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.

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

1 Answer 1


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.

$$\underbrace{\mathsf E(f_k\mid B^1\ldots B^{k-1})~f_k}_{\text{function of }B^1\ldots B^k}\qquad\underbrace{\mathsf E(f_kf_k\mid B^1\ldots B^{k-1})}_{\text{function of }B^1\ldots B^{k-1}}$$


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