[proofplan] We center the process and express the mean-square error of the sample mean as the variance of a finite average. Weak stationarity converts each covariance term in the double sum into a function only of the lag $s-t$. Grouping the double sum by lags produces the triangular weights $1-|h|/n$, and the assumed averaged absolute summability condition forces the resulting variance to vanish. [/proofplan]

[step:Center the sample mean and reduce mean-square convergence to a variance estimate] For each $t\in\mathbb Z$, define the centered [random variable](/page/Random%20Variable) $Y_t:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ by \begin{align*} Y_t=X_t-\mu. \end{align*} For each $n\in\mathbb N$, define the centered sample mean $\bar Y_n:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ by \begin{align*} \bar Y_n=\bar X_n-\mu=\frac{1}{n}\sum_{t=1}^{n}Y_t. \end{align*} Linearity of expectation gives $\mathbb E[Y_t]=0$ for every $t\in\mathbb Z$, hence $\mathbb E[\bar Y_n]=0$ for every $n\in\mathbb N$. For a square-integrable real-valued random variable $Z:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$, define its variance by \begin{align*} \operatorname{Var}(Z)=\mathbb E[(Z-\mathbb E[Z])^2]. \end{align*} Applying this definition to $Z=\bar Y_n$ and using $\mathbb E[\bar Y_n]=0$, we obtain \begin{align*} \mathbb E[(\bar X_n-\mu)^2]=\mathbb E[\bar Y_n^2]=\operatorname{Var}(\bar Y_n). \end{align*} It remains to prove that $\operatorname{Var}(\bar Y_n)\to 0$. [/step]

[step:Expand the variance as a double covariance sum]Since each $X_t$ is square-integrable, each $Y_t$ is square-integrable. For every $s,t\in\{1,\dots,n\}$, the [Cauchy-Schwarz inequality](/theorems/432) gives \begin{align*} \mathbb E[|Y_sY_t|]\le \mathbb E[Y_s^2]^{1/2}\mathbb E[Y_t^2]^{1/2}<\infty, \end{align*} so each product $Y_sY_t$ is integrable. The sum defining $\bar Y_n$ is finite: \begin{align*} \bar Y_n=\frac{1}{n}\sum_{t=1}^{n}Y_t. \end{align*} Therefore bilinearity of covariance applies to this finite linear combination, and \begin{align*} \operatorname{Var}(\bar Y_n)=\mathbb E\left[\left(\frac{1}{n}\sum_{t=1}^{n}Y_t\right)^2\right]. \end{align*} Expanding the square and using linearity of expectation over the finite double sum gives \begin{align*} \operatorname{Var}(\bar Y_n)=\frac{1}{n^2}\sum_{s=1}^{n}\sum_{t=1}^{n}\mathbb E[Y_sY_t]. \end{align*} By weak stationarity, for every $s,t\in\{1,\dots,n\}$, \begin{align*} \mathbb E[Y_sY_t]=\mathbb E[(X_s-\mu)(X_t-\mu)]=\gamma(s-t). \end{align*} Therefore \begin{align*} \operatorname{Var}(\bar Y_n)=\frac{1}{n^2}\sum_{s=1}^{n}\sum_{t=1}^{n}\gamma(s-t). \end{align*}[/step]

