I am currently reading "Measure, Integral and Probability" by Capinski, Marek (see p179). It includes some motivation for the definition of the conditional expectation. For example, given two random variables $X,Y$ with joint density $f_{(X,Y)}$ (and so the marginal and conditional densities), we want to show that for any set $A \subset \Omega, A=X^{-1}(B), B$ Borel, that $$\int_A\mathbb{E}(Y|X)dP= \int_A \mathbb{E}(Y)dP.$$ This is one of the defining condition of an conditional expectation. The book shows the following calculation, \begin{align} \int_A\mathbb{E}(Y|X)dP &= \int_\Omega 1_B(X)\mathbb{E}(Y|X)dP\\ &= \int_\Omega 1_B(X(\omega))\left(\int_\mathbb{R}yf_{Y|X}(y|X(\omega))dy\right)dP(\omega)\\ &=\int_\mathbb{R}\int_\mathbb{R}1_B(x)yf_{(Y|X)}(y|x)dy f_X(x)dx\\ &=\int_\mathbb{R}\int_\mathbb{R}1_B(x)yf_{X,Y}(x,y)dxdy\\ &= \int_\Omega 1_A(X)YdP\\ &= \int_A YdP. \end{align} What I don't understand is the second to last equality immediately above, i.e. $$ \int_\mathbb{R} y \int_\mathbb{R}1_B(x)f_{X,Y}(x,y)dxdy = \int_\Omega 1_A(X)YdP .$$ I think it is a typo since $X\in \mathbb{R}$ and $A \subset \Omega$ --- however, I cant figure the correction either!
1 Answer
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5 There's a typo, the author certainly intended to write $\int_\Omega 1_B(X)YdP $.
Indeed $$\int_\mathbb{R} \int_\mathbb{R}1_B(x)yf_{X,Y}(x,y)dxdy = \int 1_B(x)y dP_{(X,Y)}(x,y) = \int 1_B(X)YdP$$
- $\begingroup$ Ah, I see! Thank you. Could you also clarify the last equality in that case $$\int_\Omega 1_B(X)YdP_{(X,Y)} = \int_\Omega 1_A Y dP_{Y} = \int_A YdP_Y?$$ My confusion is how the measure goes from $P_{(X,Y)}$ to $P_{(Y)}$, if that is even correct? $\endgroup$D.Toe– D.Toe2019-05-22 19:59:34 +00:00Commented May 22, 2019 at 19:59
- $\begingroup$ $\int_\Omega 1_B(X)YdP_{(X,Y)} = \int_\Omega 1_A Y dP_{Y}$ need not hold. $\endgroup$Gabriel Romon– Gabriel Romon2019-05-22 20:03:56 +00:00Commented May 22, 2019 at 20:03
- $\begingroup$ I mean how is $\int 1_B(X)YdP = \int_A Y dP$ --- specifically in regards to the measure $P$ is it the joint or the marginal distribution (the notation isn't very good for this part of the book)? Surely, there must be some change in the measure? $\endgroup$D.Toe– D.Toe2019-05-22 20:13:14 +00:00Commented May 22, 2019 at 20:13
- 1$\begingroup$ Simply note that for all $w$, $1_B(X(w)) = 1_A(w)$. $\endgroup$Gabriel Romon– Gabriel Romon2019-05-22 20:23:24 +00:00Commented May 22, 2019 at 20:23
- 1$\begingroup$ In $\int 1_B(X)YdP$ and $\int 1_A YdP$, $P$ is the measure on the original measure space, usually introduced as $(\Omega,\Sigma, P)$. $\endgroup$Gabriel Romon– Gabriel Romon2019-05-22 20:25:59 +00:00Commented May 22, 2019 at 20:25