8
$\begingroup$

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.

$\endgroup$
  • 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 '17 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 '17 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 -Reinstate Monica Mar 26 '17 at 0:26
  • $\begingroup$ It appears it should be $1/(n-h)$ rather than $1/n$ $\endgroup$ – Alecos Papadopoulos Mar 26 '17 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 '17 at 2:51
4
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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 '17 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 $Y_{n-1}Y_n$. $\endgroup$ – Dilly Minch Mar 26 '17 at 2:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.