Let X , Y , Z be integrable random variables on the same probability space. Is i

Emmy Dillon

Emmy Dillon

Answered question

2022-06-24

Let X , Y , Z be integrable random variables on the same probability space.
Is it true that σ ( E [ X | Y ] , E [ X | Z ] ) σ ( E [ X | Y , Z ] ) ?
My intuition says yes: the approximation of X on σ ( Y , Z ) is richer in information than the one on σ ( Y ) and the one on σ ( Z ). How to prove this?
If it is not true, what if we add σ ( Y ) σ ( Z ) ? Does this make the above true?

Answer & Explanation

Marlee Guerra

Marlee Guerra

Beginner2022-06-25Added 25 answers

No, not even with the added condition. The issue is that conditioning on a smaller σ-field may be able to "split apart" two events that X does not otherwise distinguish.
Consider a probability space with 4 outcomes, Ω = { a , b , c , d }, each having probability 1/4. We take F = 2 Ω so all sets are measurable.
Let Y = 1 { a , b } and let Z be injective, so that σ ( Z ) = F and in particular σ ( Y ) σ ( Z ). Define X as follows:
ω X ( ω ) a 0 b 2 c 2 d 8
We have E [ X Z ] = E [ X Y , Z ] = X. So σ ( E [ X Y , Z ] ) = σ ( X ) is generated by the events { a } , { b , c } , { d } and is not all of F . In particular { b } σ ( E [ X Y , Z ] ).
However, you may check that E [ X Y ] is given by
ω E [ X Y ] ( ω ) a 1 b 1 c 5 d 5
Thus { a , c } σ ( E [ X Y ] ). So σ ( E [ X Y ] , E [ X Z ] ) contains both { a , b } and { b , c } so it is all of F . In particular, it contains { b } = { E [ X Z ] = 2 } { E [ X Y ] = 1 }.

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