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20065ee131A_1_EE131AF06HW6Solution

# 20065ee131A_1_EE131AF06HW6Solution - Solution of Assignment...

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Solution of Assignment 6, EE 131A Due: Monday November 27, 2006 Problem 1. If a random variable X is such that E [( X - 1) 2 ] = 10 and E [( X - 2) 2 ] = 6 , find E [ X ] and Var [ X ] . Solution: We have E [( X - 1) 2 ] = E [ X 2 ] - 2 E [ X ] + 1 = 10 , E [( X - 2) 2 ] = E [ X 2 ] - 4 E [ X ] + 4 = 6 . From these equations it follows that E [ X ] = 7 2 and E [ X 2 ] = 16 ; therefore Var [ E ] = 16 - 49 4 = 15 4 . Problem 2. A random variable X has the density function f X ( x ) = 1 2 e −| x | , -∞ < x < . (a) Find the distribution function of X . (b) Find the mean and variance of X . (c) Find P [ | X | > 2] . (d) Use Chebyshev’s inequality to obtain an upper bound on P [ | X | > 2] and compare with result in (c). Solution: (a) For x 0 we have F X ( x ) = 1 2 integraldisplay x −∞ e t dt = 1 2 e x . For x > 0 we have F X ( x ) = F X (0) + 1 2 integraldisplay x 0 e t dt = 1 - 1 2 e x . Therefore, F X ( x ) = 1 2 e x if x < 0 , 1 - 1 2 e x if x 0 . (b) Since f X ( x ) is symmetric about the point m = 0 , thus E [ X ] = 0 . Now using the formula integraltext x 2 e a x dx = e a x ( a 2 x 2 - 2 a x + 2) /a 3 , we have E [ X 2 ] = 1 2 integraldisplay 0 −∞ x 2 e x dx + 1 2 integraldisplay 0 x 2 e x dx = 2 .

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Therefore, Var [ X ] = 2 . (c) P [ | X | > 2] = P [ X < - 2] + P [ X > 2] = 1 2 integraldisplay 2 −∞ e x dx + 1 2 integraldisplay 2 e x dx = e 2 0 . 135 . (d) Using Chebyshev’s inequality, we have P [ | X | > 2] Var [ X ] 4 = 0 . 5 . Problem 3. Suppose that X is a random variable with mean and variance both equal to 20. Find a lower bound for P [0 X 40] . Solution: First note that P [0 X 40] = 1 - P [ | X - 20 | ≥ 20] . Now, by Chebyshev’s inequality, we have P [ | X - 20 | ≥ 20] 20 20 2 = 0 . 05 . Thus, P [0 X 40] 1 - 0 . 05 = 0 . 95 . Problem 4. Suppose that X is non-negative integer-valued discrete random variable. Show that E [ X ] = summationdisplay k =0 P [ X > k ] .
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