Let $X: \Omega \to \mathbb{R}$ be a random variable with $\mathbb{E}[|X|]< \infty$, and suppose that $Y: \Omega \to \mathbb{R}^d$ is a discrete random vector.

Define $$f(y) = \begin{cases}\frac{1}{P(Y=y)} \int_{\{Y=y\}}XdP \quad P(Y=y) > 0\\ 0 \quad P(Y=y) =0\end{cases}$$

I'm trying to show that $f$ is a version of $E[X|Y=y]$. For this, I'm trying to prove that for $B$ a borel part of $\mathbb{R}^d$:

$$E[XI_{\{Y \in B\}}] = \int_B f(y) P_Y(dy)$$

I tried to work out both sides. Define $S:= \{y \in \mathbb{R}^d: P(Y=y) > 0\}$. This is by assumption at most countable, thus a Borel subset.

$$E[XI_{\{Y \in B\}}] = \int_{\{Y \in B\}} X dP$$

$$\int_B f(y) P_Y(dy) = \int_{B \cap S} \left(\frac{1}{P(Y=y)} \int_{\{Y=y\}}XdP\right)P_Y(dy)$$

Now, I'm stuck as to how to continue. I tried things like Fubini and applying the formula

$$\int_{A} gdP_Y = \int_{\{Y \in A\}} g \circ Y dP$$

but I could not find anything useful.

Any help will be appreciated!


1 Answer 1


Let $A:=\left\{ y\mid P\left(Y=y\right)>0\right\} $.

Then $A$ is countable, so that:

$$\mathbb{E}X\mathbf{1}_{Y\in B}=\mathbb{E}X\mathbf{1}_{Y\in A\cap B}=\mathbb{E}\sum_{y\in A\cap B}X\mathbf{1}_{Y=y}=\sum_{y\in A\cap B}\mathbb{E}X\mathbf{1}_{Y=y}=$$$$\sum_{y\in A\cap B}f\left(y\right)P\left(Y=y\right)=\sum_{y\in A}f\left(y\right)\mathbf{1}_{y\in B}P\left(Y=y\right)=\mathbb{E}f\left(Y\right)\mathbf{1}_{Y\in B}$$This proves that $\mathbb E[X\mid Y]=f(Y)$.

  • $\begingroup$ Thanks for your answer! How do you justify swapping expectation and series? $\endgroup$
    – user661541
    Commented Apr 13, 2019 at 15:16
  • $\begingroup$ If $\int\sum_{n=1}^{\infty}\left|f_{n}\right|<\infty$ and $s_{n}=\sum_{k=1}^{n}f_{k}$ and $s=\sum_{n=1}^{\infty}f_{n}$ then $\left|s_{n}\right|\leq\int\sum_{n=1}^{\infty}\left|f_{n}\right|<\infty$ and by dominant convergence theorem: $\sum_{k=1}^{\infty}\int f_{k}=\lim_{n\to\infty}\int s_{n}=\int s=\int\sum_{k=1}^{\infty}f_{k}$ This can be applied on $X\mathbf{1}_{A\cap B}=\sum_{y\in A\cap B}X\mathbf{1}_{Y=y}$ where $A\cap B$ is countable. Here it is used that $\mathbb{E}\left|X\right|<\infty$ $\endgroup$
    – drhab
    Commented Apr 13, 2019 at 15:36
  • $\begingroup$ Thank you! It seems that I can't upvote your answer (yet?) though. $\endgroup$
    – user661541
    Commented Apr 13, 2019 at 16:22
  • $\begingroup$ You are welcome anyway ;-) $\endgroup$
    – drhab
    Commented Apr 13, 2019 at 18:15
  • $\begingroup$ I can upvote now! Thanks again! $\endgroup$
    – user661541
    Commented Apr 14, 2019 at 9:17

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