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I'm taking this example from a paper I'm reading. I'm having trouble understanding the logic, and I'm hoping someone can help me.

Let $ v_{t}=a \ x_{t}+u_{t} $

where $a$ is some constant and $u$ is a mean zero random variable such that $E[u_{t}x_{t}]=0$ for all $t$.

If $E[x_{t-1} \, x_{t}]=0$ and $E[u_{t} \, x_{t-1}]=0$ then $E[x_{t-1} \, v_{t}] = 0$ This part I can clearly see.

Now Suppose $x_t= p\, x_{t-1} +e_t$ where $e$ is a mean zero white noise term and $|p|<1$.

Now we have $E[x_{t-1} \, v_{t}] = a \, p\, {\rm Var}(x_t) \neq 0$ even if $E[u_{t}\, x_{t-1}]=0$ (Still good)

But, then it goes on to say that if: $ E[x_{t-1}\, u_{t} | x_t]=0$, then we get that $E[x_{t-1}\, v_{t} | x_t] = 0$ and I do not see where that comes from, we still have the same problem as before.

$E[x_{t-1}\, v_{t} | x_t] = E[x_{t-1} (a \, x_{t} + u_{t})|x_t] = a \, x_t x_{t-1} \neq 0$

Am I missing something?

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$E[x_{t-1}v_{t}] = ap\operatorname{var}(x_t)$ seems wrong. I get $$E[x_{t-1}v_{t}] =E[x_{t-1}(ax_t+u_t)]= E[x_{t-1}(a(px_{t-1}+e_t)+u_t)]=apE[x_{t-1}^2]$$ (assuming $E[x_{t-1}e_t]=0$). That will generally not be $\operatorname{var}(x_t)$. – joriki Oct 16 '12 at 5:43
@joriki Which equals $ap\mathrm{var}(x_t)$ if the process $(x_t)_t$ is stationary. This hypothesis seems to be missing. As a matter of fact, I suggest that the OP rewrites the question to make clear the exact list of hypotheses which hold when one asks for $E(x_{t-1}v_t\mid x_t)$. – Did Oct 17 '12 at 7:04
Edited to add $|p|<1$, which I believe makes the process stationary. All other assumptions/hypotheses given to me are in the question. – Greg Oct 17 '12 at 17:19
Are you sure you got the indexes right? Are you conditioning $x_{t-1} u_t$ on the "future" ? – leonbloy Oct 17 '12 at 17:46
up vote 1 down vote accepted

When you are given $x_t$, you must treat it as a constant. Then

$$E[x_{t−1}(a x_t+u_t)|x_t] = a x_t E[x_{t-1} | x_{t}] + E[x_{t-1} u_t | x_t]$$

The second term is zero by hypothesis. But the other gives $a\, p \, x_t^2$ (assuming a process that is not only stationary but also time-reversible - which is the case e.g. is the white noise is gaussian). So, it seems there is some error

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