# Univariate Normal Converted to Multivariate Normal: Covariance Derivation

I am reading the paper available at this link: https://drive.google.com/file/d/0B2_rKFnvrjMARnU1QjB4anR3RDA/edit?usp=sharing

I am having trouble understanding section 5.1 (page 2741).

Essentially it says the following:

$\theta_{ABi} \sim \mathrm{N}(\mu_{AB}, \tau^2)$

$\theta_{ACi} \sim \mathrm{N}(\mu_{AC}, \tau^2)$

$\theta_{BCi} \sim \mathrm{N}(\mu_{BC}, \tau^2)$

$\mu_{BC} = \mu_{AC}-\mu_{AB}$

implies

$\begin{pmatrix}\theta_{ABi} \\ \theta_{ACi}\end{pmatrix} \sim \mathrm{N} \left(\begin{pmatrix}\mu_{ABi} \\ \mu_{ACi}\end{pmatrix}, \begin{pmatrix}\tau^2 & \tau^2/2 \\ \tau^2 /2 & \tau^2\end{pmatrix} \right)$

I do not understand how $\mathrm{Cov} \left[ \theta_{ABi}, \theta_{ACi} \right] = \tau^2 /2$ ? Could someone please explain this?

Variable $\theta_{BC}$ is defined as $\theta_{AC} - \theta_{AB}$ as can be seen from the definition of $E(\theta_{BC}) = \mu_{BC}$. That being said: \begin{align} \textrm{Var}(\theta_{BC})&=\textrm{Var}(\theta_{AC} - \theta_{AB})\\ \textrm{Var}(\theta_{BC})&=\textrm{Var}(\theta_{AC}) + (-1^2)\cdot\textrm{Var}(\theta_{AB}) + 2(1)(-1)\textrm{Cov}(\theta_{AC}, \theta_{AB})\\ \end{align} Substituting, we get: \begin{align} \tau^2 &= \tau^2 + \tau^2 - 2\textrm{Cov}(\theta_{AC}, \theta_{AB})\\ 2\textrm{Cov}(\theta_{AC}, \theta_{AB}) &= \tau^2\\ \textrm{Cov}(\theta_{AC}, \theta_{AB}) &= \frac{\tau^2}{2} \end{align}