# Variance of Sample Mean in Time Series

The following is a proposition which is important for deriving the distribution of the sample mean for a covariance stationary process.

Proposition:

Let $\{x_{t}\}$ be a mean zero covariance stationary time series with $\sum_{k=0}^{\infty}|\gamma_{x}(k)|<\infty$ ( where $\gamma_{x}(k)$ is the autocovariance function at lag $k$).

Then $\frac{1}{T} \mathbb{V} \left[ \sum_{t=1}^{T} x_{t} \right] = \gamma_{x}(0) + 2 \sum_{k=1}^{\infty}\gamma_{x}(k)$

Proof:

My proof thus far is:

\begin{align*} \frac{1}{T} \mathbb{V} \left[ \sum_{t=1}^{T} x_{t} \right] &= \frac{1}{T}\left[ \sum_{t=1}^{T} \mathbb{V}[x_{t}] + \mathop{\sum^T\sum^T}_{i\neq j} Cov(x_{i}, x_{j}) \right]\\ &= \frac{1}{T}\left[ \sum_{t=1}^{T} \mathbb{V}[x_{t}] + 2 \mathop{\sum^T\sum^T}_{i < j} Cov(x_{i}, x_{j}) \right]\\ &= \frac{1}{T}\left[ T\cdot \gamma_{x}(0) + 2 (T-1) \cdot \gamma_{x}(1) + 2(T-2) \cdot \gamma_{x}(2) + \ldots + 4 \gamma_{x}(T-2) + 2 \gamma_{x}(T-1) \right]\\ &= \frac{1}{T}\left[ T\cdot \gamma_{x}(0) + 2 \sum_{i=1}^{T-1} (T-i) \cdot \gamma_{x}(i) \right]\\ &= \gamma_{x}(0) + 2 \sum_{i=1}^{T-1} \frac{(T-i)}{T} \cdot \gamma_{x}(i) \\ \end{align*} Now all I need to show is: \begin{align*} \lim_{T\to\infty} \gamma_{x}(0) + 2 \sum_{i=1}^{T-1} \frac{(T-i)}{T} \cdot \gamma_{x}(i) = \gamma_{x}(0) + 2 \sum_{i=1}^{\infty} \gamma_{x}(i) \\ \end{align*}

However I am having trouble with this limit and am worried it does not converge.

Context:

This proposition was stated by my professor in a time series class and has been used to prove many other results. However it is possible I have approached the proof in the wrong way. Any input will help.

• – Taylor Apr 29 '17 at 21:34

The upper limit of your sum should be $T-1$ (since the $t=T$ term is zero). We can write
$$\sum_{i=1}^{T-1}\frac{T-i}T\gamma_x(i)=\sum_{i=1}^{T-1}\left(1-\frac iT\right)\gamma_x(i):=a_T.$$ Clearly $a_T$ need not converge as $T\to\infty$. However, $a_T$ converges if and only if $$\sum_{i=-\infty}^\infty \gamma_x(i)<\infty$$ (from my time series notes, a proof of this statement is escaping me at the moment). When that limit exists, your result holds.
• I have incorporated this minor change to the limit of the sum. However, the issue of the convergence of $a_{T}$ is at the crux of the proof; not only must $a_{T}$ converge, but it must converge to the desired limit above. Your comment gives me hope, but the proof of your comment is needed to resolve the issue. – möbius Feb 22 '15 at 14:45
• Actually, elementary computations show that the result holds if and only if $$\frac1T\sum_{i=1}^Ti\gamma(i)\to0,$$ a condition which is implied by the convergence of the series $$\sum_{i}\gamma(i).$$ – Did Feb 22 '15 at 16:32
• Rewrite the first sums in terms of the rests $\Gamma(i)=\sum\limits_{k\geqslant i}\gamma(k)$ of the second series, which, by hypothesis, are finite and converge to zero. – Did Feb 22 '15 at 19:01