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For two real centered time series $X_t,Y_t$, require joint weak stationarity: all auto- and cross-covariances depend only on the time lag. Individual weak stationarity alone does not impose this condition on their joint behavior. With the convention ${\gamma }_{XY}(h)=\mathbb{E}[X_t{Y}_{t+h}]$, define the cross-spectrum byAssume absolute summability of the auto- and cross-covariance sequences, so the displayed Fourier series and the two individual spectral densities are continuous. Reality and joint weak stationarity giveConsequently the spectral matrix $F(\lambda )=\left(\begin{array}{cc}{f}_{XX}& f_{XY}\\ f_{YX}& f_{YY}\end{array}\right)$ is a Hermitian matrix. It is positive semidefinite, because its quadratic forms are nonnegative spectral densities of complex linear combinations of the jointly stationary series. Its nonnegative determinant therefore yields $|f_{XY}(\lambda ){|}^2\le f_{XX}(\lambda )f_{YY}(\lambda )$.

The imaginary part records directional lag information. For white noise $X_t$ of variance $v$ and $Y_t=X_{t-1}$, only ${\gamma }_{XY}(1)=v$ is nonzero. Hence $f_{XY}(\lambda )=v{e}^{-i\lambda }/(2\pi )$, although both individual spectral densities equal $v/(2\pi )$. The cross-spectrum is generally complex, rather than real and even like a real series' own spectral density.

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 $(b{\theta }_{12}-a{\theta }_{21})\sin \lambda /(2\pi )$.