AR1 and Law Of Iterated Expectations : No serial correlation In the AR(1) model $y_{t}=\beta_{0}+\beta_{1}y_{t-1}+u_{t}$, assuming $E(u_{t-1}|y_{t-1},y_{t-2}...)=0$, how does the law of iterated expectations ensure that the errors must be uncorrelated: $E(u_{t}u_{s}|x_{t},x_{s})=0$?
 A: Consider $s \neq t$. The law of iterated expectations tells us that
$$\begin{equation*} \mathbb{E}[u_su_t \mid y_{s-1}, y_{t-1}] =
\mathbb{E}_{u_s}\left[\mathbb{E}[u_su_t \mid u_s, y_{s-1}, y_{t-1}]\right] =
\mathbb{E}_{u_s}\left[u_s\mathbb{E}[u_t \mid u_s, y_{s-1}, y_{t-1}]\right]. \end{equation*} $$
Suppose that $s < t$, with the opposite case following by symmetry.
The AR(1) process has the property that, for $y_t = \beta_0 + \beta_1 y_{t-1} + u_t$, $\mathbb{E}[u_t \mid y_{t-1}, y_{t-2}, \dots] = \mathbb{E}[u_t \mid y_{t-1}] = 0$. In words, if I know what happened last period, then nothing else in the past can help me guess what happens today.
Rewriting our AR(1) equation reveals that $u_s = y_s - \beta_0 - \beta_1 y_{s-1}$; there is a one-to-one transformation betweent $u_s$ and $y_s$. As a result, conditioning on $u_s$ is equivalent to conditioning on $y_s$.
Using this relationship and our fact about the AR(1) process, we find that
$$\begin{equation*} \mathbb{E}[u_t \mid u_s, y_{s-1}, y_{t-1}] = 
\mathbb{E}[u_t \mid y_s, y_{s-1}, y_{t-1}] =
\mathbb{E}[u_t \mid y_{t-1}] = 0. \end{equation*} $$
Plugging this in to the LIE result above gives that
$$\begin{equation*}
\mathbb{E}_{u_s}\left[u_s\mathbb{E}[u_t \mid u_s, y_{s-1}, y_{t-1}]\right] =
\mathbb{E}_{u_s}\left[u_s \times 0\right] = 0. \end{equation*} $$
