Factorized Conditional Expectation w.r.t. a smaller domain

Question

Let $X$ and $Y$ be real-valued random variables with joint density denoted by $f:=f^{(X,Y)}$ for sake of brevity. Prove that $$ E[X \mid \{Y=y\}] := \frac{1}{f^Y\!(y)} \int_\mathbb{R} x \cdot f(x,y) \, \lambda(dx) $$ for all $y\in\mathbb{R}$ with $f^Y\!(y)>0$ is a factorized conditional $E[ X \mid Y = \cdot \,]$ with domain $\{f^Y > 0\}$.

I cannot understand the concept of Factorized Conditional Expectation. Here, the exercise asks to show that via the given formula a factorized conditional expectation is defined, not on the entire real numbers but on the smaller domain $\{f^Y>0\}$, where the formula actually makes sense.

How can I do that?

Answer 1

This is indeed slightly weird.

Denote by $\mathcal{Y}$ the domain $\{f^Y > 0\}$. Clearly, $\mathcal{Y}$ is non-empty: otherwise, the probability density function $f^Y$ would not integrate to one. Fix $y_0 \in \mathcal{Y}$. Define a new random variable $\tilde{Y}$ according to \begin{align*} \tilde{Y}(\omega) = Y(\omega) \boldsymbol{1}_{\{Y(\omega) \in \mathcal{Y}\}} + y_0 \boldsymbol{1}_{\{Y(\omega) \notin \mathcal{Y}\}}. \end{align*} Then $\tilde{Y}$takes values in the smaller domain $\mathcal{Y}$. As far as I understand the question, it asks us to prove that the expression is a factorized conditional expectation of $X$ given $\tilde{Y}$. In other words, we have to show that $(Q_y)_{y\in\mathcal{Y}}$ given by \begin{align*} Q_y(E) = \frac{1}{f^Y\!(y)} \int_E f(x,y) \, \lambda(\mathrm{d}x) \end{align*} is a $(\mathcal{Y},\mathbb{B}(\mathcal{Y}))$-Markov kernel on $(\mathbb{R},\mathbb{B}(\mathbb{R}))$ which satisfies $$ \mathbb{P}(X \in E, \tilde{Y} \in K) \overset{\star}{=} \int_K Q_y(E) \, \tilde{Y}(\mathbb{P})(\mathrm{d}y). $$ for $E \in \mathbb{B}(\mathbb{R})$, $K \in \mathbb{B}(\mathcal{Y})$.

Checking that $(Q_y)$ is a Markov kernel is straightforward. The following argument verifies that $\star$ holds: Denote by $\tilde{f}$ the joint probability density function of $(X,\tilde{Y})$. By definition, $\tilde{f}^Y = f^Y$ and $f = \tilde{f}$ almost everywhere (on $\mathcal{Y}$ and $\mathcal{Y}\times\mathbb{R}$ w.r.t. the Lebesgue measure), thus \begin{align*} \int_K Q_y(E) \, \tilde{Y}(\mathbb{P})(\mathrm{d}y) &= \int_K \frac{\tilde{f}^Y\!(y)}{f^Y\!(y)} \int_E f(x,y) \, \lambda(\mathrm{d}x)\, \lambda(\mathrm{d}y) \\ &= \int_{K \times E} \tilde{f}(x,y) \, \lambda(\mathrm{d}(x,y)) = \mathbb{P}(X \in E, \tilde{Y} \in K) \end{align*} as desired.

Answer 2

I guess that by "factorized conditional expectation" of $X$ given $Y$ you mean a Borel measurable function $h:\Bbb R\to\Bbb R$ such that $h(Y)$ is integrable and a version of $E[X\mid\sigma\{Y\}]$. The latter means that $$ E[h(Y)\cdot 1_B(Y)] = E[X\cdot 1_B(Y)]\qquad(\dagger) $$ for each Borel subset $B$ of $\Bbb R$. One is then is inclined to interpret $h(y)$ as $E[X\mid Y=y]$.

Assuming $X$ is integrable, the numerator in your expression is well defined for $\lambda$-a.e. $y$; more precisely, if $y\in G:=\{y:\int_{\Bbb R} |x|f(x,y)\lambda(dx)<\infty\}$. The ratio itself is well defined if, in addition, $f_Y(y)>0$. Define $H:=G\cap\{f_Y>0\}$, noting that $P[Y\in H]=1$, and then $$ h(y):=\cases{{\int_{\Bbb R} x\cdot f(x,y)\lambda(dx)\over f_Y(y)},& $y\in H$;\cr 0,& $y\notin H$.\cr} $$ Now computing as did @Furrer, it's straightforward to check that ($\dagger$) holds with this choice of $h$.