# Calculating Mean Vector and Covariance Matrix of Mixture of Multivariate Normal Distributions [duplicate]

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.