# Substitute Concrete Value in Conditional Expectation

• Let $$(\Omega, \mathcal{G}, \mathbb{P})$$ be a probability space.
• Let $$X, Y : \Omega \rightarrow \mathbb{R}$$ be random variables.
• Furthermore, let
$$f: \mathbb{R}^2 \rightarrow \mathbb{R}$$ be a $$\mathcal{B}(\mathbb{R}^2)/\mathcal{B}(\mathbb{R})$$-measurable function such that, for all $$y \in \mathbb{R}$$, the random variables $$f(X,y)$$ and $$f(X,Y)$$ have finite expectation.

Now let $$y \in \mathbb{R}$$ be arbitrary. Under the above assumptions, the expected value $$\mathbb{E}[f(X,y)]$$ and a $$\mathbb{P}$$-unique conditional expectation $$\mathbb{E}[f(X,Y) \mid Y]$$ do exist.

Furthermore, since $$\mathbb{E}[f(X,Y) \mid Y]$$ is $$\sigma(Y)/\mathcal{B}(\mathbb{R})$$-measurable, there exists a $$\mathbb{P}_Y$$-unique $$\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$$-measurable function $$\varphi : \mathbb{R} \rightarrow \mathbb{R}$$ such that $$\varphi(Y) = \mathbb{E}[f(X,Y) \mid Y]$$.

Under which circumstances does it hold, that $$\varphi$$ can be chosen such that $$\varphi (y) = \mathbb{E}[f(X,y)]$$ and why?

• The notion of a regular conditional probabilit (sometimes called a transition kernel) is usually helpful with this sort of thing. Do you have a particular set of circumstances where you'd like your $\varphi$ to exist?
– DCM
Apr 12, 2019 at 17:40
• *probability, not 'probabilit'...
– DCM
Apr 12, 2019 at 17:50


$$E(f(X,Y)|Y)=\vpi(Y)$$ for all $$f$$ such that $$f(X,Y)$$ has a finite expectation

holds iff $$X$$ and $$Y$$ are independent.

Indeed, if $$X$$ and $$Y$$ are independent, then for each Borel subset $$B$$ of $$\R$$ \begin{align*} EI\{Y\in B\}\vpi(Y)&=\int P(Y\in dy)I\{y\in B\}\vpi(y) \\ &=\int P(Y\in dy)I\{y\in B\}Ef(X,y) \\ &=\int P(Y\in dy)I\{y\in B\}\int P(X\in dx)f(x,y) \\ &=\int P(X\in dx,Y\in dy)I\{y\in B\}f(x,y) \\ &=EI\{Y\in B\}f(X,Y), \end{align*} where $$I$$ is the indicator. So, $$E(f(X,Y)|Y)=\vpi(Y)$$.

Vice versa, suppose that $$E(f(X,Y)|Y)=\vpi(Y)$$ for all $$f$$ such that $$f(X,Y)$$ has a finite expectation, where $$\vpi(Y)=Ef(X,y)$$ for all real $$y$$. Take any Borel subsets $$A$$ and $$B$$ of $$\R$$, and let $$f(x,y):=I\{x\in A\}I\{y\in B\}$$ for all real $$x,y$$. Then for each $$y\in\R$$ $$$$\vpi(y)=Ef(X,y)=EI\{X\in A\}I\{y\in B\}=P(X\in A)I\{y\in B\}$$$$ and \begin{align*} P(X\in A)P(Y\in B)&=EP(X\in A)I\{Y\in B\} \\ &=EI\{Y\in B\}\vpi(Y) \\ &=EI\{Y\in B\}f(X,Y) \\ &=EI\{X\in A\}I\{Y\in B\}=P(X\in A,Y\in B), \end{align*} so that $$P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$$.

• Iosif, if $f(x,y) = I\{x \in A\} I\{y \in B\}$, then $\varphi(y)$ need not be equal $P(X \in A) I\{y \in B\}$! You are assuming independence here... Apr 13, 2019 at 11:25
• What happens $P_X$ and $P_Y$ supported on finite (or compact) sets and let $f(x,y)=1$?
– DCM
Apr 13, 2019 at 12:24
• I read this question as being about the functional $\mu_{X,Y,f}:\psi\longmapsto \int\int\psi(y)f(x,y)(dP^{X,Y}(x,y) - dP^X(x)dP^Y(y))$ for a fixed choice of $f$.
– DCM
Apr 13, 2019 at 12:25
• I was going to use the word distribution instead of `functional' there for a second...
– DCM
Apr 13, 2019 at 12:28
• @MateuszKwaśnicki : Mateusz, I don't know why you said "need not". I was certainly not assuming any independence at that point. If $f(x,y)=I\{x\in A\}I\{y\in B\}$ for all real $x,y$, then for each $y\in\mathbb R$ we have $\varphi(y)=Ef(X,y)=EI\{X\in A\}I\{y\in B\}=P(X\in A)I\{y\in B\}$. Here, $y$ is just a real number (not a random variable), and hence $I\{y\in B\}$ is also just a real number. I have added this little detail to the answer. Please let me know whether this convinces you. Apr 14, 2019 at 1:26

This isn't an answer, just a (hopefully correct and useful) reformulation of the question

Let $$f$$ be as stipulated (and assumed fixed throughout). The equation $$E(f(X,Y)|Y) = u(Y)$$ is just a way of saying that $$u$$ satisfies

$$\int f(x,y)\psi(y)dP^{X,Y}(x,y) = \int u(y)\psi(y)dP^Y(y)$$

for all $$\psi$$ in a suitably large class of test functions. The question here is under what circumstances is the function $$u$$ defined by

$$u(y) =\int f(x,y)dP^X(x)$$

a solution to all these equations. That is, when is it true that

$$\int\int f(x,y)\psi(y)dP^{X,Y}(x,y) = \int \left(\int f(x,y)dP^X(x)\right)\psi(y)dP^Y(y)$$

for all test functions $$\psi$$? Re-arranging this equation slightly, our question becomes: when is it true that

$$\int \int \psi(y)f(x,y)\left( dP^{X,Y}(x,y) - dP^X(x)dP^Y(y)\right) = 0$$

for all test functions $$\psi$$?.

Edit: I don't disagree with anything in Iosif's answer, but my initial feeling (before you accepted his answer) was that you were asking a slightly different question. Whether anything interesting can be said for my question is perhaps a matter for another occasion.