# Help understanding conditional expectation identity

• I
• psie
psie
TL;DR Summary
I'm reading a proof on conditional probabilities and there is an identity involving conditional expectation which I'm stuck on.
Let ##(\Omega,\mathcal{F},P)## be a probability space, and let us define the conditional expectation ##{\rm E}[X\mid\mathcal{G}]## for integrable random variables ##X:\Omega\to\mathbb{R}##, i.e. ##X\in L^1(P)##, and sub-sigma-algebras ##\mathcal{G}\subseteq\mathcal{F}##.

Definition 1: The conditional expectation ##{\rm E}[X\mid\mathcal{G}]## of ##X## given ##\mathcal{G}## is the random variable ##Z## having the following properties:
(i) ##Z## is integrable, i.e. ##Z\in L^1(P)##.
(ii) ##Z## is (##\mathcal{G},\mathcal{B}(\mathbb{R}))##-measurable.
(iii) For any ##A\in\mathcal{G}## we have $$\int_A Z\,\mathrm dP=\int_A X\,\mathrm dP.$$

Definition 2: If ##X\in L^1(P)## and ##Y:\Omega\to\mathbb{R}## is any random variable, then the conditional expectation of ##X## given ##Y## is defined as $${\rm E}[X\mid Y]:={\rm E}[X\mid\sigma(Y)],$$ where ##\sigma(Y)=\{Y^{-1}(B)\mid B\in\mathcal{B}(\mathbb{R})\}## is the sigma-algebra generated by ##Y##.

If ##\mathcal{G}=\sigma(Y)##, then (iii) in definition 1 says that $${\rm E}[\mathbf{1}_A{\rm E}[X\mid Y]]={\rm E}[\mathbf{1}_AX],\quad \forall A\in\sigma(Y).\tag1$$

Now, in a proof I'm reading currently, there are three random variables ##U,S,T## and the following computation appears in the proof: $$\int_{T^{-1}(B)} U\,\mathrm dP={\rm E}[\mathbf{1}_B(T)U]={\rm E}[\mathbf{1}_B(T){\rm E}[U\mid S,T]].$$I simply do not comprehend the last equality, that is ##{\rm E}[\mathbf{1}_B(T)U]={\rm E}[\mathbf{1}_B(T){\rm E}[U\mid S,T]]##. How does this follow from the definitions above and the identity ##(1)##? I'm grateful for any help on this.

Dale
I think I understand the identity in question now. First, from Durret's book, we have

If ##X## is ##\mathcal G##-measurable and ##E|Y|,E|XY|<\infty##, then $$E[XY|\mathcal{G}]=XE[Y|\mathcal{G}].$$

Second, we need the Tower property or the law of the iterated expectation, that is ##E[Y|\mathcal{H}]= E\big[E[Y\mid \mathcal G]\mid \mathcal H\big]##, where ##\mathcal H\subset\mathcal G##.

By the latter property, we have $$E[\mathbf1_B(T)U]=E[E[\mathbf1_B(T)U\mid S,T]].$$ Now, by the theorem in Durret, ##\mathbf1_B(T)=\mathbf1_{T^{-1}(B)}## is ##\sigma(T)##-measurable, and this is a subset of ##\sigma(S,T)=\sigma(\sigma(S)\cup\sigma(T))##. So we can "pull it out", and we are left with $$E[\mathbf1_B(T)U]=E[\mathbf1_B(T)E[U\mid S,T]],$$which is the desired identity.

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