proof of mean square convergence of the sample mean of a stationary process

n\operatorname{var}(\bar{X}_{n})=\frac{1}{n}\sum_{i=1}^{n}\sum_{j=1}^{n}% \operatorname{cov}(X_{i},X_{j})\\ =\sum_{|h|<n}(1-\frac{|h|}{n})\gamma(h)\leq\sum_{|h|<n}|\gamma(h)|

If $\gamma(n)\to 0$ as $n\to\infty$ then $\lim_{n\to\infty}\frac{1}{n}\sum_{|h|<n}|\gamma(h)|=2\lim_{n\to\infty}|\gamma(% n)|=0$ , whence $\operatorname{var}[\bar{X}_{n}]\to 0$ .
If $\sum_{h=-\infty}^{\infty}|\gamma(h)|<\infty$ then the dominated Convergence theorem gives

\lim_{n\to\infty}\sum_{|h|<n}(1-\frac{|h|}{n})\gamma(h)=\sum_{h=-\infty}^{% \infty}\gamma(h)

Title	proof of mean square convergence of the sample mean of a stationary process
Canonical name	ProofOfMeanSquareConvergenceOfTheSampleMeanOfAStationaryProcess
Date of creation	2013-03-22 15:22:19
Last modified on	2013-03-22 15:22:19
Owner	georgiosl (7242)
Last modified by	georgiosl (7242)
Numerical id	6
Author	georgiosl (7242)
Entry type	Proof
Classification	msc 60G10