As above p x 1 2 and z 1 r 1 1 2 1 x 2 dx 5 24 so p x

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As above, P ( X > 1 / 2 and Z 1) = R 1 1 / 2 1 - x 2 dx = 5 / 24, so P ( X > 1 / 2 | Z 1) = 5 / 24 2 / 3 = 5 16 . (d) [8 points] (Deleted from exam.) Find the c.d.f. of Z . Solution: For z (0 , 1], F Z ( z ) = P ( Z z ) = Z z 0 z - x 2 dx = 2 z 3 / 2 3 . Page 4

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Math 431: Exam 2 For z (1 , 2), F Z ( z ) = P ( Z z ) = Z z - 1 0 dx + Z 1 z - 1 z - x 2 dx = z - 2( z - 1) 3 / 2 + 1 3 . 6. [12 points] Let X have exponential distribution with rate λ = 3. Find the density of Y = 1 /X . Solution: Note that Y has range (0 , ). The c.d.f. of Y is F Y ( y ) = P ( Y y ) = P (1 /X y ) = P ( X 1 /y ) = Z 1 /y 3 e - 3 t dt. (We could integrate, but since we are about to differentiate to find the density, it is not necessary.) The density of Y is f Y ( y ) = d dy F Y ( y ) = d dy Z 1 /y 3 e - 3 t dt = - 3 e - 3(1 /y ) d dy (1 /y ) = 3 e - 3 /y y 2 . Alternate Solution: We could also apply the change of variable formula from the text (instead of essentially re-deriving it). Again, not that the range of Y is (0 , ). We have Y = g ( x ) where g ( x ) = 1 /x . If y > 0, then y = g ( x ) = 1 /x for some x > 0. Then dy/dx = - 1 /x 2 = - y 2 and x = 1 /y , so f Y ( y ) = f X ( x ) | dy/dx | = f X (1 /y ) y 2 = 3 e - 3 /y y 2 . 7. [15 points] Assume that three points are picked independently and uniformly at random from a disk of radius one. Let R 1 , R 2 and R 3 be the distances of these points from the center of the disk. Let U = min { R 1 , R 2 , R 3 } and V = max { R 1 , R 2 , R 3 } . Find the joint density of U and V . Solution: Let 0 < u, v < 1 and let Δ u and Δ v be very small. Then P ( U ( u, u + Δ u ) , V ( v, v + Δ v )) 3 · 2 P ( R 1 ( u, u + Δ u ) , R 2 ( u, v ) , R 3 ( v, v + Δ v )) = 6 P ( R 1 ( u, u + Δ u )) P ( R 2 ( u, v )) P ( R 3 ( v, v + Δ v )) , Page 5
Math 431: Exam 2 where the 3 · 2 counts the number of ways to choose the least and greatest radius.
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