6], [7], [8])
• $$\boldsymbol\varSigma_{11}$$ (the first entry of the covariance matrix)
• $$\lambda_1(\boldsymbol\varSigma)$$ (the first eigenvalue of $$\boldsymbol\varSigma$$), this is called the spectral norm and is discussed in @Aksakal answer below.

Among these, $$S=|\boldsymbol\varSigma|^{1/k}$$ is the only scaling function for which the Fisher Information matrix for the corresponding estimates of scale and shape, in locally asymptotically normal families, are block diagonal (that is the scale and shape components of the estimation problem are asymptotically orthogonal) [0]. This means, among other things, that the scale functional $$S=|\boldsymbol\varSigma|^{1/k}$$ is the only choice of $$S$$ for which the non specification of $$\sigma_S$$ does not cause any loss of efficiency when performing inference on $$\boldsymbol V_S$$.

I do not know of any comparably strong optimality characterization for any of the many possible choices of $$S$$ that satisfy (1).

• [0] Paindaveine, D., A canonical definition of shape, Statistics & Probability Letters, Volume 78, Issue 14, 1 October 2008, Pages 2240-2247. Ungated link
• [1] Dumbgen, L. (1998). On Tyler’s M-functional of scatter in high dimension, Ann. Inst. Statist. Math. 50, 471–491.
• [2] Ollila, E., T.P. Hettmansperger, and H. Oja (2004). Affine equivariant multivariate sign methods. Preprint, University of Jyvaskyla.
• [3] Tyler, D.E. (1983). Robustness and efficiency properties of scatter matrices, Biometrika 70, 411–420.
• [4] Dumbgen, L., and D.E. Tyler (2005). On the breakdown properties of some multivariate M-Functionals, Scand. J. Statist. 32, 247–264.
• [5] Hallin, M. and D. Paindaveine (2008). Optimal rank-based tests for homogeneity of scatter, Ann. Statist., to appear.
• [6] Salibian-Barrera, M., S. Van Aelst, and G. Willems (200 6). Principal components analysis based on multivariate MM-estimators with fast and robust bootstrap, J. Amer. Statist. Assoc. 101, 1198–1211.
• [7] Taskinen, S., C. Croux, A. Kankainen, E. Ollila, and H. O ja (2006). Influence functions and efficiencies of the canonical correlation and vector estimates based on scatter and shape matrices, J. Multivariate Anal. 97, 359–384.
• [8] Tatsuoka, K.S., and D.E. Tyler (2000). On the uniqueness of S-Functionals and M-functionals under nonelliptical distributions, Ann. Statist. 28, 1219–1243.
• Also, $\varSigma_{11}$ is a strange choice for the scale component because it is not rotation-invariant... – amoeba Jul 26 '16 at 0:02
• Thanks for the deliberated answer! it'll take me some time to fully understand it though :) – dontloo Jul 26 '16 at 1:57
• @amoeba: $\boldsymbol\varSigma$ applied to $\pmb X$. I drop the $\pmb X$ in the rest of the answer because there is no confusion possible. I agree it is a bit clumsy so I now use $\boldsymbol\varSigma(\pmb X)$. I agree with your second comment. By the same tocken $\lambda_1(\boldsymbol\varSigma)$ is not invariant to rescaling. In this sense the homogeneity constraint placed on $S$ is a very low bar. – user603 Jul 26 '16 at 7:03
• Wait; why would one want or expect the scale component to be invariant to rescaling?? – amoeba Jul 26 '16 at 8:15
• Sorry, I meant if you use $\lambda_1(\boldsymbol\varSigma)$ as scaling function then the resulting shape matrix is not equivariant to rescaling. – user603 Jul 26 '16 at 8:26

The variance of a scalar variable is defined as the squared deviation of the variable from its mean:

$$\operatorname{Var}(X) = \operatorname E\left[\left(X - \operatorname E\left[X\right]\right)^2\right]$$

