part12

# part12 - ECE 863 Analysis of Stochastic Systems Fall 2001...

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ECE 863 1 ECE 863 – Analysis of Stochastic Systems Fall 2001 Solutions for Homework Set #2 Michigan State University Department of Electrical & Computer Engineering Problem 1 3.2 S 1 ,2 ,3 ,4 Y = lq PY 1 P a 1 2 == = PY 2 1 4 PY 3 1 8 PY 4 1 8 Problem 2 3.4 (a) SY : 0 Y 1 Y £ (b) Y y point is inside circle of radius y £= l q (c) PY y y 1 y 2 2 2 = p p

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ECE 863 2 Problem 3 3.7 Fx X bg 1 - - 7 8 -------------------------------------- 3 4 -------------------------- 1 2 ------------ 1 2 3 4 x X is a discrete random variable. Problem 4 3.9 F y PY y Y from problem (3.4), PY y y y 1 2 £= " £ , 1 1 y Y is a continuous random variable. F Y (y)
ECE 863 3 Problem 5 3.13 PX 1 3 1PX 1 3 1F 1 3 11 0 X > L N M O Q P =- £ L N M O Q P F H G I K J =-= PX 1 PX 1 X 1 ≥= £ - » lq l q s i n c e X1 £- and are mutually exclusive events, f≥ = £ + + = PX 1 PX 1 PX 1 F10 1 3 X bg 1 3 1P 2 3 X 4 3 -< L N M O Q P < < L N M O Q P s i n c e F x X is continuous at x 2 3 and x 4 3 = f- < L N M O Q P = F H G I K J -- F H G I K J =-∑ F H G I K J = 1 3 4 3 F 2 3 1 2 3 1 3 25 27 XX 2 Problem 6 3.14 (a) X is a random variable of the mixed type (b) 1 2 F 1 2 0 X <- L N M O Q P F H G I K J F H G I K J = -

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ECE 863 4 3.14 (b) – Continued: PX 0 F 0 0 X <= = - di 1 4 X £= = bg ()   ≤< = = =     - - XX 11 1 5 3 PX 1 F 1 - F 44 2 1 6 1 6 P 1 4 X1 F1 F 1 4 11 16 ££ L N M O Q P =- F H G I
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part12 - ECE 863 Analysis of Stochastic Systems Fall 2001...

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