10 solution we want to maximize the objective

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Unformatted text preview: xjy? 10 Solution: We want to maximize the objective function fN y , xfX x = p1 e, y,x 2 = p1 e, x, 2 2  3  2  52 =2 5 : This is equivalent to minimizing gx = y , x + x , 5 = 1 , x + x , 5 = 2x , 12x + 26: 2 2 2 2 2 This is achieved at x = 3 and therefore the best estimate for the signal is x = 3 . 6. Random variable X has the following double exponential p.d.f. ,1 x 1: fX x = 3 e, jxj; 2 5 3 Double exponential pdf a Find the probability PrX 0. b Find the conditional probabilities PrX 1jX ,1 and PrX c If Y = eX , nd the variance Y of Y . d Find the characteristic function c.f. X u = E ejuX of X . 2 Solution: a Since fX x is symmetric w.r.t. x = 0, we have PrX b First, we nd PrX 1 = 1 Z 1 fX x dx = Z 1 1 3 ,3x e 2 1jX  ,1. 0 = 0:5 . 4 14 12 10 4 1 , 1 dx = 3 ,3 e, x = e2 = 0:0249: 2 3 3 1 3 Due to the symmetry in fX x, we have PrX ,1 = PrX 1 = 0:0249 and hence PrX ,1 = 1 , PrX ,1 = 0:9751. 3 Conditional probability: PrX 1 PrX 1jX ,1 = PrX X1; X,1,1 = PrX ,1 = 0::0249 = 0:0255 Pr 0 9751 Since X 1 and X  ,1 are mutually exclusive events, we have PrX 1jX  ,1 = PrX 1; X  ,1 = 0 PrX  ,1 4 3 5 c We rst nd the mean of Y = e : X Z EY = 1x e ,1 fX x dx 13 = ex 3 e x dx + ex 2 e, x dx ,1 2 3 e x dx + 1 3 e, x dx = 2 ,1 2 3 1 + 3 1 = 9 = 1:125: = 24 22 8 Z0 Z 3 3 0 Z Z0 4 2 0 5 We then nd the 2nd-order moment of Y ; i.e., E Y = E e X . 2 EY 2 1 2x e ,1 2 Z fX x dx 1 3 = e x 2 e x dx + e x 3 e, x dx 2 ,1 3 e x dx + 1 3 e,x dx = 2 ,1 2 = 3 1 + 3 = 1:8: 25 2 = Z0 Z 2 3 Z0 2 3 0 Z 5 0 5 Therefore the variance of Y is 2 Y = E Y , E Y  = 1:8 , 1:125 = 0:5344 : 2 2 2 2 d The c.f. of X : Y u = = = = = Z 1 ,1 fX x dx 1 3 ejux 2 e x dx + ejux 3 e, x dx 2 ,1 3 e ju x dx + 1 3 e ju, x dx 2 ,1 2 3 1 + 3 ,1 2 ju + 3 2 ju , 3 =9: ju + ,ju 9+u Z0 Z 3 Z0 3 Z  0 +3  3 0 3 2 1 3+ 1 3 2 10 5...
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