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# HW5 - Probability Theory and Statistics(EE/TE 3341 Homework...

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Probability Theory and Statistics (EE/TE 3341) Homework 5 Solutions Problem Solutions : Yates and Goodman, 3.1.1 3.2.1 3.2.4 3.3.2 3.3.4 3.3.6 3.4.3 3.4.5 3.4.11 and 3.5.4 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. 1

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Problem 3.2.1 Solution f X ( x ) = braceleftbigg 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. integraldisplay 2 0 cx dx = 2 c = 1 (2) Therefore c = 1 / 2. (b) P [0 X 1] = integraltext 1 0 x 2 dx = 1 / 4 (c) P [ - 1 / 2 X 1 / 2] = integraltext 1 / 2 0 x 2 dx = 1 / 16 (d) The CDF of X is found by integrating the PDF from 0 to x .
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