Theorem 24 suppose that y 1 y n are random variables

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Theorem 2.4. Suppose that Y 1 ,...,Y n are random variables with joint pdf or pmf f ( y 1 ,...,y n ) . Let { i 1 ,...,i k } ⊂ { 1 ,...,n } , and let f ( y i 1 ,...,y i k ) be the marginal pdf or pmf of Y i 1 ,...,Y i k with support Y Y i 1 ,...,Y i k . Assume that E [ h ( Y i 1 ,...,Y i k )] exists. Then E [ h ( Y i 1 ,...,Y i k )] = integraldisplay -∞ · · · integraldisplay -∞ h ( y i 1 ,...,y i k ) f ( y i 1 ,...,y i k ) dy i 1 · · · dy i k = integraldisplay · · · integraldisplay Y Y i 1 ,...,Y i k h ( y i 1 ,...,y i k ) f ( y i 1 ,...,y i k ) dy i 1 · · · dy i k if f is a pdf, and E [ h ( Y i 1 ,...,Y i k )] = summationdisplay y i 1 · · · summationdisplay y i k h ( y i 1 ,...,y i k ) f ( y i 1 ,...,y i k ) = summationdisplay ( y i 1 ,...,y i k ) ∈Y Y i 1 ,...,Y i k h ( y i 1 ,...,y i k ) f ( y i 1 ,...,y i k ) if f is a pmf. Proof. The proof for a joint pdf is given below. For a joint pmf, replace the integrals by appropriate sums. Let g ( Y 1 ,...,Y n ) = h ( Y i 1 ,...,Y i k ) . Then E [ g ( Y )] = integraldisplay -∞ · · · integraldisplay -∞ h ( y i 1 ,...,y i k ) f ( y 1 ,...,y n ) dy 1 · · · dy n = integraldisplay -∞ · · · integraldisplay -∞ h ( y i 1 ,...,y i k ) bracketleftbiggintegraldisplay -∞ · · · integraldisplay -∞ f ( y 1 ,...,y n ) dy i k +1 · · · dy i n bracketrightbigg dy i 1 · · · dy i k = integraldisplay -∞ · · · integraldisplay -∞ h ( y i 1 ,...,y i k ) f ( y i 1 ,...,y i k ) dy i 1 · · · dy i k since the term in the brackets gives the marginal. QED
CHAPTER 2. MULTIVARIATE DISTRIBUTIONS 42 Example 2.5. Typically E ( Y i ) ,E ( Y 2 i ) and E ( Y i Y j ) for i negationslash = j are of pri- mary interest. Suppose that ( Y 1 ,Y 2 ) has joint pdf f ( y 1 ,y 2 ) . Then E [ h ( Y 1 ,Y 2 )] = integraldisplay -∞ integraldisplay -∞ h ( y 1 ,y 2 ) f ( y 1 ,y 2 ) dy 2 dy 1 = integraldisplay -∞ integraldisplay -∞ h ( y 1 ,y 2 ) f ( y 1 ,y 2 ) dy 1 dy 2 where -∞ to could be replaced by the limits of integration for dy i . In particular , E ( Y 1 Y 2 ) = integraldisplay -∞ integraldisplay -∞ y 1 y 2 f ( y 1 ,y 2 ) dy 2 dy 1 = integraldisplay -∞ integraldisplay -∞ y 1 y 2 f ( y 1 ,y 2 ) dy 1 dy 2 . Since finding the marginal pdf is usually easier than doing the double integral, if h is a function of Y i but not of Y j , find the marginal for Y i : E [ h ( Y 1 )] = integraltext -∞ integraltext -∞ h ( y 1 ) f ( y 1 ,y 2 ) dy 2 dy 1 = integraltext -∞ h ( y 1 ) f Y 1 ( y 1 ) dy 1 . Similarly, E [ h ( Y 2 )] = integraltext -∞ h ( y 2 ) f Y 2 ( y 2 ) dy 2 . In particular , E ( Y 1 ) = integraltext -∞ y 1 f Y 1 ( y 1 ) dy 1 , and E ( Y 2 ) = integraltext -∞ y 2 f Y 2 ( y 2 ) dy 2 . Suppose that ( Y 1 ,Y 2 ) have a joint pmf f ( y 1 ,y 2 ) . Then the expectation E [ h ( Y 1 ,Y 2 )] = y 2 y 1 h ( y 1 ,y 2 ) f ( y 1 ,y 2 ) = y 1 y 2 h ( y 1 ,y 2 ) f ( y 1 ,y 2 ) . In particular , E [ Y 1 Y 2 ] = summationdisplay y 1 summationdisplay y 2 y 1 y 2 f ( y 1 ,y 2 ) . Since finding the marginal pmf is usually easier than doing the double summation, if h is a function of Y i but not of Y j , find the marginal for pmf for Y i : E [ h ( Y 1 )] = y 2 y 1 h ( y 1 ) f ( y 1 ,y 2 ) = y 1 h ( y 1 ) f Y 1 ( y 1 ) . Similarly, E [ h ( Y 2 )] = y 2 h ( y 2 ) f Y 2 ( y 2 ) . In particular , E ( Y 1 ) = y 1 y 1 f Y 1 ( y 1 ) and E ( Y 2 ) = y 2 y 2 f Y 2 ( y 2 ).

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