Let X and Y be integrable, real-valued, i.i.d. random variables on the probability space ( Ω , F , P ) and let A ∈ F . Is it true that E ( X 1 A ) = E ( Y 1 A ), or, more generally, are X 1 A and Y 1 A i.i.d.? For any B ∈ B ( R ) we have P ( X 1 A ∈ B ) = P ( ( { X 1 A ∈ B } ∩ A ) ∪ ( { X 1 A ∈ B } ∩ A c ) ) = P ( { X ∈ B } ∩ A ) + P ( { 0 ∈ B } ∩ A c ) P ( Y 1 A ∈ B ) = … = P ( { Y ∈ B } ∩ A ) + P ( { 0 ∈ B } ∩ A c ) . This shows that X 1 A and Y 1 A are i.i.d. if A is independent of σ ( X ) and σ ( Y ). Is it possible to drop this condition? If not, is it possible to conclude the equality for the expectations? If necessary, one can assume that A ∈ σ ( X + Y ).

Question

Let   X and   Y be integrable, real-valued, i.i.d. random variables on the probability space   (  Ω  ,      F    ,  P  ) and let   A  ∈      F  . Is it true that   E  (  X      1    A    )  =  E  (  Y      1    A    ), or, more generally, are   X      1    A   and   Y      1    A   i.i.d.? For any   B  ∈      B    (      R    ) we have                    P        (        X                  1          A                ∈        B        )                            =        P        (        (        {        X                  1          A                ∈        B        }        ∩        A        )        ∪        (        {        X                  1          A                ∈        B        }        ∩                  A          c                )        )        =        P        (        {        X        ∈        B        }        ∩        A        )        +        P        (        {        0        ∈        B        }        ∩                  A          c                )                            P        (        Y                  1          A                ∈        B        )                            =        …        =        P        (        {        Y        ∈        B        }        ∩        A        )        +        P        (        {        0        ∈        B        }        ∩                  A          c                )        .            This shows that   X      1    A   and   Y      1    A   are i.i.d. if A is independent of   σ  (  X  ) and   σ  (  Y  ). Is it possible to drop this condition? If not, is it possible to conclude the equality for the expectations? If necessary, one can assume that   A  ∈  σ  (  X  +  Y  ).

ryancameron52 · Accepted Answer

If A is in   σ  (      X    +      Y    ) then             1        A    =  f  (      X    +      Y    ), a.s., for a Borel measurable function   f  :      R    →      R  . So       E    [      X              1        A    ]  =      E    [      X    f  (      X    +      Y    )  ] and       E    [      Y              1        A    ]  =      E    [      Y    f  (      X    +      Y    )  ]. Since       X   and       Y   are iid that means that the joint distribution of   (      X    ,      Y    ), or the distribution of this pair as a vector in             R        2   is the same as that of   (      Y    ,      X    ). So the expectations are equal.

Avah Knapp · Accepted Answer

If   X  ,  Y are i.i.d then   E  X      1    A    =  E  Y      1    A   need not hold. If   X  ,  Y are i.i.d.   N  (  0  ,  1  ) and   A  =  {  X  &amp;gt;  0  } then   E  X      1    A    &amp;gt;  0 wheras   E  Y      1    A    =  (  E  Y  )  P  (  X  &amp;gt;  0  )  =  0.The question in the title seems to have a typo. It is very different from the question whether  E  X      1    A    =  E  Y      1    A   holds.

Let X and Y be integrable, real-valued, i.i.d. random variables on the probability space

Answered question

Answer & Explanation

New Questions in Pre-Algebra