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The scalar process has zero mean, because white noise has zero mean. Expanding products and retaining only matching noise indices gives its autocovariance function:The finite second moment and covariance depending only on the lag prove weak stationarity. With the two-sided angular-frequency convention, the spectral density of a stationary process isIt is nonnegative because the bracket is $|1+\theta e^{i\lambda }{|}^2$, and it is even because the cosine is even. Hence $f(-\lambda )=f(\lambda )$. No restriction on the real parameter $\theta$ is needed for stationarity here; invertibility is a separate condition.

For the bivariate calculation let $a={\sigma }_u^2$, $b={\sigma }_v^2$, $D=\mathrm{diag}(a,b)$, and $\Theta =({\theta }_{ij})$. Uncorrelated white noise processes here means zero cross-covariances at every pair of times. Therefore the vector noise hasThe filtered vector is the multivariate moving-average model $Z_t=W_t+\Theta W_{t-1}$. Linearity gives $\mathbb{E}{X}_t=\mathbb{E}{Y}_t=0$. Put $\Gamma (h)=\mathbb{E}[Z_t{Z}_{t+h}^T]$, with exactly the orientation of the printed covariance definition. At lag zero, the present and past noise terms are uncorrelated, giving $D+\Theta D{\Theta }^T$. At lag one the only shared noise is $W_t$, contributing $D{\Theta }^T$; at lag minus one it contributes $\Theta D$. Thus the covariance and cross-spectrum of a vector MA(1) are determined byExplicitly,In particular the off-diagonal positive- and negative-lag covariances need not coincide. Reading the upper-right entries and taking their Fourier series gives the requested cross-spectrum:Every coefficient is real, so replacing $\lambda$ by $-\lambda$ conjugates both complex exponentials. Therefore $f_{XY}(-\lambda )=\overline{f_{XY}(\lambda )}$. Unlike the scalar auto-spectrum, this cross-spectrum may be complex and need not be even: its imaginary part is $(b{\theta }_{12}-a{\theta }_{21})\sin \lambda /(2\pi )$.