# Finding the conditional PDF of the conditional expectation

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

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

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

• 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? – Fam May 3 at 5:05
• 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 $Y$. – Fam May 3 at 5:26
• Finally, in the (2) answer I would like to show that the pdf of the $E(X | Y)$ is the conditional distribution $f_{X|Y}(x|Y)$. – Fam May 3 at 5:29
• @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. $\Lambda(B)=\int_{\{\omega:Y(\omega)\in B\}}XdP$. – XIAODA QU May 3 at 14:02
• @Fam $\int_{Y(G)}d\Lambda=\Lambda(Y(G))=\int_{Y^{-1}(Y(G))}E(X|Y)dP$. – XIAODA QU May 3 at 20:40