= Solution
For a <weakly stationary process>, every observation has the same finite <expectation> $\mu$. Linearity of <expectation>, without any assumption of <independence>, gives
$$
\boxed{\mathbb E\overline Y=\frac1n\sum_i\mathbb E Y_i=\mu.}
$$
The <variance> is a sum of all pairwise <covariances>:
$$
\operatorname{Var}(\overline Y)=\frac1{n^2}\sum_{i,j=1}^n\gamma(i-j).
$$
For a real-valued <weakly stationary process>, the <autocovariance> is even. There are $n$ pairs at lag zero and $n-h$ pairs in each direction at positive lag $h$. Therefore
$$
\boxed{\operatorname{Var}(\overline Y)=\frac{\gamma(0)}n+\frac2{n^2}\sum_{h=1}^{n-1}(n-h)\gamma(h).}
$$
Positive <autocovariances> inflate the uncertainty relative to the independent-observation formula; negative terms may reduce it. An <unbiased estimator> need not have <variance> tending to zero: a process consisting of the same random variable at every time is weakly stationary, but averaging cannot reduce its <variance>.
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