Autocorrelation of signal with offset
Let me just write an answer because what I wrote above about the expectation is not correct and extremely confusing because I didn't show how the $b^2$ drops out.

By definition:

$$\operatorname{cov}(X, Y) \triangleq E(XY) - E(X)E(Y)$$

So, $$\begin{align}
\operatorname{cov}\big((X_{t+\tau} + b),(X{_t} + b)\big) 
 &amp;= E\big((X_{t+\tau} + b)(X_{t} + b)\big) - E\big(X_{t+\tau} + b\big)\times E\big(X_{t} + b\big) \\
 &amp;= E\big(X_{t+\tau} \times X_{t}\big) + b^2 - \Big(E\big(X_{t+\tau}\big) \times E\big(X_{t}) + b^2\Big) \\
 &amp;= E\big(X_{t+\tau} \times X_{t}\big)  - \Big(E\big(X_{t+\tau}\big) \times E\big(X_{t}\big)\Big) \\
 &amp;= \operatorname{cov}\big(X_{t+\tau}, X_{t}\big) \\
 &amp;= \gamma(\tau) \\
\end{align}$$

Apologies for confusion.