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oldfinalsol - UCSD ECE 153 Handout#25 Prof Young-Han Kim...

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Unformatted text preview: UCSD ECE 153 Handout #25 Prof. Young-Han Kim Friday, December 5, 2008 Solutions to Old Final Exam 1. Breaking a stick (20 points). Take a stick of length 1. Break it at a location chosen uniformly ar random. Throw away the longer part and take the shorter one, the length of which will be denoted by a random variable X . (a) (5 points) What is the expected length EX of the remaining stick? (Hint: Let U denote the uniform location of break. Then X = min( U, 1- U ) . ) (b) (5 points) Find the pdf of X . Break the remaining stick once again and take the shorter one of length Y . (c) (5 points) What is the expected length EY of the remaining stick? (d) (5 points) Find the MMSE estimate of X given Y . (Hint: Find the conditional pdf f X | Y ( x | y ) first. Recall that integraltext 1 t dt = ln t + c .) X Y Solution: First note that X cannot be larger than 1 / 2. From the provided hint, for ≤ x ≤ 1 / 2, Pr( X ≥ x ) = Pr( x ≤ U ≤ 1- x ) = 1- 2 x, or equivalently, Pr( X ≤ x ) = 2 x , so that X is ∼ U(0 , 1 / 2). 1 (a) EX = 1 / 4 . (b) f X ( x ) = braceleftBigg 2 , ≤ x ≤ 1 / 2 , , otherwise. (c) Similarly, conditioned on { X = x } , Y is a uniform r.v. on [0 ,x/ 2]. Thus, E ( Y | X ) = X/ 4, and EY = EE ( Y | X ) = E ( X/ 4) = 1 / 16 . (d) First note that for 0 ≤ y ≤ x/ 2 ≤ 1 / 4, f X,Y ( x,y ) = f X ( x ) f Y | X ( y | x ) = 2 · 2 x = 4 x . Thus, for 0 ≤ y ≤ x/ 2 ≤ 1 / 4, f Y ( y ) = integraldisplay f X,Y ( x,y ) dx = integraldisplay 1 2 2 y 4 x dx = 4(ln(1 / 2)- ln(2 y )) = 4 ln(1 / 4 y ) , and f X | Y ( x | y ) = f X,Y ( x,y ) f Y ( y ) = 1 x ln(1 / 4 y ) . Therfore, E ( X | Y = y ) = integraldisplay 1 2 2 y xf X | Y ( x | y ) dx = integraldisplay 1 2 2 y 1 ln(1 / 4 y ) dx = 1 / 2- 2 y ln(1 / 2)- ln(2 y ) , and E ( X | Y ) = 1 / 2- 2 Y ln(1 / 2)- ln(2 Y ) . 2. Iocane or Sennari: Return of the chemistry professor (20 points)....
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oldfinalsol - UCSD ECE 153 Handout#25 Prof Young-Han Kim...

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