Say for example I have the white noise process $Y_t\sim\text{WN}(\mu,\sigma^2)$. Is it true that $\mathbb{E}[Y_t]=\mathbb{E}[Y_{t-h}]$, where $h\in\mathbb{N}?$
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2$\begingroup$ Yes. This is true for both the strong and weak version of stationarity. Be aware, that in many cases when working with time-series analysis, we only assume weak stationarity (also called covariance stationarity), ie. the mean and variance do not vary with time. Nevertheless, a strictly stationary process with finite mean and variance is also weakly stationary. I'd advise you to read the wiki page. $\endgroup$– PlebCommented Apr 17, 2021 at 17:02
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$\begingroup$ @Pleb - Thanks. But how come then according to definition, if $X_t$ is a stochastic process with mean function $\mu_X(t)$, then the covariance function is given by $\text{Cov}(X_r,X_s)=E(X_rX_s)-\mu_X(r)\mu_X(s)?$ Shouldn't by your comment $\mu_X(r) = \mu_X(s) = \mu$ so we can just set $-\mu^2?$ $\endgroup$– ParsevalCommented Apr 17, 2021 at 17:44
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2$\begingroup$ For a weak sense stationary process we can indeed do what you say and have $-\mu^2$ in the covariance formula. $\endgroup$– nbbo2Commented Apr 17, 2021 at 18:20
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