{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

oldfinalsol

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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)....
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online