1

Setup:

Consider an integrable random variable $X$ defined on a probability space $(\Omega, \mathscr A, P)$. Suppose that there are disjoint events $B_1, B_2,\ldots \in \mathscr A$ such that $\cup_{n=1}^\infty B_n = \Omega$. Let $\mathscr C = \sigma(\left\{ B_1, \ldots, B_n\right\})$ be the sigma algebra generated by those events.

Let $$Y=\sum_{n=1,\, P(B_n)>0}^\infty \int_{B_n}X\,dP \frac{1}{P(B_n)}1_{B_n}$$

I want to show that $$E\left ( X|\mathscr C\right ) = Y\, a.s.$$

In order to show that, I have to show that $Y$ is $\mathscr C$-measurable and that for any $C \in \mathscr C$ it holds that $\int_C Y\, dP = \int_C X\, dP$.

Proof and questions:

My professor gave a proof for this assertion which starts like this:

$Y$ is $\mathscr C$-measurable, since all the $B_n$ are $\mathscr C$-measurable.

Question 1: Why is this enough? Surely for any $N\in \mathbb N$, the finite sum of $\mathscr C$-measurable functions $$\sum_{n=1,\, P(B_n)>0}^N \int_{B_n}X\,dP \frac{1}{P(B_n)}1_{B_n}$$ will be $\mathscr C$-measurable. But does this also hold for an infinite sum, when the summands are not necessarily non-negative?

He continues:

Let $C\in \mathscr C$. Then $C$ can be written as $C=\cup_{i\in I}B_i$, for some $I\subset \mathbb N$.

Then $$\int_C Y \,dP = \sum_{i\in I} \int_{B_i} Y \, dP \tag{1}$$

Question 2: Why does the last equality hold? According to our definition of "$\int_C$": $$\int_C Y\, dP = \int Y 1_C\, dP = \int Y \sum_{n=1}^\infty 1_{B_n}\, dP = \int \sum_{n=1}^\infty 1_{B_n}Y\, dP$$ Why can the sum be taken outside of the integral? If $Y$ was non-negative, this would follow from Levi's theorem. If $Y$ was integrable, one could arrive at (1) via, the dominated convergence theorem, see here. Since $Y$ is non necessarily non-negative, I guess one should show that $Y$ is integrable. But how can I do that without using that $Y = E\left ( X|\mathscr C\right )$, (which is what I want to show!)?

I don't have any questions about the rest of the proof, but for completeness, here it is:

$$\sum_{i\in I} \int_{B_i} Y \, dP = \sum_{i\in I, P(B_i)>0} \int_{B_i} Y \, dP \\ = \sum_{i\in I, P(B_i)>0} \int_{B_i}X\,dP \frac{P(B_i)}{P(B_i)} \\ = \sum_{i\in I} \int_{B_i}X\,dP\\ = \int_C X \, dP$$

Thank you very much for your help!

Bernard
  • 175,478
Epiousios
  • 3,194

1 Answers1

1

  1. A pointwise limit of $\mathcal C$-measurable functions is $\mathcal C$-measurable, no matter whether the functions are non-negative or not. If you wish, in any case, you can separate the positive and negative part.

  2. Indeed, $Y$ is integrable since for any $N$, $$\mathbb E\left[ \sum_{n=1,\, P(B_n)>0}^N\int_{B_n}\left\lvert X\right\rvert \,dP \frac{1}{P(B_n)}1_{B_n} \right] \\ = \sum_{n=1,\, P(B_n)>0}^N\int_{B_n}\left\lvert X\right\rvert \,dP\mathbb E\left[ \frac{1}{P(B_n)}1_{B_n} \right]= \sum_{n=1,\, P(B_n)>0}^N\int_{B_n}\left\lvert X\right\rvert \,dP\leqslant \mathbb E\left\lvert X\right\rvert .$$

Davide Giraudo
  • 172,925
  • Thank you! Just to add some detail to the second one:

    Let $Y_N = \sum_{n=1,, P(B_n)>0}^N\int_{B_n}\left\lvert X\right\rvert ,dP \frac{1}{P(B_n)}1_{B_n}$. Then $|Y| \leq \lim_{N\rightarrow \infty} Y_N$. $Y_N$ is integrable, as you show. And since $Y_N \uparrow \lim_{N\rightarrow \infty} Y_N$ one has $\int Y_N dP \rightarrow \int \lim_{N\rightarrow \infty} Y_N dP$ (by Levy).

    Therefore $$\int |Y| dP \leq \int \lim_{N\rightarrow \infty} Y_N dP = \lim_{N\rightarrow \infty} \int Y_N dP \leq E|X|,$$ so that $Y$ is integrable.

    Correct?

     – Epiousios Sep 05 '17 at 05:27
  • 1
    @1524 Yes. $$ $$ – Davide Giraudo Sep 05 '17 at 08:37