Auto-covariance function

Question

For the auto covariance function I'm not to understand how the simplification from the second equals to sign to the third has been done.

Any help would be much appreciated.

Answer 1

@Math1000 pretty much explains the whole situation, in case you still struggle, remember that the expectation of constant is the constant itself, so for $a\in \mathbb R$ and $X$ a random variable we have $$ E(a)=a \text{ and } E(aX)=aE(X) $$ in your case, we have $E(X_t)=\mu_t$ and $E(X_s)=\mu_s$ with $\mu_s,\mu_t\in\mathbb R$ so $E(\mu_s)=\mu_s,E(\mu_t)=\mu_t$. This means you get by expanding the product \begin{align} E((X_t-\mu_t)(X_s-\mu_s))=&E(X_tX_s+\mu_t\mu_s-X_t\mu_s-X_s\mu_t)\\ =&E(X_tX_s)+E(\mu_t\mu_s)-E(X_t\mu_s)-E(X_s\mu_t)\\ =&E(X_tX_s)+\mu_t\mu_s-\mu_sE(X_t)-\mu_tE(X_s)\\ =&E(X_tX_s)+\mu_t\mu_s-\mu_s\mu_t-\mu_t\mu_s\\ =&E(X_tX_s)-\mu_t\mu_s \end{align} and you're done.