Checking the equivalence of conditional expectations

Question

Let $X,X^1,\dots,X^{k-1},X^k\in\mathbb{R}$ be random variables, and let us define the conditional expectations as $$ f_{k-1}=\mathbb{E}[X|X^1,\dots,X^{k-1}]; \quad f_k=\mathbb{E}[X|X^1,\dots,X^k]. $$ Is it true that $$ \mathbb{E}\Big[\mathbb{E}[f_k\,|\,X^1,\dots,X^{k-1}]f_k\Big] =\mathbb{E}[f_kf_k]. $$ My attempt: \begin{align*} \mathbb{E}\Big[\mathbb{E}[f_k\,|\,X^1,\dots,X^{k-1}]f_k\Big] &=\mathbb{E}\bigg[\mathbb{E}\Big[\mathbb{E}[f_k|X^1,\dots,X^{k-1}]f_k\Big|X^k\Big]\bigg] \\ &=\mathbb{E}\bigg[\mathbb{E}\Big[\mathbb{E}[1|X^1,\dots,X^{k-1}]f_kf_k\Big|X^k\Big]\bigg] \\ &=\mathbb{E}\Big[\mathbb{E}[f_kf_k|X^k]\Big] =\mathbb{E}[f_kf_k]. \end{align*} Is this correct? If not, then where did my logic break? Thanks.

Answer 1

I suspect the identity should be (to avoid confusing your notation "$X^k$" with the conventional exponent notation, I will use $X_k$ instead in my answer below):
\begin{align*} E[E[f_k|X_1, \ldots, X_{k - 1}]f_k] = E[f_{k - 1}^2]. \tag{1}\label{1} \end{align*}

To show $\eqref{1}$, we first show that $\{f_k\}$ is a martingale with respect to the natural filtration $\mathscr{F}_k = \sigma(X_1, \ldots, X_k)$, i.e., $E[f_k|X_1, \ldots, X_{k - 1}] = f_{k - 1}$, which follows directly from the tower property of conditional expectation:
\begin{align*} & E[f_k|X_1, \ldots, X_{k - 1}] \\ =& E[E[X|X_1, \ldots, X_k]|X_1, \ldots, X_{k - 1}] \\ =& E[X|X_1, \ldots, X_{k - 1}] \\ =& f_{k - 1}. \end{align*}

Hence $\eqref{1}$ is equivalent to $E[f_kf_{k - 1}] = E[f_{k - 1}^2]$. To show this identity, apply the law of iterative expectations and the martingale property just proved: \begin{align*} & E[f_kf_{k - 1}] \\ =& E[E[f_kf_{k - 1}|\mathscr{F}_{k - 1}]] \tag{2}\label{2} \\ =& E[f_{k - 1}E[f_k|\mathscr{F}_{k - 1}]] \tag{3}\label{3} \\ =& E[f_{k - 1}f_{k - 1}] \\ =& E[f_{k - 1}^2]. \end{align*} From $\eqref{2}$ to $\eqref{3}$, $f_{k - 1}$ can be pulled out from the inner conditional expectation because it is $\mathscr{F}_{k - 1}$-measurable by definition. This completes the proof.