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I know that if a process is weakly stationary, the mean of the process will be time-independent and the covariance will be a function of the time difference.

My question: if I have a time-independent mean and a covariance, which is a function of the time difference, can I always construct a weakly stationary process (with those mean and covariance function)?

Kindly cite appropriate material, if possible. Thanks!

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So long as you have a mean $\mu$ and an auto-covariance function $\gamma$, then for any $n \in \mathbb{N}$ you can define the observable vector $\mathbf{X} = (X_1,...,X_n)$ by:

$$\mathbf{X} \sim \text{N} ( \mu \mathbf{1}, \mathbf{\Sigma} ) \quad \quad \quad \mathbf{\Sigma} \equiv \begin{bmatrix} \gamma(0) & \gamma(1) & \cdots & \gamma(n-1) \\ \gamma(1) & \gamma(0) & \cdots & \gamma(n-2) \\ \vdots & \vdots & \ddots & \vdots \\ \gamma(n-1) & \gamma(n-2) & \cdots & \gamma(0) \\ \end{bmatrix}.$$

This can be extended to arbitrary $n$ so it is sufficient to form the stationary process $\{ X_t | t \in \mathbb{Z} \}$ according to the designated mean and auto-covariance function.

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