[guided]The reason for centering is that the sample mean error is itself an average of centered variables. Define $Y_t=X_t-\mu$ for each $t\in\mathbb Z$. Then $\mathbb E[Y_t]=\mathbb E[X_t]-\mu=0$, and \begin{align*} \bar X_n-\mu=\frac{1}{n}\sum_{t=1}^{n}(X_t-\mu)=\frac{1}{n}\sum_{t=1}^{n}Y_t. \end{align*} Because only finitely many square-integrable random variables are being summed, the square of the sum expands into finitely many products. Each product is integrable: for $s,t\in\{1,\dots,n\}$, the Cauchy-Schwarz inequality gives \begin{align*} \mathbb E[|Y_sY_t|]\le \mathbb E[Y_s^2]^{1/2}\mathbb E[Y_t^2]^{1/2}<\infty. \end{align*} Thus we may expand the square and move expectation through the finite sum: \begin{align*} \mathbb E[(\bar X_n-\mu)^2]=\mathbb E\left[\left(\frac{1}{n}\sum_{t=1}^{n}Y_t\right)^2\right]. \end{align*} Expanding the product gives one term for every ordered pair $(s,t)\in\{1,\dots,n\}^2$: \begin{align*} \mathbb E[(\bar X_n-\mu)^2]=\frac{1}{n^2}\sum_{s=1}^{n}\sum_{t=1}^{n}\mathbb E[Y_sY_t]. \end{align*} Weak stationarity is used precisely here. Since $Y_s=X_s-\mu$ and $Y_t=X_t-\mu$, the covariance between $X_s$ and $X_t$ depends only on the lag $s-t$: \begin{align*} \mathbb E[Y_sY_t]=\mathbb E[(X_s-\mu)(X_t-\mu)]=\gamma(s-t). \end{align*} Substituting this into the double sum yields \begin{align*} \mathbb E[(\bar X_n-\mu)^2]=\frac{1}{n^2}\sum_{s=1}^{n}\sum_{t=1}^{n}\gamma(s-t). \end{align*} Thus the mean-square error has been converted into a purely deterministic finite sum involving the autocovariance function.[/guided]

[step:Group the double sum by lag] For a fixed $n\in\mathbb N$, define the lag-counting function $N_n:\mathbb Z\to\{0,1,\dots,n\}$ by \begin{align*} N_n(h)=\#\{(s,t)\in\{1,\dots,n\}^2:s-t=h\}. \end{align*} If $|h|\ge n$, then $N_n(h)=0$. If $0\le h<n$, then $s=t+h$ and $t$ ranges over $\{1,\dots,n-h\}$, so $N_n(h)=n-h$. If $-n<h<0$, then $t=s-h$ and $s$ ranges over $\{1,\dots,n+h\}$, so $N_n(h)=n+h$. Hence, for every $h\in\mathbb Z$ with $|h|<n$, \begin{align*} N_n(h)=n-|h|. \end{align*} Grouping the ordered pairs $(s,t)$ by their lag $h=s-t$ gives \begin{align*} \sum_{s=1}^{n}\sum_{t=1}^{n}\gamma(s-t)=\sum_{|h|<n}(n-|h|)\gamma(h). \end{align*} Therefore \begin{align*} \operatorname{Var}(\bar Y_n)=\frac{1}{n}\sum_{|h|<n}\left(1-\frac{|h|}{n}\right)\gamma(h). \end{align*} [/step]

[step:Bound the triangular covariance average by the assumed absolute average] For every integer $h$ with $|h|<n$, the triangular weight satisfies \begin{align*} 0\le 1-\frac{|h|}{n}\le 1. \end{align*} Taking absolute values and using the triangle inequality for the finite sum, we obtain \begin{align*} |\operatorname{Var}(\bar Y_n)|\le \frac{1}{n}\sum_{|h|<n}\left(1-\frac{|h|}{n}\right)|\gamma(h)|. \end{align*} Since each weight is at most $1$, \begin{align*} |\operatorname{Var}(\bar Y_n)|\le \frac{1}{n}\sum_{|h|<n}|\gamma(h)|. \end{align*} By hypothesis, the right-hand side tends to $0$ as $n\to\infty$. Hence $\operatorname{Var}(\bar Y_n)\to 0$. [/step]

[step:Conclude mean-square convergence of the sample mean] From the first step, \begin{align*} \mathbb E[(\bar X_n-\mu)^2]=\operatorname{Var}(\bar Y_n). \end{align*} The previous step proves that the right-hand side tends to $0$. Therefore \begin{align*} \mathbb E[(\bar X_n-\mu)^2]\to 0 \end{align*} as $n\to\infty$, which is exactly the assertion that $\bar X_n\to\mu$ in mean square. [/step]