HW11_Sol - ECEN 646 Homework 11 Solutions 1 Additional...

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ECEN 646 Homework 11 Solutions 1 Additional Problem The receiver that maximizes the probability of correct decision is the maximum a posteriori (MAP) estimator. This is given by the value of m that maximizes the pmf, p M | R ( m | r ) as a function of r . This is equivalent to Fnding the m that maximizes p M,R ( m,r ) as a function of r . ±irst, the prior distributions p R | M ( r | m = 0) and p R | M ( r | m = 1) are p R | M ( r | m = 0) = 0 . 125 , r = 1 0 . 75 , r = 0 0 . 125 r = 1 p R | M ( r | m = 1) = 0 . 125 , r = 0 0 . 75 , r = 1 0 . 125 r = 2 (1) So the joint distribution p R,M ( r,m ) is given by p R,M ( = 0) = 1 / 32 , r = 1 3 / 16 , r = 0 1 / 32 r = 1 p R,M ( = 1) = 3 / 32 , r = 0 9 / 16 , r = 1 3 / 32 r = 2 (2) As a check, these 6 numbers add up to 1. So the MAP estimator is ˆ m ( r ) = arg max m p R,M ( ) = arg max(1 / 32 , 0) , r = 1 arg max(3 / 16 , 3 / 32) , r = 0 arg max(1 / 32 , 9 / 16) , r = 1 arg max(0 , 3 / 32) , r = 2 = 0 , r = 1 0 , r = 0 1 , r = 1 1 , r = 2 2 Additional Problem (a) The MMSE estimate of I is given by ˆ I = i -∞ Ip I | Y ( I | y ) dI = i -∞ I p Y | I ( y | I ) p I ( I ) p Y ( y ) dI = I -∞ Y | I ( y | I ) p I ( I ) dI I -∞ p Y | I ( y | I ) p I ( I ) dI (3) Note that p independent observations are included in Y . Letting y = [ y 1 ,y 2 ,... ,y p ], the pmf of Y = y conditioned on I is P b Y = y | I B = p p i =1 ( αI ) y i e - αI y i ! = e - αIp ( αI ) P p i =1 y i P p i =1 y i ! , for I > 0 , y i 0 1 This study resource was shared via CourseHero.com
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Thus the integral in the denominator of (3), p Y ( y ), is given by p Y ( y ) = p p P i =1 y i ! ± - 1 ¯ I - 1 i 0 e - Iαp - I/ ¯ I ( αI ) P p i =1 y i dI Using this result from an integral table: I 0 x n e - μx dx = n !
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