Is E(A+B|C) Equal to E(A|C) + E(B|C)?

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Discussion Overview

The discussion revolves around the property of conditional expectation, specifically whether the equation E(A+B|C) = E(A|C) + E(B|C) holds true. Participants explore this concept through definitions and mathematical reasoning, with a focus on the implications of conditional expectations in probability theory.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants express uncertainty about the validity of the equation E(A+B|C) = E(A|C) + E(B|C) and seek verification or disproof.
  • One participant clarifies that if E(A+B|C) is interpreted as E((A+B)|C), then the equation may hold true.
  • Another participant provides a definition of conditional expectation in the discrete case, suggesting that the sum of individual expectations may not equal the expectation of the sum, but questions the correctness of their own reasoning.
  • Further mathematical manipulation is presented, attempting to break down the expression E((A+B)|C=c) into components that relate to E(A|C=c) and E(B|C=c), leading to a detailed summation process.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the validity of the equation. There are competing views on whether the property holds, with some participants questioning the assumptions and definitions involved.

Contextual Notes

Participants note that the discussion is framed within the context of discrete cases of conditional expectation, and there are unresolved questions regarding the correctness of the mathematical steps taken in the reasoning.

CantorSet
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Hi everyone,

I have a feeling the following property is true but I can't find it stated in any textbook/online reference. Maybe it's not true... Can someone verify/disprove this equation?

[itex]E(A+B|C) = E(A|C) + E(B|C)[/itex]
 
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CantorSet said:
Hi everyone,

I have a feeling the following property is true but I can't find it stated in any textbook/online reference. Maybe it's not true... Can someone verify/disprove this equation?

[itex]E(A+B|C) = E(A|C) + E(B|C)[/itex]

Is this a homework question?

Regardless of your answer, what do you know about the definition of expectation and in particular conditional expectation?
 
If you mean E((A+B)|C) by E(A+ B|C) , yes.
 
chiro said:
Is this a homework question?

Regardless of your answer, what do you know about the definition of expectation and in particular conditional expectation?

It's not a homework question.

By definition of conditional expectation, we have in the discrete case
[itex]E(A|C=c) = \sum_{a} a P(A=a|C=c)[/itex]

[itex]E(B|C=c) = \sum_{b} b P(B=b|C=c)[/itex]

[itex]E((A+B)|C=c) = \sum_{a,b} (a+b) P(A=a,B=b|C=c)[/itex]

It doesn't seem like the sum of the first two should equal the last. But maybe my sum formula for the last one is wrong.
 
We should be able to make progress in simplifying:
CantorSet said:
[itex]E((A+B)|C=c) = \sum_{a,b} (a+b) P(A=a,B=b|C=c)[/itex]
because proving [itex]E(A+B) = E(A) + E(B)[/itex] would involve dealing with a similar equation.

[itex]\sum_{a,b}(a+b) P(A=a,B=b|C=c) = \sum_{a,b}a P(A=a,B=b|C=c) + \sum_{a,b} b P(A=a,B=b|C=c)[/itex]

[itex]= \sum_a \sum_b a P(A=a,B=b|C=c) + \sum_a \sum_b b P(A=a,B=b|C=c)[/itex]

[itex]= \sum_a a \sum_b P(A=a,B=b|C=c) = \sum_b b \sum_a P(A=a,B=b|C=c)[/itex]

[itex]= \sum_a a P(A=a|C=c) + \sum_b b P(B=b|C=c)[/itex]
 
Stephen Tashi said:
We should be able to make progress in simplifying:

because proving [itex]E(A+B) = E(A) + E(B)[/itex] would involve dealing with a similar equation.

[itex]\sum_{a,b}(a+b) P(A=a,B=b|C=c) = \sum_{a,b}a P(A=a,B=b|C=c) + \sum_{a,b} b P(A=a,B=b|C=c)[/itex]

[itex]= \sum_a \sum_b a P(A=a,B=b|C=c) + \sum_a \sum_b b P(A=a,B=b|C=c)[/itex]

[itex]= \sum_a a \sum_b P(A=a,B=b|C=c) = \sum_b b \sum_a P(A=a,B=b|C=c)[/itex]

[itex]= \sum_a a P(A=a|C=c) + \sum_b b P(B=b|C=c)[/itex]
Thanks.
 

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