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Let $X$ be a random variable. Then its variance (dispersion) is defined as $D(X)=E((X-E(X))^2)$. As I understand it, this is supposed to be a measure of how far off from the average we should expect to find the value of $X$.

This would seem to suggest that the natural generalization of variance to the case where $X = (X_1,X_2,\ldots,X_n)$ is random vector, should be $D(X)=E((X-E(X))^T(X-E(X)))$. Here vectors are understood to be columns, as usual. This generalization would again, quite naturally, measure how far off from the average (expectation) we can expect to find the value of vector $X$.

The usual generalization, however, is $D(X)=E((X-E(X))(X-E(X))^T)$, the variance-covariance matrix which, as I see it, measures the correlation of components.

Why is this the preferred generalization? Is $E((X-E(X))^T(X-E(X)))$ also used and does it have a name?

The variance-covariance matrix does seem to contain more information. Is this the main reason or is there something deeper going on here?

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Note that your value is the trace of the variance-covariance matrix, i.e. the sum of the variances. It would seem natural to call this something like the total variance, but that term doesn't seem to be in widespread use. –  joriki Jan 16 '12 at 13:43
    
Note: the covariance matrix contains covariances, not correlations (i.e. the elements are not constrained to $[-1,1]$). Further note: the diagonals of the covariance matrix are variances. –  Mike Wierzbicki Jan 16 '12 at 14:42
    
Here's a hint: Say you wanted to calculate the variance of an arbitrary linear combination of the coordinates of $X$. How would you do it? –  cardinal Jan 16 '12 at 14:46
    
Hmmm ... Yes, it does seem immensely more useful for studying inter-relations between random variables. But do probabilists never study random vectors as such? The quantity seems natural enough to deserve a name of its own, I think =) Anyway, thanks for the suggestions so far. –  Dejan Govc Jan 16 '12 at 15:12
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2 Answers

up vote 3 down vote accepted

If $X \in \mathbb{R}^{n\times1}$ is a column-vector-valued random variable, then $$V=E((X-E(X))(X-E(X))^T)$$ is the variance of $X$ according to the definition given in Feller's famous book. But many authors call it the covariance matrix because its entries are the covariances between the scalar components of $X$.

It is the natural generalization of the $1$-dimensional case. For example, the $1$-dimensional normal distribution has density proportional to $$ \exp\left( \frac{-(x-\mu)^2}{2\sigma^2} \right) $$ where $\sigma^2$ is the variance. The multivariate normal has density proportional to $$ \exp\left( -\frac12 (x-\mu)^T V^{-1} (x-\mu) \right) $$ with $V$ as above.

The variance satisfies the identity $$ \operatorname{var}(AX) = A\Big(\operatorname{var}(X)\Big) A^T. $$ The matrix $A$ need not be $n\times n$. It could be $k\times n$, so that $AX$ is $k\times1$ and then both sides of this identity are $k\times k$.

It follows from the (finite-dimensional) spectral theorem that every non-negative-definite real matrix is the variance of some random vector.

Look at these:

The last-listed article above has a very elegant argument. The trick of considering a scalar to be the trace of a $1\times 1$ matrix is very nice.

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Ok, this is beginning to make sense. So, if I understand correctly, the usual definition naturally arises from the multidimensional central limit theorem? This certainly would explain why probabilists consider it to be the natural generalization. –  Dejan Govc Jan 16 '12 at 20:46
    
The multivariate central limit theorem is certainly one way in which the multivariate normal distribution arises. –  Michael Hardy Jan 16 '12 at 23:07
    
Thanks, this does put things in perspective. –  Dejan Govc Jan 16 '12 at 23:19
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The covariance matrix has more information, indeed: it has the variance of each component (in the diagonal), and also the cross-variances. Your value is the sum of the variances of each component. This is not often a very useful measure. For one thing, the components might correspond to entirely different magnitudes, and hence it would little or no sense to sum the variances of each one. Think for example $X = (X_1,X_2,X_3)$ where $X_1$ height of a man, measured in meters, $X_2$ his waist circumference in centimeters, $X_3$ his weight, in kilograms...

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Thanks, this indeed explains why applied statisticians would prefer the usual definition. (I thought of the components as plain numbers before.) I can't imagine why the other definition wouldn't be useful, though. I believe physicists, for example, study particles in three-dimensions. Here I think the most natural thing to ask oneself would be: how far from my predictions will the particle be found? (Which is exactly what the sum of variances describes.) –  Dejan Govc Jan 16 '12 at 16:55
    
Your value can be useful, what is hardly useful is to give it a special name, instead of just "sum of variances", or "total energy/power" or "trace of the covariance matrix". See eg the convergence analysis of LMS en.wikipedia.org/wiki/Least_mean_squares_filter –  leonbloy Jan 16 '12 at 17:31
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