{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midPDC03-sol

midPDC03-sol - Solution Problem 1(a H = 0(telling truth H =...

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

View Full Document Right Arrow Icon
Solution Problem 1 (a) H = 0 (telling truth): f Y | 0 ( y ) = αe - αy , y 0 H = 1 (telling lie): f Y | 1 ( y ) = βe βy , y 0 The detector decides for ˆ H = 0 if (1 p ) αe - αy pβe - βy . After re-arranging we obtain the following test: ˆ H = 1 iff y φ, where φ = 1 α - β ln bracketleftBig α β (1 - p ) p bracketrightBig is the threshold. (b) If φ 0 then P L | T = 0. Otherwise P L | T = integraldisplay φ 0 αe - αy dy = 1 e - αφ . (c) P T | L = integraldisplay φ βe - βy dy = e - βφ . (d) H = 0 : f Y | 0 ( y ) = α n e - α ( y 1 + ... + y n ) = α n e - αy H = 1 : f Y | 1 ( y ) = β n e - β ( y 1 + ... + y n ) = β n e - βy , where y = n i =1 y i . With this new definition of y the test remains the same but with the new threshold φ = 1 α - β ln bracketleftBigparenleftBig α β parenrightBig n (1 - p ) p bracketrightBig . Solving the problem to this point was sufficient to obtain full credit. P L | T = integraldisplay φ 0 f Y | 0 ( y | 0) dy but given H = 0, Y is no longer exponentially distributed. (It can be easily verified that f Y | 0 ( y | 0) is the Erlang density α n ( n - 1)!
Image of page 1

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

View Full Document Right Arrow Icon
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