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.

  • 5
    $\begingroup$ Hint: by subtracting a constant from all the $X_t$, which will change none of the $\hat\gamma(h)$, you may assume $0=\sum_{t=1}^nX_t$. Square that and look for pieces that match your two sums. $\endgroup$
    – whuber
    Mar 25, 2017 at 22:04
  • $\begingroup$ Thanks for the reply. I understand that subtracting a constant does not affect any of the $\hat{\gamma}(h)$, but I don't see why it allows me to assume that the sum of the series is equal to 0. $\endgroup$
    – Ernesto
    Mar 25, 2017 at 23:04
  • $\begingroup$ Subtract exactly the constant that makes $\sum X_t$ equal to 0. Now your $\hat\gamma$ is simplified (because the new $X_t$'s have mean 0) and the terms are much easier to play with (but without loss of generality). $\endgroup$
    – Glen_b
    Mar 26, 2017 at 0:26
  • $\begingroup$ It appears it should be $1/(n-h)$ rather than $1/n$ $\endgroup$ Mar 26, 2017 at 2:20
  • 1
    $\begingroup$ @AlecosPapadopoulos I believe both versions are valid estimators of the autocovariance function with the same asymptotic properties but I read somewhere that $1/n$ is preferred. (The reason is that the matrix $\hat{\gamma}(i-j)$ is positive semi-definite, I am not a mathematician so I can't really explain this reason!) $\endgroup$
    – Ernesto
    Mar 26, 2017 at 2:51

1 Answer 1


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!

  • $\begingroup$ Thank very much, I'll accept this answer in a moment, I just have one final question. Everything is clear to me except this part: $\sum_{t=1}^{n}Y_t^{2} = \left(\sum_{t=1}^{n}Y_t\right)^2 - 2\sum_{h=1}^{n-1} \sum_{t=1}^{n-h}Y_t Y_{t+h}$. I don't understand how we are able to include the double summation here, I assume it is a property or identity of the summation? $\endgroup$
    – Ernesto
    Mar 26, 2017 at 2:44
  • 3
    $\begingroup$ To see this, try to expand $(\sum_{t=1}^n Y_t)^2$. You get the sum of $Y_t^2$, then the rest of the terms are of type $Y_iY_j$ for $i\neq j$, each of which occurs twice in the expansion due to symmetry. Now, the double summation comes from enumerating these pairs in the following way: For $Y_1$, we count $Y_2, Y_3$, etc. For $Y_2$, we count $Y_3,Y_4$ etc., until we reach $Y_{n-1}$ for the final pair $Y_{n-1}Y_n$. $\endgroup$ Mar 26, 2017 at 2:59

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