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Problem Solutions – Chapter 3 Problem 3.1.1 Solution The CDF of X is F X ( x ) = 0 x < 1 ( x + 1 )/ 2 1 x < 1 1 x 1 (1) Each question can be answered by expressing the requested probability in terms of F X ( x ) . (a) P [ X > 1 / 2] = 1 P [ X 1 / 2] = 1 F X ( 1 / 2 ) = 1 3 / 4 = 1 / 4 (2) (b) This is a little trickier than it should be. Being careful, we can write P [ 1 / 2 X < 3 / 4] = P [ 1 / 2 < X 3 / 4] + P [ X =− 1 / 2] P [ X = 3 / 4] (3) Since the CDF of X is a continuous function, the probability that X takes on any specific value is zero. This implies P [ X = 3 / 4 ]= 0 and P [ X 1 / 2 0. (If this is not clear at this point, it will become clear in Section 3.6.) Thus, P [ 1 / 2 X < 3 / 4] = P [ 1 / 2 < X 3 / 4] = F X ( 3 / 4 ) F X ( 1 / 2 ) = 5 / 8 (4) (c) P [ | X | ≤ 1 / 2] = P [ 1 / 2 X 1 / 2] = P [ X 1 / 2] P [ X < 1 / 2] (5) Note that P [ X 1 / 2 F X ( 1 / 2 ) = 3 / 4. Since the probability that P [ X 1 / 2 0, P [ X < 1 / 2 P [ X 1 / 2 ] . Hence P [ X < 1 / 2 F X ( 1 / 2 ) = 1 / 4. This implies P [ | X | ≤ 1 / 2] = P [ X 1 / 2] P [ X < 1 / 2] = 3 / 4 1 / 4 = 1 / 2 (6) (d) Since F X ( 1 ) = 1, we must have a 1. For a 1, we need to satisfy P [ X a ] = F X ( a ) = a + 1 2 = 0 . 8 (7) Thus a = 0 . 6. Problem 3.1.2 Solution The CDF of V was given to be F V (v) = 0 v< 5 c (v + 5 ) 2 5 7 1 v 7 (1) (a) For V to be a continuous random variable, F V (v) must be a continuous function. This oc- curs if we choose c such that F V (v) doesn’t have a discontinuity at v = 7. We meet this requirement if c ( 7 + 5 ) 2 = 1. This implies c = 1 / 144. 80
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(b) P [ V > 4] = 1 P [ V 4] = 1 F V ( 4 ) = 1 81 / 144 = 63 / 144 (2) (c) P [ 3 < V 0] = F V ( 0 ) F V ( 3 ) = 25 / 144 4 / 144 = 21 / 144 (3) (d) Since 0 F V (v) 1 and since F V (v) is a nondecreasing function, it must be that 5 a 7. In this range, P [ V > a ] = 1 F V ( a ) = 1 ( a + 5 ) 2 / 144 = 2 / 3 (4) The unique solution in the range 5 a 7is a = 4 3 5 = 1 . 928. Problem 3.1.3 Solution In this problem, the CDF of W is F W (w) = 0 w< 5 (w + 5 )/ 8 5 3 1 / 4 3 3 1 / 4 + 3 (w 3 )/ 83 5 1 w 5 . (1) Each question can be answered directly from this CDF. (a) P [ W 4] = F W ( 4 ) = 1 / 4 (2) (b) P [ 2 < W 2] = F W ( 2 ) F W ( 2 ) = 1 / 4 1 / 4 = 0 (3) (c) P [ W > 0] = 1 P [ W 0] = 1 F W ( 0 ) = 3 / 4 (4) (d) By inspection of F W (w) , we observe that P [ W a ]= F W ( a ) = 1 / 2 for a in the range 3 a 5. In this range, F W ( a ) = 1 / 4 + 3 ( a 3 )/ 8 = 1 / 2 (5) This implies a = 11 / 3. Problem 3.1.4 Solution (a) By definition, d nx e is the smallest integer that is greater than or equal to . This implies ≤d e≤ + 1. 81
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(b) By part (a), nx n d e n + 1 n (1) That is, x d e n x + 1 n (2) This implies x lim n →∞ d e n lim n →∞ x + 1 n = x (3) (c) In the same way, b c is the largest integer that is less than or equal to . This implies 1 ≤b c≤ . It follows that 1 n b c n n (4) That is, x 1 n b c n x (5) This implies lim n →∞ x 1 n = x lim n →∞ d e n x (6) Problem 3.2.1 Solution f X ( x ) = ± cx 0 x 2 0 otherwise (1) (a) From the above PDF we can determine the value of c by integrating the PDF and setting it equal to 1.
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