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 $x_{t-1}$?


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 $t$.

  • 1
    $\begingroup$ I've made some changes, the general thrust of the answer is ok. Note in general exogeneity is defined via conditional expectactions. $\endgroup$ – mpiktas Feb 19 '14 at 7:48
  • $\begingroup$ 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. $\endgroup$ – random_guy Feb 19 '14 at 7:51
  • $\begingroup$ Yes it is perfectly ok to do that, hence the upvote. $\endgroup$ – mpiktas Feb 19 '14 at 8:54
  • $\begingroup$ @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. $\endgroup$ – Aksakal Dec 3 '14 at 4:54
  • $\begingroup$ 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 $\endgroup$ – random_guy Dec 3 '14 at 9:24

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