stat450_sfinal_19_253bf1f6821ca9607a69a9b8a1fb1289.pdf - STAT 450/MAST 672/MAST 881C(2019 Sample questions for final with solutions 1(a Let X1 X2 have

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STAT 450/MAST 672/MAST 881C (2019) Sample questions for final with solutions 1. (a) Let X 1 , X 2 have the joint pdf f ( x 1 , x 2 ) = ( x 1 + x 2 ) - 1 e - ( x 1 + x 2 ) I ( x 1 > 0 , x 2 > 0). Find the joint pdf of T 1 = X 1 + X 2 , T 2 = X 1 /X 2 . Solution : We have: X 1 = T 1 T 2 / (1 + T 2 ) , X 2 = T 1 / (1 + T 2 ) , J = T 2 / (1 + T 2 ) T 1 / (1 + T 2 ) 2 1 / (1 + T 2 ) - T 1 / (1 + T 2 ) 2 and det( J ) = - T 1 / (1 + T 2 ) 2 . Hence joint pdf of ( T 1 , T 2 ) : e - t 1 t 1 t 1 (1 + t 2 ) 2 I ( t 1 > 0 , t 2 > 0) = e - t 1 (1 + t 2 ) 2 I ( t 1 > 0 , t 2 > 0) . (b) Let X 1 , X 2 have the joint pdf f ( x 1 , x 2 ) = ( x 1 + x 2 ) - 1 I ( x 1 > 0 , x 2 > 0 , x 1 + x 2 1) . Find the joint pdf of T 1 = X 1 + X 2 , T 2 = X 1 - X 2 , and their marginal pdf’s. Solution : We have X 1 = ( T 1 + T 2 ) / 2 , X 2 = ( T 1 - T 2 ) / 2 , J = 1 / 2 1 / 2 1 / 2 - 1 / 2 and det( J ) = - 1 / 2 . Hence joint pdf of ( T 1 , T 2 ): 1 2 t 1 I ( t 1 + t 2 > 0 , t 1 - t 2 > 0 , t 1 1) = 1 2 t 1 I ( | t 2 | ≤ t 1 1) , and the marginal pdf’s: f T 1 ( t 1 ) = 1 2 t 1 (∫ t 1 - t 1 dt 2 ) I (0 t 1 1) = I (0 t 1 1) , f T 2 ( t 2 ) = ( 1 | t 2 | 1 2 t 1 dt 1 ) I ( | t 2 | ≤ 1) = - ln | t 2 | 2 I ( | t 2 | ≤ 1) . 2. (a) Let X 1 , X 2 have the joint pdf f ( x 1 , x 2 ) = ( x 1 / 2) e - x 1 | x 2 | I (0 x 1 1). Find the joint pdf of T 1 = X 1 , T 2 = X 1 | X 2 | and the marginal pdf of T 2 . Solution : 1
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On B 1 := { ( x 1 , x 2 ) | 0 x 1 1 , x 2 < 0 } , x 1 = t 1 , x 2 = - t 2 /t 1 , J 1 = 1 0 t 2 /t 2 1 - 1 /t 1 , det( J 1 ) = - 1 /t 1 , and on B 2 := { ( x 1 , x 2 ) | 0 x 1 1 , x 2 > 0 } , x 1 = t 1 , x 2 = t 2 /t 1 , det( J 1 ) = 1 /t 1 . Hence joint pdf of ( T 1 , T 2 ) : g ( t 1 , t 2 ) = 2 × ( t 1 / 2) e - t 2 (1 /t 1 ) I (0 t 1 1 , t 2 > 0) = e - t 2 I (0 t 1 1 , t 2 > 0) , so that T 2 is Exponential (1). (b) Let X 1 , X 2 have the joint pdf f ( x 1 , x 2 ) = (2 2 π ) - 1 e -| x 1 |- 2 - 1 x 2 2 . Find the joint pdf of T 1 = | x 1 | +2 - 1 x 2 2 , T 2 = X 2 , and the marginal pdf of T 1 . Solution : On B 1 := { ( x 1 , x 2 ) | x 1 < 0 } , x 1 = - t 1 + ( t 2 2 / 2) , x 2 = t 2 , J 1 = - 1 t 2 0 1 , det( J 1 ) = - 1 , and on B 2 := { ( x 1 , x 2 ) | x 1 > 0 } , x 1 = t 1 - ( t 2 2 / 2) , x 2 = t 2 , det( J 1 ) = 1 . Hence joint pdf of ( T 1 , T 2 ) : g ( t 1 , t 2 ) = 2 × (2 2 π ) - 1 e - t 1 I ( t 1 - ( t 2 2 / 2) > 0 ) = 1 2 π e - t 1 I ( t 1 > ( t 2 2 / 2) ) , so that the marginal pdf of T 1 is g 1 ( t 1 ) = e - t 1 2 π t 1 > ( t 2 2 / 2) dt 2 = 2 t 1 / 2 1 e - t 1 π I ( t 1 > 0) . 3. (a) Let X 1 , X 2 have the joint pdf f ( x 1 , x 2 ) = ( x 1 + x 2 ) I (0 x 1 1 , 0 x 2 1). Find the joint pdf of T 1 = min( X 1 , X 2 ) , T 2 = max( X 1 , X 2 ) and the marginal pdf’s of T 1 , T 2 . Solution : Split the support of ( X 1 , X 2 ) into B 1 = { 0 x 1 < x 2 1 } and B 2 = { 0 x 2 < x 1 1 } . On B 1 , X 1 = T 1 , X 2 = T 2 , on B 2 , X 1 = T 2 , X 2 = T 1 , and the corresponding Jacobians have | det( J i ) | = 1 , i = 1 , 2 . Thus the joint pdf of ( T 1 , T 2 ) is g ( t 1 , t 2 ) = 2( t 1 + t 2 ) I (0 t 1 < t 2 1) , so that g 1 ( t 1 ) = ( 1 t 1 2( t 1 + t 2 ) dt 2 ) I (0 t 1 1) = (2 t 1 (1 - t 1 ) + 1 - t 2 1 ) I (0 t 1 1); g 2 ( t 2 ) = ( t 2 0 2( t 1 + t 2 ) dt 1 ) I (0 t 2 1) = 3 t 2 2 I (0 t 2 1) .
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