Can I convert a covariance matrix into uncertainties for variables?

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}

(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?


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:

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

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

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

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?

Source : Link , Question Author : Dang Khoa , Answer Author : Corvus

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