3
$\begingroup$

Say I have a time series $y_t = \frac{1}{n} \sum_{i=1}^n x_t^{(i)}$. For example, $y_t$ can be the returns of the S&P 500 index at time $t$ and the $x^{(i)}_t$ is return of the $i^{th}$ company in the S&P 500 at time $t$ (notice this means $n = 500$). Now suppose instead of imposing a simple autoregressive structure on $y_t$, such as

$$y_t = \phi_0 + \phi_1 y_{t-1} + \phi_2 y_{t-2} \dots + e_t$$

for white noise process $e_t$, we instead have the following

$$y_t = \phi_0 + \sum_{i=1}^n \phi^{(i)}_1 x^{(i)}_{t-1} + \sum_{i=1}^n \phi^{(i)}_2 x^{(i)}_{t-2} \dots + e_t$$

meaning each component series $x^{(i)}_t$ is allowed its own own parameter. My intuition tells me that, since the process $y_t$ is endogenous (in fact fully parametrized by) the $x^{(i)}_t$, this can't be considered a regression on exogenous regressors. So is this an autoregression? If it is an autoregression, what are the constraints that the $\phi$ parameters must satisfy for the model to be invertible?

Also, a follow up question: Suppose that $y_t = f\left(x_t^{(1)}, x_t^{(2)},\dots, x_t^{(n)}\right)$ where $f$ is not necessarily linear, but is invertible (meaning all the $x^{(i)}_t$ play a role in explaining $y_t$). Does the answer change?

$\endgroup$
  • $\begingroup$ Can you clarify the superscript, subscript notation? I presume these aren't polynomial trends. $\endgroup$ – AdamO May 18 '17 at 17:27
  • $\begingroup$ The $t$ subscript indexes time, $(i)$ superscript indexes the endogenous $x$ component series. $\endgroup$ – Mustafa S Eisa May 18 '17 at 18:11
2
$\begingroup$

This type of model

\begin{equation} E[Y_t|, X_t, X_{t-1}, \ldots] = \beta_0 + \sum_t \beta_t X_t + \epsilon_t \end{equation}

is technically considered a distributed lag model, at least as far as the fixed effects are concerned. Whether it's auto-regressive depends on the assumptions you make about $\epsilon_t$ conditional upon the history of $x_{s:s<t}$.

Autocorrelation means that errors are temporally correlated unconditional on any lagged effects. This temporal correlation usually follows such a format that $cor(\epsilon_{t},\epsilon_{s}) < cor(\epsilon_{t},\epsilon_{r})$ if $|t-s| < |t-r|$. Variograms are excellent visual tools to inspect the nature of the trend. Auto-regressive models use fixed lagged effects or random effects in an attempt to create independent errors.

AR-1 autoregressive trends have a correlation of $cor(\epsilon_{t},\epsilon_{s}) = \rho^{|t-s|}$. It turns out that additionally adjusting for a single lagged effect (of $y$) suffices to produce conditionally independent errors. Or you can use a mixed model or GEE.

If, even after adjustment for the covariate history, you find residual errors which are temporally correlated, no amount of "history" of $x$ will suffice to produce conditionally independent errors. This is because the $x,y$ relationship is not a deterministic one. You will need to add a lagged outcome as a covariate to actually produce independent errors.

In the last expression of your question, you do not have autoregressive effects when the model is correctly specified. If you omit some components of $x$ in the model, a backdoor autoregressive effect is introduced because past $y$ is predicting unmeasured $x$.

$\endgroup$
  • $\begingroup$ Adam, thanks for taking the time to write this answer. With regards to your comment "the $x,y$ relationship is not a deterministic one:" For the SP500 example, there is a direct deterministic, contemporaneous relationship between $x$ and $y$. Were you referring to something else? Also, an assumption I'm making here is that the errors $\epsilon_t$ are white noise after adjusting for lags...meaning $Cov(e_t, e_{t-\delta}) = 0$ for all $\delta > 0$. Does that change your comment about the error term? $\endgroup$ – Mustafa S Eisa May 17 '17 at 21:52
  • $\begingroup$ Also, regarding your comment about distributed-lag models and the expression you stated in the beginning of your answer: Isn't this something other than a distributed lag model since we know that, say, $y_t = \sum_i x^{(i)}_tt$? $\endgroup$ – Mustafa S Eisa May 17 '17 at 21:55
  • $\begingroup$ @MustafaSEisa if there is a deterministic relationship, there is no error term. Yet you have written one. $\endgroup$ – AdamO May 17 '17 at 21:55
  • 1
    $\begingroup$ @MustafaSEisa I think you mean the third expression. I was referring to equations 1 and 2. The functional form of $f$ is irrelevant as is the incorporation or omission of an error term. The basic point is that if an outcome is a combination of a history of covariates, it is a DL Model. The key point is that the outcome $y$ and covariates $x$ are different factors: like mortage rates and property sales. Autoregressive models consider $y_t = f(y_{s:s<t})$ which is a different concept altogether, $\endgroup$ – AdamO May 17 '17 at 22:01
  • 2
    $\begingroup$ I think the statement Auto-regressive models mean that errors are temporally correlated can be misunderstood, especially if taken out of context. By itself it actually charactarizes moving average models, not autoregressive models. Could you perhaps rephrase to remove any ambiguities? $\endgroup$ – Richard Hardy May 18 '17 at 6:45

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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