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Unformatted text preview: 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 112 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 (nonrandom) value and can be pulled out of the expectation. 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 g ( x ) h ( y ) p X,Y  Y ( x,y  y ) = X x,y g ( x ) h ( y ) P [ X = x,Y = y ,Y = y ] P [ Y = y ] , where P [ X = x,Y = y ,Y = y ] = ‰ P [ X = x,Y = y ] if y = y , otherwise.otherwise....
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 Winter '10
 GFung

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