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:

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

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 $\Sigma$?

Could you provide any reference to look for information about this topic?

  • $\begingroup$ 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? $\endgroup$
    – mhdadk
    Oct 24, 2022 at 14:17
  • $\begingroup$ Is the estimate of $Z$ independent of those for $X$ and $Y$ (e.g. $\sigma_{xz} = \sigma_{yz} = 0$)? $\endgroup$
    – Eoin
    Oct 25, 2022 at 10:59
  • $\begingroup$ @mhdadk yes that is correct. $\endgroup$
    – Cruz
    Oct 25, 2022 at 15:10
  • $\begingroup$ @Eoin, yes Z is independent of X and Y. $\endgroup$
    – Cruz
    Oct 25, 2022 at 15:10
  • 1
    $\begingroup$ 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 $\operatorname{Var}(\epsilon) = \sigma_x^2$. $\endgroup$
    – Zhanxiong
    Feb 11, 2023 at 2:20

1 Answer 1


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}.$


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