solution_set3

# solution_set3 - EE 511 Solutions to Problem Set 3 1(a FY(y...

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EE 511 Solutions to Problem Set 3 1. (a) F Y ( y | X = 1 / 4) = P [ Y y | X = 1 / 4] = P [ X + N y | X = 1 / 4] = P [1 / 4 + N y | X = 1 / 4] = P [ N y - 1 / 4 | X = 1 / 4]. Since X and N are independent, we have P [ N y - 1 / 4 | X = 1 / 4] = P [ N y - 1 / 4] and F Y ( y | X = 1 / 4) = F N ( y - 1 / 4) . Therefore, f Y ( y | X = 1 / 4) = f N ( y - 1 / 4) i.e., uniform over (-1/4,3/4). Similarly, we get f Y ( y | X = - 1 / 4) = f N ( y + 1 / 4) i.e., uniform over (-3/4,1/4). Now, f Y ( y ) = P [ X = 1 / 4] f Y ( y | X = 1 / 4) + P [ X = - 1 / 4] f Y ( y | X = - 1 / 4) 1 f (y|X = 1/4) Y -1/4 3/4 1 f (y|X = -1/4) Y f (y) Y -1/4 1/4 -3/4 1/4 0.4 1 0.6 -3/4 3/4 (b) Let C denote the event that we make a correct decision. We have P [ C ] = Z -∞ P [ C | Y = y ] f Y ( y ) dy Since the integrand in the above equation is always positive, maximizing P [ C ] is the same as maximizing P [ C | Y = y ] for each y . Therefore, the optimal rule is to choose Decision = 1 / 4 if P [ X = 1 / 4 | Y = y ] > P [ X = - 1 / 4 | Y = y ] Decision = - 1 / 4 if P [ X = - 1 / 4 | Y = y ] > P [ X = 1 / 4 | Y = y ] This can also be written as Decision = 1 / 4 if f Y [ y | X = 1 / 4] P [ X = 1 / 4] f Y ( y ) > f Y [ y | X = - 1 / 4] P [ X = - 1 / 4] f Y ( y ) Decision = - 1 / 4 if f Y [ y | X = - 1 / 4] P [ X = - 1 / 4] f Y ( y ) > f Y [ y | X = 1 / 4] P [ X = 1 / 4] f Y ( y ) i.e., Decision = 1 / 4 if f Y [ y | X = 1 / 4] P [ X = 1 / 4] > f Y [ y | X = - 1 / 4] P [ X = - 1 / 4] Decision = - 1 / 4 if f Y [ y | X

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solution_set3 - EE 511 Solutions to Problem Set 3 1(a FY(y...

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