I have a question regarding the following bolded claim made on Wikipedia
Consider the following:
- $(\Omega, \mathcal{F}, P)$ is a probability space.
- $X: \Omega \rightarrow \mathbb{R}^{n}$ is a random variable on that probability space with finite expectation.
- $\mathcal{H} \subseteq \mathcal{F}$ is a sub-\sigma-algebra of $\mathcal{F}$.
Since $\mathcal{H}$ is a sub $\sigma$-algebra of $\mathcal{F}$, the function $X: \Omega \rightarrow \mathbb{R}^{n}$ is usually not $\mathcal{H}$-measurable, thus the existence of the integrals of the form $\left.\int_{H} X d P\right|_{\mathcal{H}}$, where $H \in \mathcal{H}$ and $\left.P\right|_{\mathcal{H}}$ is the restriction of $P$ to $\mathcal{H}$, cannot be stated in general. However, the local averages $\int_{H} X d P$ can be recovered in $\left(\Omega, \mathcal{H},\left.P\right|_{\mathcal{H}}\right)$ with the help of the conditional expectation. A conditional expectation of $X$ given $\mathcal{H}$, denoted as $\mathrm{E}(X \mid \mathcal{H})$, is any $\mathcal{H}$ measurable function $\Omega \rightarrow \mathbb{R}^{n}$ which satisfies:
$$
\int_{H} \mathrm{E}(X \mid \mathcal{H}) \mathrm{d} P=\int_{H} X \mathrm{~d} P
$$
Why is a random variable on a sigma algebra not necessarily measurable on a sub sigma algebra? If a sub-sigma algebra is a collection of subsets from the original sigma algebra which forms a sigma algebra in its own right, why wouldn't this enable any random variable to continue to be measurable on some sub-sigma algebra?
Best Answer
The sequence of comments above was getting a bit long, so I'll convert it into an answer.
As noted in @William's answer, making the domain's $\sigma$-algebra smaller while keeping the co-domain's $\sigma$-algebra fixed makes it harder for a function to be measurable.
As an example, if we take any non-constant random variable $X$ on a $\sigma$-algebra $\mathcal{F}$, and then set $\mathcal{G} = \{\emptyset, \Omega\} \subset \mathcal{F}$, then we note that $X$ cannot be measurable, as only constants are $\mathcal{G}$-measurable.
As an example for computing a conditional expectation, set $\Omega = [0,1]$, $\mathcal{F} = \mathcal{B}[0,1]$, and $P = \lambda$ (Lebesgue measure). We now set $\mathcal{G} = \{\emptyset , [0,1], [0,0.5), [0.5, 1]\}$ and relativise $\lambda$ to $\mathcal{G}$ (i.e. we assign Lebesgue measure to the sets in $\mathcal{G}$). Consider the random variable $X : \Omega \to \mathbb{R}$ given by $X(\omega ) = \omega$. This is $\mathcal{F}$-measurable, but it is not $\mathcal{G}$-measurable.
Notice that $\int_{[0,0.5)} X dP = \frac{1}{8}$ and $\int_{[0.5, 1]} X dP = \frac{3}{8}$. We now wish to find a $\mathcal{G}$-measurable function that integrates to these same values. For a function to be $\mathcal{G}$-measurable, it must be constant on $[0,0.5)$ and $[0.5,1]$. Thus, we may set $$Y(\omega) = \begin{cases} \frac{1}{4} & 0 \leq \omega < 0.5 \\ \frac{3}{4} & 0.5 \leq \omega \leq 1 \end{cases}$$ Notice that $Y$ is $\mathcal{G}$-measurable and $\int_{[0,0.5)} Y dP = \frac{1}{8}$ and $\int_{[0.5, 1]} Y dP = \frac{3}{8}$. Thus $E(X | \mathcal{G}) = Y$.