Suppose I have two bivariate random vectors $\tilde{W}, \bar{W}$ distributed as follows $$ \bar{W}\sim \mathcal{N}(\begin{pmatrix}1\\-1 \end{pmatrix}, \begin{pmatrix} 1 & \rho_{\bar{W}}\\ \rho_{\bar{W}} & 1 \end{pmatrix}) $$ $$ \tilde{W}\sim \mathcal{N}(\begin{pmatrix}-1\\1 \end{pmatrix}, \begin{pmatrix} 1 & \rho_{\tilde{W}}\\ \rho_{\tilde{W}} & 1 \end{pmatrix}) $$ with $\bar{\rho}_W, \tilde{\rho}_w \in [-1,1]$. Take a bivariate random vector $W$ distributed as $\frac{1}{2}\bar{W}+\frac{1}{2}\tilde{W}$ (equally weighted mixture).
Could you help me to derive $E(W)$ and $Var(W)$? I found some directions for univariate normal (e.g., here) but I couldn't find the analysis for the multivariate case.
To calculate the expectation of $W=0.5 \tilde{W} + 0.5\overline{W}$, you should use the linearity property of expectations ($\mathbb{E}(X+Y)=\mathbb{E}(X)+\mathbb{E}(Y)$) and to calculate the variance you essentially need to calculate $$ \mathbb{E}(WW') = \mathbb{E}((0.5 \tilde{W} + 0.5 \overline{W})(0.5 \tilde{W} + 0.5 \overline{W})') = 0.25\mathbb{E}\left( \tilde{W} \tilde{W}' + \tilde{W} \overline{W}' + \overline{W} \tilde{W}' + \overline{W}\overline{W}'\right)$$ Here, you should use again the linearity property of the expectation and that $\tilde{W}$ and $\overline{W}$ are independent (thus the expectation of the product, is the product of the expectations). Then invoke the assumptions, i.e., you know for example $\mathbb{E}(\tilde{W}\tilde{W}') = \text{Var}(\tilde{W}) + \mathbb{E}(\tilde{W})\mathbb{E}(\tilde{W}')$.
If you work this out, you will get the result. If you need more information on the matter, I can provide you with it. I hope this helps!
edit: changed the notation to make it clearer.