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It can be shown by contradiction that exogeneity fails to hold for an AR(1) model.
Is there any proof that contemporaneous exogeneity does not fail to hold?
All I've come across is assuming it does hold and showing that the contradiction for the strict exogeneity case does not apply to this case, which isn't really a proof?

Assuming that,

Say we have a weakly stationary and weakly dependent AR(1) model: $y_t=\beta_0+\beta_1y_{t-1}+u_t$ with
$E[u_t|y_{t-1}]=0$
$E[u_t]=0$
$E[u_t^2]=\sigma^2$

Since we have $E[u_t|\beta_0+\beta_1y_{t-2}+u_{t-1}]=0$ and so on, does this imply:
$E[u_t|y_{t-1},y_{t-2},...,u_{t-1},u_{t-2},...]=0$ ?

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The reason you mostly encounter contemporaneous exogeneity as an assumption is - in my view - that it amounts to assuming that the model is not dynamically misspecified.

By way of example, let us consider the case that the process is in fact generated as an AR(2) process, $$ y_t=\phi_1y_{t-1}+\phi_2y_{t-2}+u_t, $$ where $u_t$ satisfies the assumptions you present, in particular that $E(u_t|y_{t-1}, y_{t-2})=0$.

Now, suppose that the statisticians fits an AR(1) model to $y_t$, $$ y_t=\phi y_{t-1}+\tilde u_t. $$ Then, $\tilde u_t=\phi_2y_{t-2}+u_t$, so that, evidently, $E(\tilde u_t|y_{t-1}, y_{t-2})\neq0$.

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