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Good day,

Currently I am working with "Probability: Theory and Examples" by Durrett and while getting familiar with conditional expectations I got to this problem:

Consider the Lebesgue probability space on the interval $[0,1)$. (I.e. the state space is $Ω = [0, 1)$, the $\sigma$-field is the set of Lebesgue measurable sets and the measure is the Lebesgue measure.) We define the random variable $X$ as: $$X(w)=\begin{cases} 2w &, 0\leq w < 1/2 \\ 2w−1 &, 1/2\leq w<1 \end{cases}$$ Compute the conditional expectation $E(Y |X)$ where $Y : [0, 1) \to \mathbb{R}$ is a measurable function.

First off $Y$ is not defined further. I am a bit confused about the term "measurable function". Of what? A measurable function of $X$? But then it would just be $E(Y|X)=Y$. So I assume $Y$ to be a random variable not necessarily independent of $X$.

Second let's define conditional expectation: $E(Y|X):=E(Y|\sigma(X))$ is a random variable $Z$ such that $Z$ is measurable w.r.t. $\sigma(X)$ and $E(1_A Y)=E(1_A Z)$ for all $A \in \sigma(X)$.

Okay, let's pick a $A \in \sigma(X)$ then there exists a $B \in \mathcal{B}(\mathbb{\mathbb{R}})$ such that $A=X^{-1}(B)$. As Graham Kemp helpfully hinted, the correct inverse of $X$ is

$$X^{-1}\{x\}~=~\{w\in[0;1):x=X(w)\}~=~\begin{cases}\{x/2, (1+x)/2\}&:& 0\leq x<1\\ \{\}&:&\text{otherwise}\end{cases}$$

Now I am not sure to do. Normally I would begin from $$E(1_A Y)= \int_A Y dP=\int_B y P_Y(dy)$$

but now I am at my end. Can someone take it from here and show me what to do? I am thankful for every help/hint.

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    $\begingroup$ Hint: $$E(Y\mid X)(\omega)=\frac{Y(\omega)+Y(\omega+\frac12)\mathbf 1_{2\omega<1}+Y(\omega-\frac12)\mathbf 1_{2\omega\geqslant1}}2=\frac12\sum_{\alpha\in B(\omega)}Y(\alpha)$$ where, for every $\omega$ in $[0,1)$, $B(\omega)=\{\alpha\in[0,1)\mid2\alpha=2\omega\bmod(1)\}$. $\endgroup$ – Did Sep 28 '16 at 13:01
  • $\begingroup$ @Did Thank you for your hint. I am a bit confused. Is this the end solution? And is $$E(Y|X)(\omega)=\frac{1}{2} \sum_{\alpha \in B(w)} Y(\alpha)$$ a general valid formula for conditional expectations? I would have preferred it via the integral. Let $$A=X^{-1}(B)= \bigcup_{ x \in B \cap [0,1)} \{ \frac{x}{2} , \frac{1+x}{2} \}$$ then $$\int_A Y dP = \sum_{ x \in B \cap [0,1)} \int_{ \{ \frac{x}{2} , \frac{1+x}{2} \} } Y dP$$ But this does not make sense. I don't know how to get to the formulation you hinted. $\endgroup$ – Fritz Sep 28 '16 at 13:48
  • $\begingroup$ Yes this the end solution (in every meaning of the term I can imagine) and yes the formula is valid (??). Your intuition might be enhanced by the remark that $B(\omega)=\{\alpha\mid X(\alpha)=X(\omega)\}$ is $X^{-1}(\{x\})$ for $x=X(\omega)$. $\endgroup$ – Did Sep 28 '16 at 14:00
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For every value of $x$ in $[0;1)$, the equation $x=X(w)$ has two solutions for $w$.

$$X^{-1}\{x\}~=~\{w\in[0;1):x=X(w)\}~=~\begin{cases}\{x/2, (1+x)/2\}&:& 0\leq x<1\\ \{\}&:&\text{otherwise}\end{cases}$$

Since the probability space is $\cal Lebesgue$ ...

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  • $\begingroup$ Thank you for writing down $X^{-1}$ correctly. Hmm I am not sure how to continue. As written how do I evaluate $\int_B y P_Y(dy)$? The measure of integration is Lebesgue, okay. I don't know how to determine $B$ if I take an arbitrary set $A$ as described in my question. $\endgroup$ – Fritz Sep 28 '16 at 12:17

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