3
$\begingroup$

I am new to time series analysis and I came across the following definition of autocovariance:

$\gamma_h(Y) = \mathbb{E}[(y_t - \mu_t)(y_{t-h}-\mu_{t-h})]$

It should tell us, how correlated are two rv's of our process at some specific $t$ and $t-h$.

My question is:

Is the expectation in this formula taken with respect to all indices of the stochastic process (then why is it not taken into account in the formula?), and if yes, why do we estimate the autocorrelation with the following formula?

$\hat{\gamma}_h = \sum_{t=h+1}^T(y_t - \bar{y})(y_{t-h}-\bar{y})$, where $\bar{y}$ is the sample average.

Doesn't the second formula estimate rather the average correlation of rv's in our process, that are $h$ time units apart, than the correlation between rv's at specific point in time $t$ and $t-h$?

$\endgroup$
3
  • $\begingroup$ Where did you encounter the definition you mention? I think that definition would only hold if the $Y$ process were stationary. A more general, albeit tougher to interpret, definition is along the lines of what you suggest. The estimator you list is even more confusing. I think the right plug-in mean-estimate is $\hat{Y}_t$ or similar. Some sources would be helpful. $\endgroup$
    – AdamO
    Commented Dec 19, 2017 at 21:04
  • $\begingroup$ The definitions come from my lecturers slides. I can not upload them without his permission. We were dealing with stationary processes. The problem with estimating the mean of $Y_t$ is that we only get one realisation - especially in time-series. Could you please provide me with more general definitions? $\endgroup$
    – Jan Vainer
    Commented Dec 19, 2017 at 21:09
  • 1
    $\begingroup$ It is a good starting place, you merely need to state the conditions for the definitions. Once you have those foundations, you can start thinking like a modeler in terms of characterizing a fixed component and a random component to a process. For an example of "stable characteristics" in a more general family of processes you can look up an ergodic process. $\endgroup$
    – AdamO
    Commented Dec 19, 2017 at 21:16

1 Answer 1

1
$\begingroup$

For index set $\mathcal{T}$, is custom in some time series analysis texts to denote by $\{\mu_t\}_{t \in \mathcal{T}}$ the time-deterministic component of your series $\{\widetilde{Y}_t\}_{t \in \mathcal{T}}$. Making this more explicit, one may write any trend-stationary $\{\widetilde{Y}_t\}_{t \in \mathcal{T}}$ as the decomposition \begin{align} \widetilde{Y}_t = \mu_t + Y_t + \varepsilon_t, \end{align} where $\mu_t:\mathcal{T} \to \mathbb{R}$ is purely deterministic (i.e., purely a function of time), the component $ Y_t$ is a purely non-deterministic (i.e., stochastic) trend term and is weakly stationary, and $\varepsilon_t$ is your iid white noise term. Now since stationarity is a property of stochastic sequences and $\mu_t$ is purely-deterministic, you don't care for it. Instead, you define $Y_t = \widetilde{Y}_t - \mu_t$, which will be purely non-deterministic. You then calculate auto-covariances for this transformed process. Since $\mu_t$ is non-stochastic, these auto-covariances will be identical to those of $\widetilde{Y}_t$.

In practice of course, you don't know $\mu_t$. Instead, you give it some functional form with parameters $\theta$, for example $\mu_t = \theta_0 + t\theta_1$ and estimate those parameters to obtain $Y_t$ from the observations $\widetilde{Y}_t$. Alternatively (and that is what the autocovariance estimator in your post is based upon), you assume that $\mu_t = \mu$ for all $t \in \mathcal{T}$, and you simply estimate it by the mean of $\widetilde{Y}_t$.

Once you have obtained the autocovariance estimates $\hat{\gamma}(h)$ for lag lengths $h$ that you deem relevant, you can also transform them into autocorrelation estimates by defining $\hat{\rho}(h) = \hat{\gamma}(h)/\hat{\gamma}(0)$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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