Conditional expectation property

Question

If:

$E[Y]=E[Z]+aE[G]+a^2E[H]$

Does the following expression hold:

$E[Y|F]=E[Z|F]+aE[G|F]+a^2E[H|F]$

?

p.s. If so, I would appreciate a reference to a proof .

Answer 1

No.   What is true is that:

• $Y=Z+aG+a^2H ~\implies~ \mathsf E(Y\mid F)=\mathsf E(Z+aG+a^2H\mid F)$

However $\mathsf E(Y)=\mathsf E(Z+aG+a^2H)$ does not necessarily infer that $Y=Z+aG+a^2H$, and so will not chain to the suggested hypothesis.

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Disproof by Counterexample:

Let $a=0$ so we can ignore $G$ and $H$.   Let the sample space be $\{(0,1), (0,2), (3,0)\}$, with these outcomes uniformly distributed, then take $Y$ as the value of the first member of the outcome, $Z$ that of the second, and $F$ their sum.

We have that $\mathsf E(Y)=1$, $\mathsf E(Z)=1$, but $\mathsf E(Y\mid F) = \begin{cases}0 &:& F\in\{1,2\}\\3&:& F=3\end{cases}$ and $\mathsf E(Z\mid F) = \begin{cases}F &:& F\in\{1,2\}\\0&:& F=3\end{cases}$.

So $\mathsf E(Y)=\mathsf E(Z)$ but $\mathsf E(Y\mid F)\neq \mathsf E(Z\mid F)$.

The existence of this witness is sufficient to disprove the hypothesis.

$\blacksquare$