HW-help-10-4

HW-help-10-4 - 4.26 with 1 / = and 1 / = . By independence,...

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Hints for homework assigned Sept. 29 Originally due Oct. 4; revised due date Oct. 6. 4.26 (Needed for 7.14 ). Here, we have random variables X and Y independent, with f X ( x ) = αe - αx and f Y ( y ) = βe - βy , Z := min { X,Y } , W , with W = 1 iF X Y , and W = 0 iF Y < X . We are asked to show that Z and W are independent. Let us calculate the joint distribution F Z,W ( z,i ) := P ( Z z W = i ) by integrating over appropriate regions in the plane: P ( Z z W = 1) = P ( X z X Y ) = i z 0 i x αe - αx βe - βy dy dx = ··· (You provide details.) ··· = α α + β p 1 e - ( α + β ) z P . Similarly P ( Z z W = 0) = ··· = β α + β p 1 e - ( α + β ) z P . Also, show by integrating over the appropriate potions of the plane: P ( W = i ) = b β α + β , if i = 0; α α + β , if i = 1. ±inally, calculate P ( Z z ) = 1 P ( X > z Y > z ) = 1 P ( X > z ) · P ( Y > z ) = ··· = 1 e - ( α + β ) z . It is now easy to verify independence of Z and W . 7.14 We can apply
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Unformatted text preview: 4.26 with 1 / = and 1 / = . By independence, the joint distribution is the pdf f Z times the pmf f W , where f Z ( z | , ) = ( + ) e-( + ) z , z (0 , ) , f W ( w | , ) = w + (1 w ) + , w = 0 , 1 . 7.19 Here, 2 is the same for all the i . Note that Y i normal( x i , 2 ), so f Y i ( y i ) = 1 2 exp ( y i x i ) 2 2 2 . (a) Begin by writing f v Y ( vy | , 2 ), and then use the factorization theorem. (b) Use the log likelihood. The MLE is a linear combination of Y i , so the expected value is easy to nd. (c) See the hint for (b)....
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