hmwk8sol2027 - Solutions to Homework 8 ISyE 2027 Fall 2006...

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Solutions to Homework 8, ISyE 2027 Fall 2006 Problem 1 Suppose X is a continuous random variable with the following probability den- sity function: f ( t ) = ± ct 2 , 0 t 1; 0 , otherwise. (a) Here 1 = Z -∞ f ( x ) dx = Z 1 0 cx 2 dx = c/ 3 so it follows that c = 3. (b) Now P ( X 2 / 3) = Z 2 / 3 0 3 x 2 dx = (2 / 3) 3 = 8 / 27 . (c) But P ( X > 2 / 3) = 1 - P ( X 2 / 3) = 1 - 8 / 27 = 19 / 27 . (d) Furthermore, P ( X = 2 / 3) = Z 2 / 3 2 / 3 3 x 2 dx = 0 . (e) Notice that P ( X 1 / 3 | X 2 / 3) = P ( X 1 / 3 ,X 2 / 3) P ( X 2 / 3) = P ( X 1 / 3) P ( X 2 / 3) = 1 / 27 8 / 27 = = 1 / 8 . (f) Similarly, we see that P ( X 2 / 3 | X 1 / 3) = 1. (g) Notice that X can only take values in the interval [0 , 1], so it’s clear that P ( X 3) = 1. (h) Also, it’s clear that P ( X ≤ - 4) = 0. Problem 2 1
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We know that P ( X t ) = R t -∞ f ( s ) ds . Notice that if t < 0, P ( X t ) = Z t -∞ 0 ds = 0 . If 0 t < 1, then P ( X t ) = Z t 0 3 s 2 ds = t 3 . Finally, if
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This note was uploaded on 02/01/2010 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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hmwk8sol2027 - Solutions to Homework 8 ISyE 2027 Fall 2006...

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