A simple conditional expectation question

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Discussion Overview

The discussion revolves around the properties of conditional expectations in probability theory. Participants explore whether certain conditions regarding expected values of a random variable hold true under specific circumstances, particularly focusing on the implications of conditional expectations given different subsets of information.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that if E(v|x1) = b and E(v|x2) = b, it follows that E(v|x1,x2) = b, seeking clarification on this implication.
  • Another participant argues against the converse, stating that E(v|x1,x2) = b does not necessarily imply E(v|x1) = b and E(v|x2) = b, suggesting that counterexamples exist.
  • A participant suggests verifying the original claim by breaking down the definitions of conditional expectations and using algebraic manipulation.
  • One participant provides a specific example involving rolling two dice, where v represents the total and x1 and x2 represent specific outcomes of the dice.
  • Subsequent posts question whether the provided example serves as a valid counterexample to the original claim.

Areas of Agreement / Disagreement

Participants express disagreement regarding the validity of the converse statement about conditional expectations. The original claim about the implications of E(v|x1) and E(v|x2) remains a point of contention, with no consensus reached.

Contextual Notes

Participants reference the definitions of conditional expectations and the need for specific examples to illustrate their points, indicating that further mathematical exploration may be necessary to fully resolve the questions posed.

michonamona
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Let v be a random variable distributed according to F(.). Let X be a set containing the objects x1 and x2. Suppose

E(v|x1) = b AND E(v|x2) = b (The expected value of v conditional on x1 is b, etc)

where b is some constant.

Does it follow that E(v|x1,x2) = b? If so, why?



Additionally, does the converse hold?

i.e. does E(v|x1,x2) = b imply E(v|x1) = b and E(v|x2) = b?


Thank you!
M
 
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Your converse is not true. Since the "If" part is always true for some b, it does not follow that every subdivision gives equal values. Find a random two cases E(v|x1) not equal to E(v|x2) any example of this would give you a counter example when you set b equal to E(v|x1,x2).

The original holds (I'm pretty sure) and you can verify it by simply breaking down the (conditional) expectation values into their definitions, sums over values times the corresponding (conditional) probabilities. You then use the P(A|B) = P(A and B)/P(B) formula and some algebra...at least that's how I'd start. There might be a lot of regrouping terms in a sum of sums but you should be able to manipulate definition of one side to a definition of the other.
 
michonamona said:
Let v be a random variable distributed according to F(.). Let X be a set containing the objects x1 and x2. Suppose

E(v|x1) = b AND E(v|x2) = b (The expected value of v conditional on x1 is b, etc)

where b is some constant.

Does it follow that E(v|x1,x2) = b?
Roll two dice. v is total, x1 is first die showing 1, x2 is second showing 1.
 
Is this a counter example?
 
michonamona said:
Is this a counter example?
That's what I'm suggesting.
 

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