# Error propagation with covariance matrix [closed]

I have two measurements $$X = x \pm \sigma_{x}$$ , $$Y= y \pm \sigma_{y}$$, where $$x,y$$ are the mean values of $$X$$ and $$Y$$ and $$\sigma_{x},\sigma_{y}$$ are their corresponding uncertainties. The covariance matrix ($$\Sigma$$) for them is the following:

$$$$\Sigma = \begin{pmatrix} \sigma_{x}^{2} & \sigma_{xy}\\ \sigma_{yx} & \sigma_{y}^{2} \end{pmatrix}$$$$

Also I have a number $$\textbf{Z} = z \pm \sigma_{z}$$.

Suppose that I create a vector $$V = \left \{ X,Y \right \}$$ and I want to multiply the vector by the constant $$\textbf{Z}$$, what do I have to do in order to obtain the correct covariance matrix for V?

I know that first I have to do this multiplication $$z\Sigma$$, but I don't know how to do the error propagation. What should I do with $$\sigma_{z}$$? Should I add in quadrature $$\sigma_{z}$$ to $$\Sigma$$?

• Just to confirm, given $V = (X,Y)$, where $X$ and $Y$ are random variables with means $x$ and $y$ and variances $\sigma_x^2$ and $\sigma_y^2$ respectively, you want to know the covariance matrix for $ZV = (ZX,ZY)$, where $Z$ is a random variable with mean $z$ and variance $\sigma_z^2$. Is this right? Oct 24, 2022 at 14:17
• Is the estimate of $Z$ independent of those for $X$ and $Y$ (e.g. $\sigma_{xz} = \sigma_{yz} = 0$)?
– Eoin
Oct 25, 2022 at 10:59
• @mhdadk yes that is correct.
– Cruz
Oct 25, 2022 at 15:10
• @Eoin, yes Z is independent of X and Y.
– Cruz
Oct 25, 2022 at 15:10
• I strongly recommend you revising the notations. For example, "$X = x \pm \sigma_x$" is better to be replaced by $X = x + \epsilon$, where $E(\epsilon) = 0$ and $\operatorname{Var}(\epsilon) = \sigma_x^2$. Feb 11, 2023 at 2:20

You just need to apply the definition of Covariance to the product of the random variables:

Let $$W = ZV$$ be the random variable you are interested in, and note that by the independence of $$Z$$ and $$V$$ you have $$w \equiv E[W]=E[Z]E[V] \equiv zv$$ where $$v=(x,y)$$.

Now calculate directly the covariance of $$W$$:

$$\Sigma_w \equiv Cov(W,W) = E[(W - E[W])(W - E[W])^T] = E[WW^T] - ww^T$$

noting that (again using the independence of $$Z$$ and $$V$$)

$$E[WW^T] = E[Z^2VV^T]=E[Z^2]E[VV^T]=(z^2+\sigma_z^2)(\Sigma + vv^T)$$

you get

$$\Sigma_w = (z^2+\sigma_z^2)\Sigma + \sigma_z^2vv^T$$

$$\;\;\;\;\;\; = (z^2+\sigma_z^2)\begin{pmatrix} \sigma_{x}^{2} & \sigma_{xy}\\ \sigma_{xy} & \sigma_{y}^{2} \end{pmatrix} + \sigma_z^2 \begin{pmatrix} x^2 & xy\\ xy & y^2 \end{pmatrix}.$$