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# hw1sol - Stochastic Processes nctuee07f Homework 1...

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Stochastic Processes nctuee07f Homework 1 Solutions 1. f X ( x ) = d dx P [ X x ] = d dx X n P [ X x | Y = n ] P [ Y = n ] (by total probability theorem) = X n d dx P [ X x | Y = n ] P [ Y = n ] = X n f X | Y ( x | Y = n ) P [ Y = n ] . 2. The probability that S = - 1, given that we have observed Y = y is P [ S = - 1 | Y = y ], which, according to the formula on page 1-12 of Topic 1, can be calculated by P [ S = - 1 | Y = y ] = P [ S = - 1] f Y | S ( y | S = - 1) P [ S = 1] f Y | S ( y | S = 1) + P [ S = - 1] f Y | S ( y | S = - 1) . With the given received signal model, the conditional density of Y given S = - 1 is just Gaussian with mean - 1 and variance σ 2 . Thus, we have f Y | S ( y | S = - 1) = 1 2 πσ 2 e - ( y +1) 2 2 σ 2 . And, similarly, f Y | S ( y | S = 1) = 1 2 πσ 2 e - ( y - 1) 2 2 σ 2 . After a little algebraic effort, it gives P [ S = - 1 | Y = y ] = e - y σ 2 e y σ 2 + e - y σ 2 . 3. (a) This identity holds true for both discrete and continuous random variables. Consider discrete X and Y . E [ E [ X | Y ]] = X y E [ X | Y = y ] p Y ( y ) = X y ˆ X x xp X | Y ( x | y ) ! p Y ( y ) = X x x ˆ X y p X | Y ( x | y ) p Y ( y ) ! = X x x ˆ X y p X,Y ( x, y ) ! = X x xp X ( x ) = E [ X ] . (b) The result of this problem is intuitively obvious. The intuition is, given that Y = y , there is nothing random about h ( Y ) | Y = y . Thus, it serves as a deterministic (non-random) value and can be pulled out of the expectation.

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But we still need to justify it mathematically. We consider the discrete random variables case here. Continuous random variables case can also be shown in a similar fashion. Let X and Y be two discrete random variables. From the rule of expected value, the conditional expectation is E h g ( X ) · h ( Y ) fl fl fl Y = y i = X x,y 0 g ( x ) h ( y 0 ) p X,Y | Y ( x, y 0 | y ) = X x,y 0 g ( x ) h ( y 0 ) P [ X = x, Y = y 0 , Y = y ] P [ Y = y ] , where P [ X = x, Y = y 0 , Y = y ] = P [ X = x, Y = y ] if y 0 = y , 0 otherwise. Thus, the conditional expectation is evaluated only when y 0 = y , which gives E h g ( X ) · h ( Y ) fl fl fl Y = y i = X x,y 0 g ( x ) h ( y 0 ) P [ X = x, Y = y 0 , Y = y ] P [ Y = y ] = X x g ( x ) h ( y ) P [ X = x, Y = y ] P [ Y = y ] (a result from y 0 = y ) = h ( y ) X x g ( x ) P [ X = x, Y = y ] P [ Y = y ] = h ( y ) ˆ X x g ( x ) p X | Y ( x
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