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14.2 Exercise 26

Avg difficulty 2.5

Use the fact that g(X,Y)=g*(X,Y,Z)g \left(X , Y\right) = g^{\star} \left(X , Y , Z\right), where g*(t,u,v)g^{\star} \left(t , u , v\right) does not vary with v. Extend property (CE10) to show

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

Pfeiffer, Applied Probability, 14.2 Problems on Conditional Expectation, Regression, exercise 26

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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.