I am confused why strict exogeneity must be violated when we have lagged time series variables. My understanding of strict exogeneity is that a variable must be uncorrelated with error terms in all periods. But isn't exogeneity always a necessary assumption for estimation? If $x_t$ and $u_t$ are uncorrelated, and $x_{t-1}$ and $u_{t-1}$ are uncorrelated, how would it violate strict exogeneity if we have a specification with $x_t$ and $x_{t-1}$?
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2 Answers
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In the most cases it is assumed that $E[\epsilon_t]=0$. Then, strict exogeneity implies that the regressors are orthogonal to the error term for all observations $s$, i. e. $E[x_s \epsilon_t]=0$. For some time series models this is violated. Consider the AR(1) model $ \ y_t=\beta y_{t-1}+ \epsilon_t \ $ with $ \ \epsilon_t \sim N(0, \sigma^2) \ $ $ \ \forall \ $ $t$. Since you regress $y_t$ on $y_{t-1}$ the error term $\epsilon_t$ is orthogonal to $y_{t-1}$, i. e. $E[y_{t-1} \epsilon_t]=0$.
However, strict exogeneity requires $y_t$ to be orthogonal to $all$ $\epsilon_t$. That does not hold for the considered model - as will be shown:
$E[y_t \epsilon_t]=E[(\beta y_{t-1}+ \epsilon_t)\epsilon_t] \qquad (by \ \ \ y_t=\beta y_{t-1}+ \epsilon_t)$
$ \quad \qquad =\beta E[y_{t-1} \epsilon_t]+E[\epsilon_t^2]$
$ \quad \qquad =E[\epsilon_t^2] \qquad \qquad \qquad \quad (by \ \ \ E[y_{t-1} \epsilon_t]=0)$.
$ \quad \qquad =\sigma^2 \qquad \qquad \qquad \quad \quad (by \quad \epsilon_t \sim N(0, \sigma^2))$.
Therefore, $y_t$ is not orthogonal to all error terms but the regressor for $y_{t+1}$. Thus, strict exogeneity is violated.
This implies, there is only strict exogeneity if $\epsilon_t = 0$ for all $t$.
answered Feb 17, 2014 at 20:42
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1$\begingroup$ I've made some changes, the general thrust of the answer is ok. Note in general exogeneity is defined via conditional expectactions. $\endgroup$– mpiktasFeb 19, 2014 at 7:48
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$\begingroup$ Hello, thanks for your edits. I am aware of this but thought I can show the violation of strict exogeneity via the implication of zero correlation if E[e]=0. $\endgroup$ Feb 19, 2014 at 7:51
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$\begingroup$ Yes it is perfectly ok to do that, hence the upvote. $\endgroup$– mpiktasFeb 19, 2014 at 8:54
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$\begingroup$ @random_guy, the weakness of the argument is in $E[y_{t-1}\varepsilon_t]=0$. You didn't explain where is this coming from. When you regress on the lag, you get automatically $E[y_{t-1}e_t]=0$, where $e_t$ is the residual from the regression, not the error. $\endgroup$– AksakalDec 3, 2014 at 4:54
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$\begingroup$ See, this is what you get when you edit your own answer in order to improve it almost 10 months after posting it: a down vote! However, in this context, strict exogeneity implies $E[y_{t-1} \epsilon_t] = 0$, orthogonality and no correlation! And the argument comes after the formulas where I say that even if $y_{t-1}$ is orthogonal to $\epsilon$, this does not hold for ${y_t}$ even though it is the regressor for $y_{t+1}$. Please read page 6 (explanation of strict exogeneity) to page 9 (str. exog. for AR(1) process) of this paper (and up vote ;)): press.princeton.edu/chapters/s6946.pdf $\endgroup$ Dec 3, 2014 at 9:24
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Strict exogeneity and lagged variables
The problem your refers on is about strict exogeneity and lagged dependent variables. The two things together are contradictory.
Indeed as shown in random guy answer if we have a model like
$y_t = \beta_1 y_{t-1} + \epsilon_t$
(we assume $[\epsilon_t]=0$)
We have that $E[y_t \epsilon_t]=\sigma^2$
However if other lagged variables as $x_{t-j}$ are included, and dependent are not, the contradiction disappear.
Moreover the main point, in my view, is that exogeneity is a causal concept (in any detailed form) and it must be write on a structural equation not a regression one. Indeed the equation above must be intended as structural. (Read here for more about: Under which assumptions a regression can be interpreted causally?)
The reply of random_guy is mathematically correct but terminology used is ambiguous and tend to mix concepts. Mainly regression equation and/or errors and exogeneity assumpion. Speaking about AR (autoregressive) model the conflation become even more evident. I read the the ref suggested in the comment (http://assets.press.princeton.edu/chapters/s6946.pdf), and It suffer from the same problem. This problem is quite common (read here: How would econometricians answer the objections and recommendations raised by Chen and Pearl (2013)?)
