Let ( Σ , F , P ) be a probability space, let X be an F-random variable, let G be a sub-sigma algebra of F, let ω ∈ Ω, let A be the intersection of all the sets in G that contain ω, and suppose that A ∈ F. Then is it true that E ( X | G ) ( ω ) = E ( X | A )?I just wrote this statement down based on my intuition of what conditional expectation means, and I want to verify if it’s actually true. Note that sigma algebras need not be closed under uncountable intersections, so A need not be an element of G. But I assumed A is at least an element of F so that E ( X | A ) is meaningful.

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Let   (  Σ  ,  F  ,  P  ) be a probability space, let   X be an   F-random variable, let   G be a sub-sigma algebra of   F, let   ω  ∈  Ω, let   A be the intersection of all the sets in   G that contain   ω, and suppose that   A  ∈  F. Then is it true that   E  (  X      |    G  )  (  ω  )  =  E  (  X      |    A  )?I just wrote this statement down based on my intuition of what conditional expectation means, and I want to verify if it’s actually true. Note that sigma algebras need not be closed under uncountable intersections, so   A need not be an element of   G. But I assumed   A is at least an element of   F so that   E  (  X      |    A  ) is meaningful.

Anika Stevenson · Accepted Answer

No, when you condition on a sigma field, there is little connection to the basic definition of conditionings. The only result I know is when you condition on a discrete random variable. In other words, let   Z be an integrable random variable and let   X be a discrete random variable. Suppose all the values   X take with positive probability are   (      x    k    )  . Then      E    (  Z  ∣  X  )  =      ∑    k        E    (  Z  ∣  X  =      x    k    )            1              {      X      =              x        k            }      and here       E    (  Z  ∣  X  =      x    k    ) is the usual conditional expectation over an event   A with positive probabiltiy       E    (  Z  ∣  A  )  =            1                        P                (        A        )                  ∫    A    Z  d      P    . (Note that when   Z  =            1              {      Y      ∈      B      }       then you have      E    (  Z  ∣  A  )  =                    P        (        Y        ∈        B        ,        X        =                  x          k                )                    P        (        X        =                  x          k                )              =  P  (  Y  ∈  B  ∣  X  =      x    k    )  .  )

Let ( <mi mathvariant="normal">Σ , F , P ) be a probability

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