Conditional expectation, Lebesgue measure

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SUMMARY

The discussion focuses on calculating the conditional expectation E(X|Y) where X(w) = 5w² and Y(w) is defined piecewise based on the interval [0,1]. The Lebesgue measure P is utilized to determine the probabilities of the sets A₁ = [0, 1/4] and A₂ = (1/4, 1]. The correct Lebesgue measure for A₁ is confirmed to be 1/4, leading to the conclusion that P(A₁) = 1/4 and P(A₂) = 3/4. This establishes the framework for calculating E(X|Y).

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  • Understanding of conditional expectation in probability theory
  • Familiarity with Lebesgue measure and its properties
  • Knowledge of Borel sets and σ-fields
  • Ability to perform integration of functions over specified intervals
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  • Study the properties of Lebesgue measure in detail
  • Learn about conditional expectations in probability theory
  • Explore integration techniques for piecewise functions
  • Investigate the relationship between Borel sets and Lebesgue measure
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spitz
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Homework Statement



Let Ω = [0,1] with the σ-field of Borel sets and let P be the Lebesgue measure on [0,1]. Find E(X|Y) if:

Homework Equations



[itex]X(w)=5w^2[/itex]

[itex]Y(w)= \left\{ \begin{array}{ll}<br /> 4 & \mbox{if $w \in [0,\frac{1}{4}]$} \\<br /> 2 & \mbox{if $w \in (\frac{1}{4},1]$} \\<br /> \end{array}<br /> \right.[/itex]

The Attempt at a Solution



For [itex]w\in A_1=[0,\frac{1}{4}][/itex]:

[itex]E(X|Y)(w)=E(X|A_1)=\frac{\int_{A_1}x\,dp}{P(A_1)}=\frac{1}{{P(A_1)}} \displaystyle\int_{0}^{1/4}5w^2\,dw[/itex]

Do I use [itex]P(A_1)=P(A_2)=\frac{1}{2}[/itex],

or [itex]P(A_1)=\frac{1}{4}[/itex], and [itex]P(A_2)=\frac{3}{4}[/itex] ?
 
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spitz said:

Homework Statement



Let Ω = [0,1] with the σ-field of Borel sets and let P be the Lebesgue measure on [0,1]. Find E(X|Y) if:

Homework Equations



[itex]X(w)=5w^2[/itex]

[itex]Y(w)= \left\{ \begin{array}{ll}<br /> 4 & \mbox{if $w \in [0,\frac{1}{4}]$} \\<br /> 2 & \mbox{if $w \in (\frac{1}{4},1]$} \\<br /> \end{array}<br /> \right.[/itex]

The Attempt at a Solution



For [itex]w\in A_1=[0,\frac{1}{4}][/itex]:

[itex]E(X|Y)(w)=E(X|A_1)=\frac{\int_{A_1}x\,dp}{P(A_1)}=\frac{1}{{P(A_1)}} \displaystyle\int_{0}^{1/4}5w^2\,dw[/itex]

Do I use [itex]P(A_1)=P(A_2)=\frac{1}{2}[/itex],

or [itex]P(A_1)=\frac{1}{4}[/itex], and [itex]P(A_2)=\frac{3}{4}[/itex] ?

You said that P was Lebesgue measure, so what do you think is the Lebesgue measure of [0,1/4]?

RGV
 
oh. 1/4
 

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