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

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Unformatted text preview: 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 ) = 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 Problem 3.2.1 Solution f X ( x ) = braceleftbigg cx x 2 otherwise (1) (a) From the above PDF we can determine the value of c by integrating the PDF and setting it equal to 1.setting it equal to 1....
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HW5 - Probability Theory and Statistics (EE/TE 3341)...

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