midterm_solutions

# A find the conditional mean and variance of l given x

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(a) Find the conditional mean and variance of L given X , i.e., E ( L | X ) and Var( L | X ). 3

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(b) Find the pdf of L . (c) Derive the optimum detector of X given L , d ( L ), that minimizes the probability of error P ( d ( L ) negationslash = X ). (d) Express the minimum probability of error as a function of σ 2 . Solution: (a) g ( y ) can be simplified as follows: g ( y ) = ln parenleftbigg f Y | X ( y | 1) f Y | X ( y | − 1) parenrightbigg = ln 1 2 πσ e - ( y - 1) 2 2 σ 2 1 2 πσ e - ( y +1) 2 2 σ 2 = ( y 1) 2 2 σ 2 + ( y + 1) 2 2 σ 2 = 2 y σ 2 . Since L = 2 σ 2 Y , E ( L | X ) = 2 σ 2 E ( Y | X ) = 2 σ 2 E ( X + Z | X ) = 2 σ 2 X and Var( L | X ) = 4 σ 4 Var ( Y | X ) = 4 σ 4 Var ( X + Z | X ) = 4 σ 4 Var( Z ) = 4 σ 2 (b) Since a linear function of a Gaussian random variable is Gaussian, L | ( X = ± 1) is Gaussian. Then from part (a), L | ( X = ± 1) ∼ N ( ± 2 σ 2 , 4 σ 2 ), i.e., f L | X ( l | x ) = σ 2 2 π e - σ 2 ( l - 2 σ 2 x ) 2 8 Therefore, the pdf of L is f L ( l ) = P { X = 1 } f L | X ( l | 1) + P { X = 1 } f L | X ( l | − 1) = σ 4 2 π e - σ 2 ( l - 2 σ 2 ) 2 8 + σ 4 2 π e - σ 2 ( l + 2 σ 2 ) 2 8 (c) Since X is equiprobable, the following ML detector is optimum. d ( l ) = braceleftBigg 1 , if f L | X ( l | 1) f L | X ( l | − 1) 1 , otherwise . 4
Since f L | X ( l | x ) = σ 2 2 π e - σ 2 ( l - 2 σ 2 x ) 2 8 as was calculated in part (b), f L | X ( l | 1) f L | X ( l | − 1) if and only if l 0. Thus, the optimum detector is d ( L ) = braceleftBigg 1 , if L 0 1 , otherwise .
• Fall '08
• staff
• Variance, Probability theory, pN |X

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