# Covariance matrix of element-wise quotient of two sets of measurements with known covariance matrices

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 $$z$$.

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

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.)