I have a GPS unit that outputs a noise measurement via covariance matrix $\Sigma$:

$\Sigma = \left[\begin{matrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{matrix}\right] $

(there's also $t$ involved but let's ignore that for a second.)

Suppose I want to tell someone else that the accuracy in each direction ($x,y,z$) is some number. $\mu_x, \mu_y, \mu_z$. That is to say, my GPS may give me a reading of $x=\bar{x}\pm\mu_x$, etc. My understanding is that $\mu$ in this case implies that all measurands are independent of each other (i.e., the covariance matrix is diagonal). Moreover, finding the vector error is as simple as as adding errors in quadrature (square root of sum of squares).

What happens if my covariance matrix is not diagonal? Is there a simple number $\mu_x^*$ that encompasses the effects of the $y$ and $z$ directions? How can I find that given a covariance matrix?

  • $\begingroup$ What do you mean by finding the vector error by adding errors in quadrature? Each of your directions is an error on a different quantity - adding errors in quadrature is for when you are combining multiple sources of error on one quantity. What do you envisage the vector error to mean? $\endgroup$
    – Corvus
    Feb 26, 2013 at 7:05
  • $\begingroup$ A side note - in multiple regression people often state the standard error of the regression coefficients but actually the estimates for the different coefficients are correlated. It is possible to produce 95% confidence ellipsoids that represent the uncertainty in multiple dimensions - very much analogous to the situation you're considering. $\endgroup$
    – Silverfish
    Feb 25, 2015 at 14:03

1 Answer 1


There is no single number that encompasses all of the covariance information - there are 6 pieces of information, so you'd always need 6 numbers.

However there are a number of things you could consider doing.

Firstly, the error (variance) in any particular direction $i$, is given by

$\sigma_i^2 = \mathbf{e}_i ^ \top \Sigma \mathbf{e}_i$

Where $\mathbf{e}_i$ is the unit vector in the direction of interest.

Now if you look at this for your three basic coordinates $(x,y,z)$ then you can see that:

$\sigma_x^2 = \left[\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\right]^\top \left[\begin{matrix} \sigma_{xx} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{yx} & \sigma_{yy} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma_{zz} \end{matrix}\right] \left[\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\right] = \sigma_{xx}$

$\sigma_y^2 = \sigma_{yy}$

$\sigma_z^2 = \sigma_{zz}$

So the error in each of the directions considered separately is given by the diagonal of the covariance matrix. This makes sense intuitively - if I am only considering one direction, then changing just the correlation should make no difference.

You are correct in noting that simply stating:

$x = \mu_x \pm \sigma_x$

$y = \mu_x \pm \sigma_y$

$z = \mu_z \pm \sigma_z$

Does not imply any correlation between those three statement - each statement on its own is perfectly correct, but taken together some information (correlation) has been dropped.

If you will be taking many measurements each with the same error correlation (supposing that this comes from the measurement equipment) then one elegant possibility is to rotate your coordinates so as to diagonalise your covariance matrix. Then you can present errors in each of those directions separately since they will now be uncorrelated.

As to taking the "vector error" by adding in quadrature I'm not sure I understand what you are saying. These three errors are errors in different quantities - they don't cancel each other out and so I don't see how you can add them together. Do you mean error in the distance?

  • $\begingroup$ Yes, I mean error in the total distance, sorry for confusion. $\endgroup$
    – Dang Khoa
    Feb 26, 2013 at 17:51
  • $\begingroup$ But the distance is not $d = x + y + z$ (unless you actually do mean Taxi-Cab distance?), so the errors won't add in quadrature will they? If we assume normality then $d^2 = x^2 + y^2 + z^2$ will have a non-central Chi-squared distribution on 3 degrees of freedom. I think the distribution of distance is going to start getting messy without some simplifying approximations. $\endgroup$
    – Corvus
    Feb 26, 2013 at 18:49
  • $\begingroup$ @Corone, when you say "Firstly, the error in any particular direction" Are you referring to the variance by saying the error? $\endgroup$
    – CroCo
    Feb 24, 2015 at 19:28
  • 1
    $\begingroup$ @croco yes that's right since what we are starting with is covariance $\endgroup$
    – Corvus
    Feb 24, 2015 at 20:37

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