Convergence in Mean Square Let $E[y_t|y_{t-j},y_{t-j-1},\cdots] \xrightarrow[m.s.]{}0$ as $j \rightarrow \infty$.Is it necessarily true that $E[y_t] = 0$?
My Attempt:
\begin{align*}
E[y_t|y_{t-j},y_{t-j-1},\cdots] \xrightarrow[m.s.]{}0 \\
& \implies E[(E[y_t|y_{t-j},y_{t-j-1},\cdots]-0)^2] \rightarrow 0 \\
& \implies E[(E[y_t|y_{t-j},y_{t-j-1},\cdots])^2] \rightarrow 0 \\ 
& \implies E[y_t|y_{t-j},y_{t-j-1},\cdots] \rightarrow 0\\
& \implies E[y_t] = E[E[y_t|y_{t-j},y_{t-j-1},\cdots]] \rightarrow 0
\end{align*}
My problem is how to get the equal sign in the last convergence result? That is to show $E[y_t] = 0$ if this is correct.
 A: As long as you're conditioning on something and you get conditional mean zero, then the unconditional mean is zero, since you're going from more information to less information.
For precisely, if $E[y|\mathcal{G}] = 0$  for any $\sigma$-subalgebra $\mathcal G$, then
$E[y] = 0$. For your problem, one needs to pass to the limit---even as your information set recedes into the infinite past, in the limit you cannot have less information than the trivial $\sigma$-algebra.
Conditional mean is projection onto a subspace, for $L^2$-random variables. What you have is a decreasing sequence of subspaces. Limit of the projections is projection onto the limit---i.e. they commute and you're done. In other words, write
$$
\mathcal{G}_j = \sigma(y_{t-j},y_{t-j-1},\cdots),
$$
as the $\sigma$-algebra generated by $\{ y_{t-j},y_{t-j-1},\cdots \}$ and $\mathcal{G} = \cap_j \mathcal{G}_j $. Then, in the $L^2$-sense
$$
\lim_j E[y | y_{t-j},y_{t-j-1},\cdots  ] = \lim_j E[y | \mathcal{G}_j ] = E[y | \mathcal{G}].
$$
By assumption $E[y | \mathcal{G}] = 0$, which proves your claim.
