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Let $(\Omega ,\mathcal F ,P) := \bigl((0,1],\mathcal B((0,1]),u \bigr)$, where $u$ is the Lebesgue measure restricted to $\mathcal B((0 ,1])$. Let $X\colon\Omega\to\mathbb R$ be defined by $X(\omega) := \omega^2$. $\mathcal F_n:=\sigma(\mathcal E_n)$. $\mathcal E_n=\{(\frac{k-1}{2^n} , \frac k{2^n}]\mid 1 \le k \le 2^n\}$

(a)Determine $H^+(\Omega,\mathcal F)$ and $E(X\mid \mathcal F_1)$

(b)for each $n$, determine a version $Y_n$ of $E[X\mid \mathcal F_n]$.

(c)Show that $E[Y_n\mid\mathcal F_m]= Y_m$ a.s for all $m\le n$.

This is a question from my recent homework. Can anyone give me a hint? By $H^+(\Omega,\mathcal F)$, I mean the set of nonnegative measurable functions .

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I don't know what $H^+$ means, but to get the conditional expectations you just average the random variable over the intervals in the partition,e.g. the first would be $2 \int _0^{\frac12}X , x < \frac 12, 2\int_{\frac 12}^1 X, x > \frac 12$ –  mike Jun 9 '12 at 18:55
    
keji: Any luck with an answer below? –  Did Jun 21 '12 at 6:21

2 Answers 2

Hint: Try to think how the $\mathcal{F}_n$'s look like. Are they countable or uncountable as sets? Then look in your definition of conditional expectation for the case you are in.

I also think that $H^{+}(\Omega,\mathcal{F})$ should read $H^{+}(\Omega,\mathcal{F}_n)$ for some $n \geq 1$ because else that is a pretty difficult set.

Another Hint: measurable functions with respect to finite $\sigma$-algebras (major hint...) have to be constant on the sets of the $\sigma$-algebra.

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This is a special case of the more general result below.

Consider a sigma-algebra $\mathcal G$ generated by a partition $(A_k)_k$ of $\Omega$. Hence, $\Omega=\bigcup\limits_kA_k$, $A_k\cap A_i=\varnothing$ for every $k\ne i$, and $A_k$ is in $\mathcal F$ for every $k$. Assume furthermore that $\mathrm P(A_k)\ne0$ for every $k$. Then, for every measurable and integrable random variable $X$ on $(\Omega,\mathcal F,\mathrm P)$,

$$ \mathrm E(X\mid \mathcal G)=\sum\limits_{k}\mathrm P(A_k)^{-1}\mathrm E(X:A_k)\cdot\mathbf 1_{A_k}=\sum\limits_{k}\mathrm E(X\mid A_k)\cdot\mathbf 1_{A_k}. $$

In the present case, $\mathcal F_n$ is generated by the sets $A(k,n)=(2^{-n}(k-1),2^{-n}k]$ for $1\leqslant k\leqslant 2^n$ and $\mathrm P(A(k,n))=2^{-n}$ for every $k$, hence $$ \mathrm E(X\mid \mathcal F_n)=2^n\sum\limits_{k=1}^{2^n}\mathrm E(X:A(k,n))\cdot\mathbf 1_{A(k,n)}. $$

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