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14.2 Exercise 27

Avg difficulty 2.5

Use the result of the exercise and properties (CE9a) and (CE10) to show that

E [ g ( X , Y ) | Z = v ] = ∫ E [ g ( t , Y ) | X = t , Z = v ] F X | Z ( d t | v ) a . s . [ P Z ] E \left[\right. g \left(X , Y\right) \left|\right. Z = v \left]\right. = \int E \left[\right. g \left(t , Y\right) \left|\right. X = t , Z = v \left]\right. F_{X \left|\right. Z} \left(\right. d t \left|\right. v \left.\right) a . s . \left[P_{Z}\right]

Pfeiffer, Applied Probability, 14.2 Problems on Conditional Expectation, Regression, exercise 27

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Adapted from Applied Probability © 2009 Paul E. Pfeiffer. Licensed under CC BY 3.0. Source: Applied Probability. Imported at revision d7a0420.