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Pitrellais

Pitrellais

Answered question

2022-05-21

Consider a probability space ( Ω , F , P ) and some sub-σ-algebra G F . Can the conditional distribution
P ( A | G ) = E P [ 1 A | G ] , for  A F
be an actual probability measure on G and be used as such? For example for a random variable X, we would have
P ( X A | G ) = P ( | G ) ( X 1 ( A ) )
and so forth?
Or let me ask in another way: Should I understand a conditional distribution as an hypothetical distribution?

Answer & Explanation

Dreforganzv

Dreforganzv

Beginner2022-05-22Added 9 answers

The value P ( A | G ) is a G -measurable function, not a number. So A P ( A | G ) is not a measure.
You might attempt a "disintegration" where, for each ω Ω, the set-function A P ( A | G ) ( ω ) is a probability measure defined on F . [Search "disintegration" to find conditions under which this can be done.]
Note, it is not very interesting on G , since when A G , we have P ( A | G ) = 1 A .
nileteenice

nileteenice

Beginner2022-05-23Added 1 answers

Oh, yes, since 1 A is G -measurable, P ( A | G ) = E [ 1 A | G ] = 1 A .

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