How do I go from $E[Z_{n+1}\mid {\mathcal {F}}_{n}] = Z_n,\; n \in \mathbb N$ to $E[Z_{n}|{\mathcal {F}}_{k}]=Z_k,\;k<n$ using the tower rule?

Question

How do I go from $E[Z_{n+1}\mid {\mathcal {F}}_{n}] = Z_n,\; n \in \mathbb N$ to $E[Z_{n}|{\mathcal {F}}_{k}]=Z_k,\;k<n$ using the tower rule?

Answer 1

$E[Z_{n+1}|F_n] = Z_n$ holds when $Z_n$ is a martingale w.r.t. filtration $F_n$. Since we want to prove $E[Z_n|F_k] = Z_k$ for some $k < n$, you need to use the tower rule: \begin{equation} E[E[Y|X]|W] = E[Y|W] \end{equation}

Starting with $n-1$, we have $E[Z_n|F_{n-1}] = Z_{n-1}$ by the martingale property. Taking the conditional expectation of both sides given $F_{n-2}$ and evoking the tower rule, we have \begin{equation} E[E[Z_n|F_{n-1}]|F_{n-2}] = E[Z_{n-1}|F_{n-2}] \implies E[Z_n|F_{n-2}] = E[Z_{n-1}|F_{n-2}] \end{equation}

But we have $E[Z_{n-1}|F_{n-2}] = Z_{n-2}$ by the martingale property, so \begin{equation} E[Z_n|F_{n-2}] = E[Z_{n-1}|F_{n-2}] \implies E[Z_n|F_{n-2}] = Z_{n-2} \end{equation}

This iterative process continues when you take $n-2, n-3, \cdots, k$ until you get $E[Z_n|F_k] = Z_k$.