ECE109 Disc7.pdf - UC San Diego J Connelly ECE 109...

Info icon This preview shows pages 1–2. Sign up to view the full content.

UC San Diego J. Connelly ECE 109 Discussion 7 Notes Problem 7.1 Suppose X is an exponential random variable with parameter λ . Let Y be the 1 ’s digit of X , e.g., if X = 11 . 27 , then Y = 1 ; if X = 4 . 82 , then Y = 4 . Is Y a discrete or continuous random variable? Determine its PMF or its PDF. Solutions Y is discrete, since Y ∈ { 0 , 1 , 2 , . . ., 9 } . If Y = 0 , then X [0 , 1) [10 , 11) [20 , 21) ∪ · · · and more generally, if k ∈ { 0 , 1 , 2 , . . ., 9 } and Y = k , then X [ k, k + 1) [10 + k, 11 + k ) [20 + k, 21 + k ) ∪ · · · so P ( Y = k ) = P ( X [ k, k + 1) [10 + k, 11 + k ) [20 + k, 21 + k ) ∪ · · · ) = P ( k X < k + 1) + P (10 + k X < 11 + k ) + P (20 + k X < 21 + k ) + · · · = summationdisplay n =0 P (10 n + k X < 10 n + 1 + k ) = summationdisplay n =0 integraldisplay 10 n +1+ k 10 n + k f X ( u ) du = summationdisplay n =0 integraldisplay 10 n +1+ k 10 n + k λ e - λu du = summationdisplay n =0 e - λ (10 n + k ) - e - λ (10 n +1+ k ) = e - λk ( 1 - e - λ ) summationdisplay n =0 e - 10 λ n = e - λk 1 - e - λ 1 - e - 10 λ To verify that this is a valid PMF, we note that 9 summationdisplay k =0 e - λk = 1 - e - 10 λ 1 - e - λ which implies 9 summationdisplay k =0 P ( X = k ) = 1 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern