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Sol-HW3 - 1 EE 7615 Solutions Problem Set 3 1 Let r =(r1 r2...

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1 EE 7615 Solutions Problem Set # 3 1) Let r = ( r 1 , r 2 ) = ( s + n 1 , s + n 1 + n 2 ) . The MAP rule for this problem maximizes p ( s | r ) . The question is whether decision based on the observation of ( r 1 , r 2 ) is equivalent to the decision made based on the observation of r 1 alone. This means whether the following conditions are equivalent p 1 p ( r 1 , r 2 | s 1 ) > p 2 p ( r 1 , r 2 | s 2 ) p 1 p ( r 1 | s 1 ) > p 2 p ( r 2 | s 1 ) Using the chain rule, we check the equivalence of the following p 1 p ( r 1 | s 1 ) p ( r 2 | r 1 , s 1 ) > p 2 p ( r 1 | s 2 ) p ( r 2 | r 1 , s 2 ) p 1 p ( r 1 | s 1 ) > p 2 p ( r 2 | s 1 ) or p ( r 1 | s 1 ) p ( r 1 | s 2 ) > p 2 p ( r 2 | r 1 , s 2 ) p 1 p ( r 2 | r 1 , s 1 ) p ( r 1 | s 1 ) p ( r 1 | s 2 ) > p 2 p 1 For these to be equivalent we have to have p ( r 2 | r 1 , s 1 ) = p ( r 2 | r 1 , s 2 ) or equivalently if p ( n 2 = r 2 r 1 | n 1 = r 1 s 1 ) = p ( n 2 = r 2 r 1 | n 1 = r 1 s 2 ) and since s 1 negationslash = s 2 , this is equivalent to n 2 being independent of n 1 . Therefore if n 2 and n 1 are independent then r 2 can be ignored, otherwise r 2 has to be used in an optimal decision scheme. A counterexample of dependent noises is the case where
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