# Joint distribution of sample means from multivariate normal

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?

• Yes, its correct Commented Oct 14, 2019 at 14:32