2
$\begingroup$

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. $E(X | Y = y) = \int x f_{X|Y}(x|Y = y)dx$?

$\endgroup$
2
$\begingroup$

(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

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

$\endgroup$
  • $\begingroup$ 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 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 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 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. $\Lambda(B)=\int_{\{\omega:Y(\omega)\in B\}}XdP$. $\endgroup$ – XIAODA QU May 3 at 14:02
  • 1
    $\begingroup$ @Fam $\int_{Y(G)}d\Lambda=\Lambda(Y(G))=\int_{Y^{-1}(Y(G))}E(X|Y)dP$. $\endgroup$ – XIAODA QU May 3 at 20:40

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.