1
$\begingroup$

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 ?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $T^-X$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.