1
$\begingroup$

I have two sets of measurements, $x_i$ and $y_i$, both with the same number of elements, $N$. For each of these sets, I have known $N\times N$ covariance matrices, $\Sigma_x$ and $\Sigma_y$, representing the uncertainties in these measurements. The two sets of measurements have independent uncertainties with respect to each other. I want to take the element-wise quotients, $z_i = x_i / y_i$ and find the covariance matrix $\Sigma_z$. I know how to do this for single elements, but not sets of $N$ elements.

I know the standard propagation of error for a quotient, $\sigma^2_z/z^2 = \sigma_x^2 / x^2+ \sigma_y^2 / y^2$ (there's no correlation term because $x$ and $y$ are uncorrelated). I'm just not sure how this generalizes to the case of covariance matrices and vectors of measurements.

My intuition is that there must be a generalization that looks like $a \Sigma_z b = c \Sigma_x d + e \Sigma_y f$, where I need to find the forms of the matrices $a, b,c,d,e,$ and $f$, which are all $N \times N$ matrices somehow related to my measurement vectors, $x,y,$ and $z$.

$\endgroup$
2
  • $\begingroup$ If everything is elementwise, why would the covariance-matrices be relevant? Just use the element-wise solution. $\endgroup$ Dec 19 '20 at 11:58
  • $\begingroup$ Because I need to know the covariance between the N elements of z. $\endgroup$
    – aetb
    Dec 19 '20 at 21:09
1
$\begingroup$

Do the same you do to find the approximate solution $$ \sigma^2_z/z^2 = \sigma_x^2 / x^2+ \sigma_y^2 / y^2 $$ which is base on approximating the function $f(x,y)= x/y $ by its first order Taylor series (around $x_0, y_0$ which we can take to be the expectations) $$ f(x,y) \approx x_0/y_0 +f_x'(x_0,y_0) (x-x_0) + f_y'(x_0,y_0) (y-y_0) \\ = x_0/y_0 + \frac1{y_0} (x-x_0) - \frac{x_0}{y_0^2} (y-y_0) $$ and calculating the variance of this approximation gives the variance formula, after some manipulations. Now do the same but with covariance, and remember the bilinearity of covariance: $$ \DeclareMathOperator{\C}{\mathbb{C}} \C(z_i, z_j)\approx \C\left( \frac1{y_{i0}} (x_i-x_{i0}) - \frac{x_{i0}}{y_{i0}^2} (y_i-y_{i0}), \frac1{y_{j0}} (x_j-x_{j0}) - \frac{x_{j0}}{y_{j0}^2} (y_j-y_{j0}) \right) $$ After some manipulation that I leave for you this gives the approximation $$\frac{\sigma_{zij}}{z_i z_j} = \frac{\sigma_{xij}}{x_i x_j} + \frac{\sigma_{yij}}{y_i y_j} $$ If you want that can be written in matrix form. Let $D_x, D_y, D_z$ be diagonal matrices with the subscript vector along the diagonal. Then $$ \Sigma_z= D_z \left\{ D_x^{-1}\Sigma_x D_x^{-1} + D_y^{-1}\Sigma_y D_y^{-1} \right\} D_z $$ (which is easy to see using the rules from Intuition for the Product of Vector and Matrices.)

$\endgroup$
1
  • 1
    $\begingroup$ I just finished working through the math myself; it all makes sense. Thanks! $\endgroup$
    – aetb
    Dec 21 '20 at 18:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.