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In an effort to better understand multivariate normal distributions, I am attempting to derive the mean vector and covariance matrix of multivariate random vector defined by a mixture distribution. This is the scenario I have concocted:

I have two 2-dimensional multivariate normal random vectors, say $\mathbf{X} = (X_1, X_2)^{T} \sim \text{N}_2{(\mu_X, \Sigma_X)}$ and $\mathbf{Y} = (Y_1, Y_2)^{T} \sim \text{N}_2{(\mu_Y, \Sigma_Y)}$, and I am defining the random vector $\mathbf{W} = (W_1, W_2)^{T}$ by its PDF of the form $$ f_{\mathbf{W}}(x,y) = \frac{1}{2} f_{\mathbf{X}}(x,y) + \frac{1}{2} f_{\mathbf{Y}}(x,y) $$ where $f_{\mathbf{X}}$ and $f_{\mathbf{Y}}$ are the joint PDFs for $\mathbf{X}$ and $\mathbf{Y}$, respectively. Is there a simple way to find the mean vector $\mu_W$ and covariance matrix $\Sigma_W$ for the random vector $\mathbf{W}$, or at least a simpler way than expanding the terms within the PDFs $f_{\mathbf{X}}$ and $f_{\mathbf{Y}}$ and taking all the integrals by hand?

Using my knowledge of non-multivariate mixture distributions, I have a hunch that it is simple as $$ \mu_W = \frac{1}{2} \mu_X + \frac{1}{2} \mu_Y $$ and finding the elements of the covariance matrix $\Sigma_W$ could be as equally simply (with some additional work needed to find variances and such). However, I am having a hard time fully convincing myself of this hunch. Any and all help is greatly appreciated.

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