I will quickly summarize the general definition of conditional expectation. Let $(\Omega, \mathcal{A}, P)$ a measure space, $X: \Omega \to \mathbb{R}$ a random variable and $\mathcal{G}$ a sigma algebra such that $\mathcal{G} \subset \mathcal{A}$. By Radon-Nikodyn theorem, we guarantee the existence of the conditional expectation $E(X|\mathcal{G})$ defined as the unique random variable satisfying the following properties:

  • $E(X|\mathcal{G})$ is $\mathcal{G}$-measurable;
  • $\int_G E(X|\mathcal{G})dP = \int_G X dP$,$\quad\forall G\in \mathcal{G}$.

If $G = \Omega$, we conclude $E[X] = E[ E(X|\mathcal{G})]$. But I would like to discuss in a little detail these expectations, especially when we use their respective densities (pdf) to express the expectation. We know that

$$E[X] = \int x f_X(x) dx$$.

Suposse $X$ and $Y$ two random variables. Considere $\sigma(X)$ and $\sigma(Y)$ the respective $\sigma$-algebra generated by $X$ and $Y$. The notation is well knowk: $E(X | Y) = E(X | \sigma(Y))$. By definition, we know that $E(X | Y)$ is $\sigma(Y)$-measurable. I have two questions:

(1) How can I argument that $E(X | Y)$ is a random variable obtained as a certain function of the $Y$ random variable?

(2) How do I prove that to calculate $E(X | Y = y)$, I have to use the conditional pdf $f_{X|Y}(x|Y = y)$? In other words. $E(X | Y = y) = \int x f_{X|Y}(x|Y = y)dx$?


1 Answer 1


(1) $$\Lambda(t):=\int_{\{\omega:Y(\omega)\leq t\}}XdP=\int_{\{\omega:Y(\omega)\leq t\}}E(X|Y)dP$$

is absolutely continuous with respect to $P\circ Y^{-1}$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$, so we take the derivative as $\lambda(t)$.

Consider $\lambda\circ Y$. $\forall G\in\mathcal{G}$, $$\int_{G}\lambda\circ YdP=\int_{Y(G)}\lambda d(P\circ Y^{-1})=\int_{Y(G)}d\Lambda=\int_GE(X|Y)dP.$$ Since $E(X|Y)$ and $\lambda\circ Y$ are both $\mathcal{G}$-measurable, $\lambda\circ Y\overset{a.s.}{=}E(X|Y).$

(2) $$E_Y(E(X|Y))=\int_{-\infty}^{+\infty}\lambda(y) [P\circ Y^{-1}](dy)=\int_\Omega \lambda\circ YdP=\int_\Omega E(X|Y)dP=\int_\Omega XdP.$$ If you see $E(X|Y)$ as a function of $Y$, and $Y$ has a density function, then

$$\int_{-\infty}^{+\infty}E(X|Y=y)f_Y(y)dy=\int_{-\infty}^{+\infty}\lambda(y)f_Y(y)dy=\int_{-\infty}^{+\infty}\lambda(y) [P\circ Y^{-1}](dy)=E(X).$$

  • $\begingroup$ I have many doubts and I would like to organize them. First of all, I would like to say that I am not an expert on the subject. Well, I suppose that $P\circ Y^{-1}$ is the push-forward measure gerated by $Y$ in $(\mathbb{R}, \mathcal{B})$, that is $P\circ Y^{-1} = P_Y$. I can not find any argument to show that $\Lambda \ll P_Y$. I would start by assuming that $P_Y(B) = 0$ and then I try to show that $\Lambda(B) = 0$. But this does not make sense. Would you have a more detailed argument? $\endgroup$
    – Fam
    May 3, 2019 at 5:05
  • $\begingroup$ I try to use the following fact: Consider measurable spaces $(\Omega_1,\mathcal{A}_1)$ and $(\Omega_2,\mathcal{A}_2)$ and a measure $\mu$ on $(\Omega_1, \mathcal{A_1})$. Any measurable function $X:\Omega_1\to \Omega_2$ induces a measure $\mu_X$ on $\Omega_2$ through $\mu_X(A_2)=\mu( X^{-1}(A_2))$ for $A_2 \in \mathcal{A}_2$, the pushforward measure. If $g:\Omega_2 \to \mathbb{R}$ is an integrable function, then $\int_{A_2} g ~\mathrm{d}(\mu_X) = \int_{X^{-1}(A_2)} g \circ X ~\mathrm{d}\mu$. In your case, you are working with $g = id$ but with two random variables $X$ and $Y$. $\endgroup$
    – Fam
    May 3, 2019 at 5:26
  • $\begingroup$ Finally, in the (2) answer I would like to show that the pdf of the $E(X | Y)$ is the conditional distribution $f_{X|Y}(x|Y)$. $\endgroup$
    – Fam
    May 3, 2019 at 5:29
  • $\begingroup$ @Fam Absolutely continuity is followed by the fact that the integration of X over any set with 0 measure ($P_Y(B)=0\Rightarrow P(Y^{-1}(B))=0$) is zero. $\Lambda(B)=\int_{\{\omega:Y(\omega)\in B\}}XdP$. $\endgroup$
    – Selene
    May 3, 2019 at 14:02
  • 1
    $\begingroup$ @Fam $\int_{Y(G)}d\Lambda=\Lambda(Y(G))=\int_{Y^{-1}(Y(G))}E(X|Y)dP$. $\endgroup$
    – Selene
    May 3, 2019 at 20:40

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