- #1

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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]

- Thread starter CantorSet
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- #1

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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]

- #2

chiro

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Is this a homework question?

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]

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

- #3

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If you mean E((A+B)|C) by E(A+ B|C) , yes.

- #4

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It's not a homework question.Is this a homework question?

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

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.

- #5

Stephen Tashi

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because proving [itex] E(A+B) = E(A) + E(B) [/itex] would involve dealing with a similar equation.[itex]E((A+B)|C=c) = \sum_{a,b} (a+b) P(A=a,B=b|C=c)[/itex]

[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]

- #6

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Thanks.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]

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