= Solution
A <weakly stationary process> has finite <second moments>, a constant <expected value> $\mu$, and <covariance> depending only on the time difference:
$$
\mathbb EX_t=\mu,\qquad\operatorname{Cov}(X_{t+h},X_t)=\gamma(h).
$$
A <strictly stationary process>, also called strongly stationary, has every finite-dimensional <probability distribution> invariant under a common shift: for every finite choice of times and every shift $h$, $(X_{t_1+h},\ldots,X_{t_m+h})$ and $(X_{t_1},\ldots,X_{t_m})$ have the same law. <strict stationarity> with finite <second moments> implies <weak stationarity>. <weak stationarity> alone does not determine the full joint law, and <strict stationarity> alone does not guarantee finite moments.
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