Why is the sum of the sample autocorrelations of a stationary series equal to -1/2? I cannot grasp my head around this property of stationary series and the autocorrelation function. I have to prove that 
\begin{align}
\sum_{h=1}^{n-1}\hat\rho(h)=-\frac{1}{2}
\end{align}
Where $\hat\rho(h)=\displaystyle\frac{\hat\gamma(h)}{\hat\gamma(0)}$ and $\hat\gamma(h)$ is the autocovariance function
\begin{align}
\hat\gamma(h) = \frac{1}{n}\sum_{t=1}^{n-h}(X_t-\bar{X})(X_{t+h}-\bar{X})
\end{align}
Hopefully someone can help me out with a proof, or at least point me in the right direction.
 A: Let's start by representing the sum $S$ using the definition of the autocorrelation function:
\begin{equation}
S = \sum_{h=1}^{n-1} \hat{\rho}(h) = \sum_{h=1}^{n-1} \left(\frac{\frac{1}{n}\sum_{t=1}^{n-h}(X_t-\bar{X})(X_{t+h}-\bar{X})}{\frac{1}{n}\sum_{t=1}^{n}(X_t-\bar{X})^2}\right)
\end{equation}
Denominator does not depend on $h$ so we can simplify and move the front $\sum$ to the numerator, which gives us:
\begin{equation}
S = \frac{\sum_{h=1}^{n-1} \sum_{t=1}^{n-h} (X_t-\bar{X})(X_{t+h}-\bar{X})}{\sum_{t=1}^{n} (X_t-\bar{X})^2}
\end{equation}
Now consider the denominator. How do we represent in so we get an expression similar to the numerator? Set $Y_t=X_t-\bar{X}$. Then $\sum_{t=1}^{n}Y_t=0.$ The denominator here is $\sum_{t=1}^{n}Y_t^{2}$.
We know that $\sum_{t=1}^{n}Y_t^{2} = \left(\sum_{t=1}^{n}Y_t\right)^2 - 2\sum_{h=1}^{n-1} \sum_{t=1}^{n-h}Y_t Y_{t+h}$, i.e. subtracting all unique pairs $\times$ 2.  Because $\sum_{t=1}^{n}Y_t=0$, it follows that $\sum_{t=1}^{n}Y_t^{2} =  - 2\sum_{h=1}^{n-1} \sum_{t=1}^{n-h}Y_t Y_{t+h}$. 
Plugging back in terms of X, the denominator becomes $- 2\sum_{h=1}^{n-1} \sum_{t=1}^{n-h}(X_t-\bar{X})(X_{t+h}-\bar{X})$. Then,
\begin{equation}
S=\frac{\sum_{h=1}^{n-1} \sum_{t=1}^{n-h}(X_t-\bar{X})(X_{t+h}-\bar{X})}{- 2\sum_{h=1}^{n-1} \sum_{t=1}^{n-h}(X_t-\bar{X})(X_{t+h}-\bar{X})}= -\frac{1}{2}
\end{equation}
Hope this helps!
