If $X$ and $Y$ are random variables defined on $(\Omega,\mathscr{F},\mathbb{P})$. Is it true that:For $\mathbb{P}$-a.s $\omega$, we have
$\mathbb{E}[X|Y](\omega) = \mathbb{E}[X|Y=Y(w)]\quad$ which $\ \mathbb{E}[X|Y=a] \stackrel{\triangle}{=} \lim\limits_{\epsilon \to 0^+}\frac{\mathbb{E}[X\cdot 1_{\{|Y-a|<\epsilon\}}]}{\mathbb{P}[|Y-a|<\epsilon]}$
If $Y$ is discrete, we can divide $\sigma (Y)$ into different atoms such as $\{Y_i\}_{i=1}^\infty$ and we have
$\mathbb{E}[X|Y](\omega) = \sum\limits_{i=1}^{\infty}\frac{\mathbb{E}[X\cdot1_{Y_i}]}{\mathbb{P}(Y_i)}1_{Y_i}(\omega) $ . So the proposition above is evidently true for discrete random variable $Y$.
But how about a general random variable $Y$?
My thought and confusion:
By Radon-Nikodym theorem, the conditional expectation $\mathbb{E}[X|Y]$ is defined as a unique random variable(in the means of a.e.) satisfies:$$\int_A \mathbb{E}[X|Y](\omega) \mathbb{P}(d\omega)=\int_A X(\omega) \mathbb{P}(d\omega)\quad \forall A \in \sigma(Y)$$
However, such a definition is related on a positive measurable set $A$, I don't know how can I use this definition to a single element $\omega$ in $\Omega$.
Could someone give me a hint about this?
Best Answer
Suppose $X$ and $Y$ are $\mathbb{R}$-valued random variables such that $X \in L^1(\mathbb{P})$. Then
$$ \mu(A) = \mathbb{P}(Y \in A) \qquad\text{and}\qquad \nu(A) = \mathbb{E}[X\mathbf{1}_{\{Y \in A\}}] $$
define finite Borel measures on $\mathbb{R}$. Moreover, it is clear that $\nu \ll \mu$. Hence by the generalized Lebesgue differentiation theorem,
$$ \frac{\mathrm{d}\nu}{\mathrm{d}\mu}(y) \mathrel{\stackrel{\triangle}=} \lim_{\varepsilon \to 0^+} \frac{\nu(B(y, \varepsilon))}{\mu(B(y, \varepsilon))} = \lim_{\varepsilon \to 0^+} \mathbb{E}[X \mid |Y - y| < \varepsilon] $$
converges for $\mu$-a.e. $y \in \mathbb{R}$. Moreover, for any $\sigma(Y)$-measurable event $B$, we can find a Borel set $A \subseteq \mathbb{R}$ such that $B = \{Y \in A\}$, hence
\begin{align*} \mathbb{E}[X \mathbf{1}_B] = \mathbb{E}[X \mathbf{1}_{\{Y \in A\}}] &= \int_{\Omega} \mathbf{1}_{A}(x) \, \nu(\mathrm{d}x) \\ &= \int_{\Omega} \frac{\mathrm{d}\nu}{\mathrm{d}\mu}(y) \mathbf{1}_{A}(y) \, \mu(\mathrm{d}y) \\ &= \mathbb{E}\left[ \frac{\mathrm{d}\nu}{\mathrm{d}\mu}(Y) \mathbf{1}_A(Y) \right] = \mathbb{E}\left[ \frac{\mathrm{d}\nu}{\mathrm{d}\mu}(Y) \mathbf{1}_B \right]. \end{align*}
This shows that
$$ \mathbb{E}[X \mid Y = y] = \frac{\mathrm{d}\nu}{\mathrm{d}\mu}(y) \qquad \text{for $\mu$-a.e. $y$} $$
and
$$ \mathbb{E}[X \mid Y](\omega) = \frac{\mathrm{d}\nu}{\mathrm{d}\mu}(Y(\omega)) \qquad \text{for $\mathbb{P}$-a.s. $\omega$.} $$