C p error x 1 p 1 n t p n t 1 φ t 1 σ 5 6076 10 5 p

Info icon This preview shows pages 4–6. Sign up to view the full content.

View Full Document Right Arrow Icon
(c) P ( error | X = +1) = P [1 + N < T ] = P [ N < T - 1] = Φ( T - 1 σ ) = 5 . 6076 × 10 - 5 P ( error | X = - 1) = P [ - 1 + N > T ] = P [ N > T + 1] = 1 - Φ( T +1 σ ) = 1 . 7569 × 10 - 5 (d) The overall probability of error is given by: P ( error ) = P ( error | X = +1) P ( X = +1) + P ( error | X = - 1) P ( X = - 1) = 1 4 Φ( T - 1 σ ) + 3 4 [1 - Φ( T +1 σ )] = 2 . 7196 × 10 - 5 9. Quadratic function of Gaussian. Problem 4.90, page 223 of ALG Solution : Note X = ± p P/R and d X d P = ± 1 2 1 RP f P ( p ) = [ f X ( x ) + f X ( - x )] | d X d P | = [ f X ( p P/R ) + f X ( - p P/R )] 1 2 RP = 1 2 π [ e - ( P R - 1) 2 / 4 + e - ( - P R - 1) 2 / 4 ] 1 2 RP = 1 4 PRπ [ e - ( P R - 1) 2 / 4 + e - ( P R +1) 2 / 4 ] 10. Lognormal RV. Problem 4.91, page 223 of ALG Solution :(a) For y 0, P [ Y y ] = 0 For y > 0, P [ Y y ] = P [ e X y ] = P [ X ln y ] = F X (ln y ) F Y ( y ) = ( 0 y 0 F X (ln y ) y 0 Therefore, for y > 0 we have f Y ( y ) = d dy F Y ( y ) = 1 y f x (ln y ) (b) If X is a Gaussian random variable, then f Y ( y ) = ( 0 y 0 e - (ln y - m ) 2 / 2 σ 2 y 2 πσ y 0 4
Image of page 4

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

View Full Document Right Arrow Icon
-1 0 1 2 3 4 5 6 7 8 9 10 -0.5 0 0.5 1 1.5 Y F Y (y) The cdf of Y when X is zero-mean with variance 1/8 -2 0 2 4 6 8 10 0 0.2 0.4 0.6 0.8 1 1.2 1.4 Y f Y (y) The pdf of Y when X is zero-mean with variance 1/8 5
Image of page 5
-1 0 1 2 3 4 5 6 7 8 9 10 -0.5 0 0.5 1 1.5 Y F Y (y) The cdf of Y when X is zero-mean with variance 8 -2 0 2 4 6 8 10 0 1 2 3 4 5 6 7 8 Y f Y (y) The pdf of Y when X is zero-mean with variance 8 6
Image of page 6
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