What I'm given:
$\bf{g_i}=\bf{x_i}$$u_i$ where $\mathbf{x_i}$ is a k-dimensional vector
and
$E(u_i|\mathbf{g_{i-1},...,g_1})=0$
I want to show that $E(u_i|u_{i-1},...,u_1)=0$
My work so far:
$E(u_i|u_{i-1},...,u_1)=E[E(u_i|\mathbf{g_{i-1},...,g_1})|u_{i-1},...,u_1]=0$ since $E(u_i|\mathbf{g_{i-1},...,g_1})=0$
The second equality is by the Law of Iterated Expectations.
Did I use the Law of Iterated Expectations correctly? I have doubts about which information set nests which...
This is not correct in general. Recall that $E(X\mid H)=E(E(X\mid G)\mid H)$ for every integrable random variable $X$ when the sigma-algebras $G$ and $H$ are such that $H\subseteq G$.
Here, $G=\sigma(g_j,j\lt i)$ and $H=\sigma(u_j,j\lt i)$, with $g_j=x_ju_j$. Every $g_j$ is a function of $u_j$ hence it is $H$-measurable and $G\subseteq H$, not the other way around.
As an extreme (counter)example, assume that $x_i=0$ for every $i$, then $E(u_i\mid g_j,j\lt i)=E(u_i)$ hence the hypothesis that $E(u_i\mid g_j,j\lt i)=0$ cannot imply that $E(u_i\mid u_j,j\lt i)=0$.
In the other direction, if one can recover $u_i$ from $g_i=x_iu_i$ then $G=H$ and, obviously, $E(X\mid H)=E(E(X\mid H)\mid H)$ holds in full generality.
