This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: APPM 4/5570 Solutions to Problems from Section 5.2 and 5.3 27. The waiting time is either X Y or Y X , depending on who arrives first. We can thus write the waiting time as  X Y  . E [  X Y  ] = Z ∞∞ Z ∞∞  x y  f X,Y ( x, y ) dy dx. Since X and Y are independent, f X,Y ( x, y ) = f X ( x ) · F Y ( y ) = 6 x 2 y, ≤ x ≤ 1 , ≤ y ≤ 1 . So, E [  X Y  ] = Z 1 Z 1  x y  · 6 x 2 y dy dx Note that  x y  = x y , if x ≥ y y x , if x ≤ y So, E [  X Y  ] = R 1 R 1  x y  · 6 x 2 y dy dx = R 1 R x ( x y ) · 6 x 2 y dy dx + R 1 R 1 x ( y x ) · 6 x 2 y dy dx = 1 / 6 + 1 / 12 = 3 / 12 = 1 / 4 . 31. a. Cov ( X, Y ) = E [ XY ] E [ X ] E [ Y ] E [ XY ] = R 30 20 xy · 3 380 , 000 ( x 2 + y 2 ) dydx = 24 , 375 38 ≈ 641 . 45 f X ( x ) = Z 30 20 3 380 , 000 ( x 2 + y 2 ) dy = 3 380 , 000 10 x 2 + 19 , 000 3 , 20 ≤ x ≤ 30 . So, E [ X ] = Z 30 20 x · 3 380 , 000 10 x 2 + 19 , 000 3 dx = 1 , 925 76 Similarly, by the symmetry of the problem, f Y ( y...
View
Full
Document
This note was uploaded on 01/23/2011 for the course APPM 5440 at Colorado.
 '08
 staff

Click to edit the document details