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# 5 - from which we obtain 3 E[X | A = 2 x 2x 2x3 dx = 5 15 3...

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from which we obtain E [ X | A ] = 3 2 x · 2 x 5 dx = 2 x 3 15 3 2 = 54 15 16 15 = 38 15 . (b) We have E [ Y ] = E [ X 2 ] = 3 1 x 3 4 dx = 5 , and E [ Y 2 ] = E [ X 4 ] = 3 1 x 5 4 dx = 91 3 . Thus, var( Y ) = E [ Y 2 ] ( E [ Y ] ) 2 = 91 3 5 2 = 16 3 . Solution to Problem 3.19. (a) We have, using the normalization property, 2 1 cx 2 dx = 1 , or c = 1 2 1 x 2 dx = 2 . (b) We have P ( A ) = 2 1 . 5 2 x 2 dx = 1 3 , and f X | A ( x | A ) = 6 x 2 , if 1 . 5 < x 2, 0 , otherwise. (c) We have E [ Y | A ] = E [ X 2 | A ] = 2 1 . 5 6 x 2 x 2 dx = 3 , E [ Y 2 | A ] = E [ X 4 | A ] = 2 1 . 5 6 x 2 x 4 dx = 37 4 , and var( Y | A ) = 37 4 3 2 = 1 4 . Solution to Problem 3.20. The expected value in question is E [Time] = ( 5 + E [stay of 2nd student] ) · P (1st stays no more than 5 minutes) + ( E [stay of 1st | stay of 1st 5] + E [stay of 2nd] ) · P (1st stays more than 5 minutes) . 41

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We have E [stay of 2nd student] = 30, and, using the memorylessness property of the exponential distribution, E [stay of 1st | stay of 1st 5] = 5 + E [stay of 1st] = 35 . Also P (1st student stays no more than 5 minutes) = 1 e 5 / 30 , P (1st student stays more than 5 minutes) = e 5 / 30 . By substitution we obtain E [Time] = (5 + 30) · (1 e 5 / 30 ) + (35 + 30) · e 5 / 30 = 35 + 30 · e 5 / 30 = 60 . 394 . Solution to Problem 3.21. (a) We have f Y ( y ) = 1 /l , for 0 y l . Furthermore, given the value y of Y , the random variable X is uniform in the interval [0 , y ]. Therefore, f X | Y ( x | y ) = 1 /y , for 0 x y . We conclude that f X,Y ( x, y ) = f Y ( y ) f X | Y ( x | y ) = 1 l · 1 y , 0 x y l , 0 , otherwise. (b) We have f X ( x ) = f X,Y ( x, y ) dy = l x 1 ly dy = 1 l ln( l/x ) , 0 x l. (c) We have E [ X ] = l 0 xf X ( x ) dx = l 0 x l ln( l/x ) dx = l 4 . (d) The fraction Y/l of the stick that is left after the first break, and the further fraction X/Y of the stick that is left after the second break are independent. Furthermore, the random variables Y and X/Y are uniformly distributed over the sets [0 , l ] and [0 , 1], respectively, so that E [ Y ] = l/ 2 and E [ X/Y ] = 1 / 2. Thus, E [ X ] = E [ Y ] E X Y = l 2 · 1 2 = l 4 . Solution to Problem 3.22. Define coordinates such that the stick extends from position 0 (the left end) to position 1 (the right end). Denote the position of the first break by X and the position of the second break by Y . With method (ii), we have X < Y . With methods (i) and (iii), we assume that X < Y and we later account for the case Y < X by using symmetry. Under the assumption X < Y , the three pieces have lengths X , Y X , and 1 Y . In order that they form a triangle, the sum of the lengths of any two pieces must exceed the length of the third piece. Thus they form a triangle if X < ( Y X ) + (1 Y ) , ( Y X ) < X + (1 Y ) , (1 Y ) < X + ( Y X ) . 42
y 1 f X , Y (x , y) = 2 f X | Y (x |y ) 1 - y 1 (a) (b) x 1 1 - y y x 1 1 - y Figure 3.1: (a) The joint PDF. (b) The conditional density of X . These conditions simplify to X < 0 . 5 , Y > 0 . 5 , Y X < 0 . 5 . Consider first method (i). For X and Y to satisfy these conditions, the pair ( X, Y ) must lie within the triangle with vertices (0 , 0 . 5), (0 . 5 , 0 . 5), and (0 . 5 , 1). This triangle has area 1 / 8. Thus the probability of the event that the three pieces form a triangle and X < Y is 1/8. By symmetry, the probability of the event that the three

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