HW29Answers(11.1) - Math331 Spring 2008 Instructor David...

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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 11.1 Homework Answers Homework: pgs. 465 - 466, #’s 1, 3, 6, 7, 9, 17. 1. We have M X ( t ) = E bracketleftbig e tX bracketrightbig = 5 summationdisplay x =1 e tx P { X = x } = 5 summationdisplay x =1 e tx 1 5 = 1 5 ( e t + e 2 t + e 3 t + e 4 t + e 5 t ) . 3. We are given that for i = 1 , 2 , 3 , . . . p ( X = i ) = p ( i ) = 2 parenleftbigg 1 3 parenrightbigg i , and zero elsewhere. Thus, M X ( t ) = bracketleftbig e tX bracketrightbig = ∞ summationdisplay i =1 e ti P ( X = i ) = 2 ∞ summationdisplay i =1 e ti parenleftbigg 1 3 parenrightbigg i = 2 ∞ summationdisplay i =1 parenleftbigg e t 3 parenrightbigg i = 2 ∞ summationdisplay i =0 parenleftbigg e t 3 parenrightbigg i- 2 . Now, the sum above is only convergent if e t < 3, or t < ln(3). Thus, for such t , M X ( t ) = 2 1 1- e t / 3- 2 = 2 3 3- e t- 2(3- e t ) 3- e t = 2 e t 3- e t ....
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HW29Answers(11.1) - Math331 Spring 2008 Instructor David...

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