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Moving Supremum inside conditional expectation

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0

$begingroup$

Let $Z_l_l in Lambda,X$ be square-integrable random-variables, and $f:mathbbR^2rightarrow mathbbR$ be a continuous function, such that

• $
  sup_l in LambdamathbbE[f(Z_l,X)]<infty
  .
  $


• $operatornameesssup_l in LambdaZ_l<infty$

Under what conditions, is it true that
$$
sup_l in LambdamathbbEleft[
f(Z_l,X)
right]
=
mathbbEleft[
f(operatornameesssup_l in LambdaZ_l,X)
right]
$$

probability probability-theory measure-theory calculus-of-variations supremum-and-infimum

share|cite|improve this question

asked Apr 1 at 15:56

AIM_BLBAIM_BLB

2,5392820

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you are talking about the normal expected value, since conditional expectation is a random variable itself. At first glance, you need monotony in $f$ for $Z_l$, i.e. $Z_ileq Z_k$ implies $mathbbE[f(Z_i,X)] leq mathbbE[f(Z_k,X)]$. The bounds guarantee you that such a $sup$ exists almost surely (inside as well as outside of the expected value), so you are fine taking a surpremum over a bounded sequence. This is just a quick guess, maybe someone else is able to provide you with a more detailed answer with more relaxed conditions.
  $endgroup$
  – F. Conrad
  Apr 1 at 22:03
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $Z_l_l in Lambda,X$ be square-integrable random-variables, and $f:mathbbR^2rightarrow mathbbR$ be a continuous function, such that

• $
  sup_l in LambdamathbbE[f(Z_l,X)]<infty
  .
  $


• $operatornameesssup_l in LambdaZ_l<infty$

Under what conditions, is it true that
$$
sup_l in LambdamathbbEleft[
f(Z_l,X)
right]
=
mathbbEleft[
f(operatornameesssup_l in LambdaZ_l,X)
right]
$$

probability probability-theory measure-theory calculus-of-variations supremum-and-infimum

share|cite|improve this question

asked Apr 1 at 15:56

AIM_BLBAIM_BLB

2,5392820

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think you are talking about the normal expected value, since conditional expectation is a random variable itself. At first glance, you need monotony in $f$ for $Z_l$, i.e. $Z_ileq Z_k$ implies $mathbbE[f(Z_i,X)] leq mathbbE[f(Z_k,X)]$. The bounds guarantee you that such a $sup$ exists almost surely (inside as well as outside of the expected value), so you are fine taking a surpremum over a bounded sequence. This is just a quick guess, maybe someone else is able to provide you with a more detailed answer with more relaxed conditions.
  $endgroup$
  – F. Conrad
  Apr 1 at 22:03
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $Z_l_l in Lambda,X$ be square-integrable random-variables, and $f:mathbbR^2rightarrow mathbbR$ be a continuous function, such that

• $
  sup_l in LambdamathbbE[f(Z_l,X)]<infty
  .
  $


• $operatornameesssup_l in LambdaZ_l<infty$

Under what conditions, is it true that
$$
sup_l in LambdamathbbEleft[
f(Z_l,X)
right]
=
mathbbEleft[
f(operatornameesssup_l in LambdaZ_l,X)
right]
$$

probability probability-theory measure-theory calculus-of-variations supremum-and-infimum

share|cite|improve this question

asked Apr 1 at 15:56

AIM_BLBAIM_BLB

2,5392820

$endgroup$

share|cite|improve this question

asked Apr 1 at 15:56

AIM_BLBAIM_BLB

2,5392820

share|cite|improve this question

asked Apr 1 at 15:56

AIM_BLBAIM_BLB

2,5392820

asked Apr 1 at 15:56

AIM_BLBAIM_BLB

2,5392820

asked Apr 1 at 15:56

AIM_BLBAIM_BLB

2,5392820