Suppose $\mathbf{Z_1}, \dots, \mathbf{Z}_n$ be iid random vectors of normal distribution s.t. $\mathbf{Z}_i \sim N(\boldsymbol{\mu},\boldsymbol{\Sigma})$ where $\mathbf{Z}_i = (X_i, Y_i)^T$, $\boldsymbol{\mu} = (\mu_1,\mu_2)^T$, and $\boldsymbol{\Sigma} = \begin{pmatrix} \sigma_{11} & \sigma_{12} \\ \sigma_{21} & \sigma_{22} \end{pmatrix} $ for all $i = 1, \dots, n$.
I'm interested in the joint distribution of say $(\bar{X},\bar{Y})^T$, where $\bar{X} = \frac{1}{n}\sum_i X_i$ and $\bar{Y} = \frac{1}{n}\sum_j Y_j$.
My intuition is that $(\bar{X},\bar{Y})^T$ is just a linear function of $\mathbf{Z_1}, \dots, \mathbf{Z}_n$, i.e.
$$\frac{1}{n}(\mathbf{Z}_1 + \dots + \mathbf{Z}_n) = \frac{1}{n}\left((X_1,Y_1)^T + \dots + (X_n, Y_n)^T\right) = (\bar{X},\bar{Y})^T.$$ Hence it is of normal distribution itself and calculating the mean and variance give that $(\bar{X},\bar{Y})^T \sim N\left(\boldsymbol{\mu}, \frac{1}{n}\boldsymbol{\Sigma}\right)$.
Is this line of thinking correct?
