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Let $X,Y,Z$ three mutually independent random variables. Let $f$ an arbitrary function.

Do we have the following ? $$ \mathbb{E}\left( f\left( X,Y\right) \big| X,Y,Z \right) = \mathbb{E}\left( f\left( X,Y\right) \big| X,Y \right) $$

The tower property of conditional expectation does not seem to help. Also, does the equality fail if we drop the independence assumption ?

anonymus
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1 Answers1

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As you depend on something that is not used in the function itself and is independent of all variables, $Z$ can be dropped. If these variables are not independent the equality is false in general.

LegNaiB
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