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I am trying to understand a proof that shows, that the sum over the autocorrelation starting with lag=1 is always equal -1/2, for a stationay time series. The sum looks like this:

$$ S_{\rm{afc}}=\sum_{h=1}^{T-1} \hat \rho(h)=\sum^{T-1}_{h=1}\Big( \frac{\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y) } {\sum^{T}_{t=1} (y_t - \overline y)^2 }\Big)$$

with $\hat \rho(h)=\frac{\hat \gamma(h)}{\hat \gamma(0)}=\frac{\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y) } {\sum^{T}_{t=1} (y_t - \overline y)^2 }$ the autocorrelation function defined in terms of an estimate $\hat \gamma = \sum^{T-h}_{h=1} (y_t - \overline y)(y_{t+h} - \overline y)$ of the the autocovariance and $\overline y= { 1 \over T} \sum^T_{t=1} y_t$ the sample mean.

The proof is short: $$ S_{\rm{afc}}=\sum^{T-1}_{h=1}\Big( \frac{\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y) } {\sum^{T}_{t=1} (y_t - \overline y)^2 }\Big)=\Big( \frac{ \sum^{T-1}_{h=1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y) } {\sum^{T}_{t=1} (y_t - \overline y)^2 }\Big)\\=\Big( \frac{ \sum^{T-1}_{h=1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y) } {\big(\sum^{T}_{t=1} (y_t - \overline y)\big)^2-2 \sum_{h=1}^{T-1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y)}\Big)$$ until here I can understand everything but the last step is a mystery to me: $$\Big( \frac{ \sum^{T-1}_{h=1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y) } {\big(\sum^{T}_{t=1} (y_t - \overline y)\big)^2-2 \sum_{h=1}^{T-1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y)}\Big)=\Big( \frac{ \sum^{T-1}_{h=1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y) } {-2 \sum_{h=1}^{T-1}\sum^{T-h}_{t=1} (y_t - \overline y)(y_{t+h} - \overline y)}\Big)$$ which is of course -1/2. Why would $\big(\sum^{T}_{t=1}(y_t - \overline y)\big)^2$ vanish? I have no earthly Idea how this might be. In case you want to look at the paper is "Sum of the sample autocorrelation function" by Hossein Hassani DOI: https://doi.org/10.1515/ROSE.2009.008

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  • $\begingroup$ Isn't $\sum (y_t - \overline y) = 0$? $\endgroup$ – Gregory Apr 28 '17 at 15:05
  • $\begingroup$ It had to be something trivial, even my collegues did not see this, thanks. ^^ make it the answer if you want to. $\endgroup$ – pindakaas Apr 28 '17 at 19:07
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$\sum y_t - \overline y = 0$ Which proves the claim.

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