part25

# part25 - ECE 863 Analysis of Stochastic Systems Fall 2001...

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1 ECE 863 – Analysis of Stochastic Systems Fall 2001 Solutions for Homework Set #4 Michigan State University Department of Electrical & Computer Engineering Problem 1 4.3 (a) 2 PX 5 , Y2 ,Z 2  <>  [] [ ] () P5X51PY21P 2Z 2 =− << −− () () ( ) XX Y Z Z F5 F 5 1F2 1 F 2 F 2    4.3 (b) ,Y 0 ,Z 1 ><= =- - -- 1F5F0 F1 F1 XY Z Z bg ch di ej 4.3 (c) Pm i nX ,Y PX 2 ,Y 2 >= > > > - - 1F2 1F2 1F2 XYZ 4.3 (d) axX ,Z 6 PX 6 ,Y 6 <= < < < = --- F6F6F6 di di

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2 Problem 2 4.9 (a) F x,y ase bte dsdt XY as 2 /2 bt 2 0 y 0 x bg = F H I K F H I K -- z z =- F H I K - F H I K - - 1e ax 2 by 2 4.9 (b) PX Y axe bte dtdx ax 2 bt 2 0 x 0 >= F H I K F H I K z z F H I K - z axe 1 e dx ax 2 bx 2 0 + 1 a ab 4.9 (c) There are a couple of ways to solve this. We will use the joint cdf F XY . Fx Lim y , y 1 e XX Y ax 2 b g = A• - f= = - fx d dx Fx a x e xX ax 2 Similarly, fy b y e Y by 2 b g = -
3 Problem 3 4.15 We need to complete the square of the argument in the exponent:  −− −ρ +  σσ σ σ  22 11 2 2 ym xm 2 If we let = σ = σ 1 2 1 2 a b => ρ + a2 a b b To complete the square for ρ 2 a 2 ab, we add (and subtract) () ρ 2 b => ρ + ρ − ρ + =− ρ + ρ 2 a b b b b ab 1 b If we divide the exponent by − ρ 2 21 , this leads to: ± 2 2 2 b 2 + For now, we focus on 2 + ρ σ ρ σ ρ 2 1 2 1 2 1 2 12 2 1 2 2 1 2 1 1

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4 () + −ρ => +  =− σ ρ +  σσ σ− ρ  2 2 1 2 2
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part25 - ECE 863 Analysis of Stochastic Systems Fall 2001...

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