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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.

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  • $\begingroup$ Are you asking whether $E(y_t)=0$ always and for finite $t$ also, or whether it converges to zero as $t$ goes to infinity? $\endgroup$
    – Alecos Papadopoulos
    Commented Apr 5, 2015 at 2:23
  • $\begingroup$ @AlecosPapadopoulos Hey, I am asking wether $E[y_t] = 0$ always. I am pretty sure this is true actually, just not sure how to prove it. $\endgroup$
    – zsljulius
    Commented Apr 5, 2015 at 2:45
  • $\begingroup$ One more clarification: you are sending $j$ to infinity. This essentially increases the distance between the conditioning set and $y_t$, do I understand correctly? $\endgroup$
    – Alecos Papadopoulos
    Commented Apr 5, 2015 at 3:20

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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.

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