One generalization to a scalar-valued variance for vector-valued random variables can be obtained by interpreting the deviation as the Euclidean distance:

$$\operatorname{Var_s}(\mathbf X) = \operatorname E\left[\left\|\mathbf X - \operatorname E\left[\mathbf X\right]\right\|_2^2\right]$$

This expression can be rewritten as

$$\begin{array}{rcl} \operatorname{Var_s}(\mathbf X) & = & \operatorname E[(\mathbf X - \operatorname E[\mathbf X ])\cdot(\mathbf X - \operatorname E[\mathbf X ])] \\ & = & \operatorname E\left[\sum_{i=1}^n(X_i - \operatorname E[X_i])^2\right] \\ & = & \sum_{i=1}^n \operatorname E\left[(X_i - \operatorname E[X_i])^2\right] \\ & = & \sum_{i=1}^n \operatorname{Var}(X_i) \\ & = & \sum_{i=1}^n C_{ii} \end{array}$$

where $$\mathbf{C}$$ is the covariance matrix. Finally, this can be simplified to

$$\operatorname{Var_s}(\mathbf X) = \operatorname{tr}(\mathbf{C})$$

which is the trace of the covariance matrix.

Although the trace of the covariance matrix, tr(C), gives you a measure of the total variance, it does not take into account the correlation between variables.

If you need a measure of overall variance which is large when your variables are independent from each other and is very small when the variables are highly correlated, you can use the determinant of the covariance matrix, |C|.

If you need just one number, then I suggest taking the largest eigenvalue of the covariance matrix. This is also an explained variance of the first principal component in PCA. It tells you how much of the total variance can be explained if you reduce the dimensionality of your vector to one. See this answer on math SE.

The idea's that you collapse your vector into a single dimension by combining all variables linearly into one series, ending up with a 1d problem.

The explained variance can be reported in terms of a percentage of the total variance. In this case you'll immediately see if there is a lot of linear correlation between series. In some applications this number can be 80% and higher, e.g. interest rate curve modeling in finance. Meaning that you can construct a linear combination of variables that explains 80% of variance of all variables.

The entropy concept from information theory seems to suit the purpose, as a measure of unpredictability of information content, which is given by $$H(X)=-\int p(x)\log p(x) dx.$$

If we assume a multivariate Gaussian distribution for $p(x)$ with mean $\mu$ and covariance $\Sigma$ derived from the data, according to wikipedia, the differential entropy is then, $$H(X)=\frac{1}{2}\log((2\pi e)^n\det(\Sigma))$$ where $n$ is the number of dimensions. Since multivariate Gaussian is the distribution that maximizes the differential entropy for given covariance, this formula gives an entropy upper bound for an unknown distribution with a given variance.

And it depends on the determinant of the covariance matrix, as @user603 suggests.

• This answer doesn't seem to be in the same spirit as the question. Covariances and variances are properties of any distribution (although they might be infinite or undefined in some cases), whereas this answer focuses on an exceedingly special case of a multivariate Normal distribution. It therefore doesn't apply to most of the situations implicitly envisioned in the question. Could you perhaps elaborate on the sense in which your answer could be construed as providing some useful guidance in the general case where the data aren't necessarily Normal? – whuber Jul 25 '16 at 13:47
• @whuber thanks for the suggestion i guess maybe i should rewrite Gaussian as "the distribution that maximizes the entropy given a variance"? then the result will become some upper bound. what do you think? – dontloo Jul 25 '16 at 14:03
• That sounds like it's going somewhere useful and more general. – whuber Jul 25 '16 at 14:30
• I guess there are many ways to skin a cat;). I actually thing the link between your answer and mine are very strong. I have a minor quibble; I think the determinant has some optimality property for the problem you try to solve (and need not just be chosen on grounds of familiarity) and I think these optimality properties extend beyond covariance matrices (they hold for the determinant of whatever scatter functional you happen to chose and there are many out there) and extend beyond the Gaussian distribution (to the whole elliptical family). – user603 Jul 25 '16 at 21:31