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I have no idea how to solve the following problem:

Consider the probability space $([0,1], \mathcal B([0,1]), \lambda),$ where $\lambda$ is the Lebesgue measure and $\mathcal B([0,1])$ stands for the Borel sets of $[0,1].$ Let $X(\omega) = \omega$ be the canonical random variable. Set $\mathcal F= \sigma([0,1/2])$ and $\mathcal G = \sigma(\mathcal B[0,1/2]).$ Compute and compare the two quantities $E[X\,\vert\,\mathcal F]$ and $E[X\,\vert\,\mathcal G].$

If someone has already an idea to how to start this one, it would help me a lot.

Thanks for your time,

Bests

Bling-Bling
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1 Answers1

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We have $\mathcal{F} =\{ \emptyset,[0,\frac{1}{2}], (\frac{1}{2},1], [0,1]\}$. So a variable $\mathcal{F}$-measurable is of the form $a1_{[0,\frac{1}{2}]}+b1_{(\frac{1}{2},1]}$.

From the definition, it comes : \begin{align*} E[X|\mathcal{F}](\omega) &= \frac{1}{\lambda([0,\frac{1}{2}])}\int_0^{\frac{1}{2}}x \, d x 1_{[0,\frac{1}{2}]}(\omega)+ \frac{1}{\lambda((\frac{1}{2},1])}\int^1_{\frac{1}{2}}x \, d x 1_{(\frac{1}{2},1]}(\omega)\\ & = \frac{1}{4} 1_{[0,\frac{1}{2}]}(\omega)+ \frac{3}{4} 1_{(\frac{1}{2},1]}(\omega) \end{align*}

Then you can verify (with the 3 axioms of a sigma-algebra) $$\mathcal{G}:=\sigma(\mathcal{B}([0,\frac12]))= \mathcal{B}([0,\frac12])\cup \{A\cup ( \frac12,1], \ A \in \mathcal{B}([0,\frac12])\} $$

It comes that a $\mathcal{G}$-measurable is of the form $f1_{[0,\frac{1}{2}]}+\alpha 1_{(\frac{1}{2},1]}$, with $f:[0,\frac12] \rightarrow \mathbb{R}$ being a $\mathcal{B}([0,\frac12])$-measurable function, $\alpha\in \mathbb{R}$.

Since, by definition, $Z:=E[X|\mathcal{F}]$ is the $\mathcal{G}$-measurable r. v. such that for every $\mathcal{G}$-measurable r. v.$U$, $$ E[XU]=E[ZU] $$ we have \begin{align*} E[X|\mathcal{F}](\omega)&= \omega 1_{[0,\frac{1}{2}]}(\omega)+ \frac{1}{\lambda((\frac{1}{2},1])}\int^1_{\frac{1}{2}}x \, d x 1_{(\frac{1}{2},1]}(\omega) \\ &= \omega 1_{[0,\frac{1}{2}]}(\omega)+\frac{3}{4} 1_{(\frac{1}{2},1]}(\omega) \\ \end{align*}

Netchaiev
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