I really cannot figure out what is wrong with the last line....

There seem to be three definitions at least for conditional expectation with respect to a random variable. I would like to know if they are equivalent in all or in whichever contexts. Here are the three definitions I've found :

$\mathbb{E}(X|Y)$ is the only $\sigma(Y)$-measurable random variable such that either :

  • $\forall Z \in \mathbb{L}^2 \left( \Omega, \sigma(Y), \mathbb{P}|_{\sigma(Y)} \right) \mathbb{E}(Z \mathbb{E}(X|Y)) = \mathbb{E}(ZX)$
  • $\forall Z$ essentially bounded on $\left( \Omega, \sigma(Y), \mathbb{P}|_{\sigma(Y)} \right) \mathbb{E}(Z \mathbb{E}(X|Y)) = \mathbb{E}(ZX)$

  • $\forall U \in \sigma(Y)$, $\mathbb{E} ( \mathbb{E}(X|Y) \mathbb{1}_U)= \mathbb{E}(X\mathbb{1}_U) $

I realised 1 implies 2 and 2 implies 3. So really my question is : does 3 imply 1 ?

  • $\begingroup$ For the last one, what is $Z$? It is $E(X\mid Y)$? $\endgroup$ – Gordon Nov 10 '16 at 21:46
  • $\begingroup$ My bad... Z is the conditional expectation of X wrt Y. Thx for spotting $\endgroup$ – James Well Nov 11 '16 at 3:57

I am showing (3) implies (1). Due to confusion in notation, let me denote by $E(X|Y)$ by $Z$ and let me write the $Z$ in condition (1) as $T$.

First take $T=1_A$ for any $A\in\sigma(Y)$. Then $E(TZ)=E(Z1_A)=E(X1_A)=E(TX)$ so the relation is true for indicators. By linearity, it's true for simple non-negative functions $T$.

Now take $T\geq0$ and take $T_n\uparrow T$ where $T_n$ are non-negative simple. Since the relation (1) is true for every $n$, $E(T_nZ)=E(T_nX)$. [Note we are taking $T,T_n$ to be $\sigma(Y)-$measurable].

Suppose first that $X\geq0$ then $T_nX\uparrow TX$ and by MCT, $E(T_nX)\uparrow E(TX)$. This implies $E(T_nZ)\uparrow E(TX)$. But since $X\geq0$, we have $Z\geq0$ (a.s.) ad therefore, $T_nZ\uparrow TZ$ implying $E(T_nZ)\uparrow E(TZ)$ by MCT. Therefore, $E(TZ)=E(TX)$.

Now take $X=X^+-X^{-}$ and apply MCT with $T,T_n\geq0$ separately to $X^+$ and $X^-$ and combine by linearity. Note also that $Z^+$ and $Z^-$ are versions of $E(X^+|Y)$ and $E(X^-|Y)$. So for any general $X$ you have $E(TZ)=E(TX)$.

Now for any general $T$, break $T=T^+$ and $T^-$. Get sequences $T_n\uparrow T^+$ and $S_n\uparrow T_n^-$ and play the same game with $T^+X$ and $T^-X$.

  • $\begingroup$ Yes I have used Z in too many places! That's perfect, nice and clear $\endgroup$ – James Well Nov 11 '16 at 10:15

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