Assign_3_Solutions_1_

Assign_3_Solutions_1_ - STAT 330 Assignment 3 Solutions 1(a...

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1 . ( a ) Since Y | 9 v POI( 9 ) , E ( Y | 9 )= 9 and since 9 v GAM( . , / ) ,E ( 9 ./ . E ( Y E [ E ( Y | 9 )] = E ( 9 ./ . Since Y | 9 v 9 ) , Var ( Y | 9 9 and since 9 v . , / ) ,Var ( 9 ./ 2 . ( Y E [ ( Y | 9 )] + [ E ( Y | 9 )] = E ( 9 )+ ( 9 ) = ./ + ./ 2 ( b ) Since Y | 9 v 9 ) and 9 v . , / ) we have f 2 ( y | 9 9 y e # 9 y ! ,y =0 , 1 ,... ; 9 > 0 and f 1 ( 9 9 . # 1 e # x/ / / . # ( . ) , 9 > 0 and by the Product Rule f ( 9 f 2 ( y | 9 ) f 1 ( 9 9 y e # 9 y ! 9 . # 1 e # x/ / / . # ( . ) , 1 ; 9 > 0 . The marginal p.f. of Y is f 2 ( y Z 4 #4 f ( 9 ) d 9 = 1 y ! / . # ( . ) Z 4 0 9 y + . # 1 e # 9 (1+1 / / ) d 9 , let x = 9 µ 1+ 1 / = 9 µ / +1 / = 1 y ! / . # ( . ) µ / / y + . Z 4 0 x y + . # 1 e # x dx = / y (1 + / ) y + . # ( y + . ) y ! # ( . ) = / y (1 + / ) y + . ( y + . # 1)( y + . # 2) ··· ( . ) # ( . ) y ! # ( . ) = µ y + . # 1 y ¶µ 1 / . µ 1 # 1 / y , y , 1 If . is a nonnegative integer then we recognize this as the p.f. of a NB ³ . , 1 1+ / ´ random variable. 1 STAT 330 Assignment 3 Solutions
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3 (a) 2
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3. (b) C 2 C t 1 C t 2 M ( t 1 ,t 2 ) = C 2 C t 1 C t 2 exp µ 9 T t + 1 2 t T ) t = C 2 C t 1 C t 2 exp µ 9 1 t 1 + 9 2 t 2 + 1 2 t 2 1 > 2 1 + t 1 t 2 => 1 > 2 + 1 2 t 2 2 > 2 2 = C C t 2 £¡ 9 1 + t 1 > 2 1 + t 2 => 1 > 2 ¢ M ( t 1 2 ) ¤ = => 1 > 2 M ( t 1 2 )+ ¡ 9 1 + t 1 > 2 1 + t 2 => 1 > 2 ¢¡ 9 2 + t 2 > 2 2 + t 1 => 1 > 2 ¢ M ( t 1 2 ) therefore E ( XY )= C 2 C t 1 C t 2 M ( t 1 2 ) | ( t 1 ,t 2 )=(0 , 0) = => 1 > 2 + 9 1 9 2 From ( a ) we know E ( X 1 9 1 and E ( X 2 9 2 . Therefore Cov ( X,Y
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This note was uploaded on 07/12/2010 for the course STAT 330 taught by Professor Paulasmith during the Winter '08 term at Waterloo.

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Assign_3_Solutions_1_ - STAT 330 Assignment 3 Solutions 1(a...

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