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Unformatted text preview: Solution Problem 1 (a) H = 0 (telling truth): f Y | ( y ) = e- y , y H = 1 (telling lie): f Y | 1 ( y ) = e y , y The detector decides for H = 0 if (1 p ) e- y pe- y . After re-arranging we obtain the following test: H = 1 iff y , where = 1 - ln bracketleftBig (1- p ) p bracketrightBig is the threshold. (b) If 0 then P L | T = 0. Otherwise P L | T = integraldisplay e- y dy = 1 e- . (c) P T | L = integraldisplay e- y dy = e- . (d) H = 0 : f Y | ( y ) = n e- ( y 1 + ... + y n ) = n e- y H = 1 : f Y | 1 ( y ) = n e- ( y 1 + ... + y n ) = n e- y , where y = n i =1 y i . With this new definition of y the test remains the same but with the new threshold = 1 - ln bracketleftBigparenleftBig parenrightBig n (1- p ) p bracketrightBig . Solving the problem to this point was sufficient to obtain full credit....
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This note was uploaded on 02/05/2012 for the course EE 132B taught by Professor Izhakrubin during the Spring '09 term at UCLA.
- Spring '09