# Validity of a conditional expectation step

Question: Let $$B,B^1,\dots,B^{k-1},B^k\in\mathbb{R}$$ be random variables, and let us define the conditional expectations as $$f_{k-1}=\mathbb{E}[B|B^1,\dots,B^{k-1}]; \quad f_k=\mathbb{E}[B|B^1,\dots,B^k],$$ and by the law of total expectation we have $$f_{k-1}=\mathbb{E}[f_k|B^1,\dots,B^{k-1}].$$ Then is the following true? $$\mathbb{E}[f_{k-1}f_k] =\mathbb{E}\Big[\mathbb{E}[f_k\,|\,B^1,\dots,B^{k-1}]f_k\Big] =\mathbb{E}[f_kf_k].$$ I am concern about whether we are allowed to push the $$f_k$$ into the inner expectation. Thanks.

Attempt: We have \begin{align*} \mathbb{E}\Big[\mathbb{E}[f_k\,|\,B^1,\dots,B^{k-1}]f_k\Big] &=\mathbb{E}\bigg[\mathbb{E}\Big[\mathbb{E}[f_k|B^1,\dots,B^{k-1}]f_k\Big|B^k\Big]\bigg] \\ &=\mathbb{E}\bigg[\mathbb{E}\Big[\mathbb{E}[f_kf_k|B^1,\dots,B^{k-1}]\Big|B^k\Big]\bigg] \\ &=\mathbb{E}\Big[\mathbb{E}[f_kf_k|B^k]\Big] =\mathbb{E}[f_kf_k]. \end{align*} Is this correct? If not, then where did my logic break? Thanks.

• it's wrong........ Nov 15, 2023 at 2:22
• I have given an attempt.
– Resu
Nov 15, 2023 at 2:25
• $\mathbb{E}\Big[\mathbb{E}[f_k\,|\,B^1,\dots,B^{k-1}]f_k\Big]$ need not be $\mathbb{E}[f_kf_k]$ Nov 15, 2023 at 2:27
• But I can show that using one extra step of the law of total expectation (as shown in my above attempt), no?
– Resu
Nov 15, 2023 at 2:29

No, these are generally not equal. $$f_k$$ is a function of the random variable sequence $$(B^1,\ldots, B^{k})$$; it is the conditional expectation of $$B$$ over the $$\sigma$$-algebra of that sequence.
$$\underbrace{\mathsf E(f_k\mid B^1\ldots B^{k-1})~f_k}_{\text{function of }B^1\ldots B^k}\qquad\underbrace{\mathsf E(f_kf_k\mid B^1\ldots B^{k-1})}_{\text{function of }B^1\ldots B^{k-1}}$$