hw4_sol - ECE 302 Fall 2009 Division 2 Homework 4 Solutions...

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Unformatted text preview: ECE 302 Fall 2009 Division 2 Homework 4 Solutions Problem 1. Let X be a random variable with PMF p X ( x ) = braceleftbigg | x | C , if x =- 4 ,- 3 ,- 2 ,- 1 , , 1 , 2 , 3 , 4 , otherwise. (a) Find C . Solution. According to the normalization axiom, the probabilities of all the experimental outcomes of a discrete random variable must add to one, i.e. 1 = 4 summationdisplay x =- 4 p X ( x ) = 1 C (4 + 3 + 2 + 1 + 0 + 1 + 2 + 3 + 4) = 20 C ⇒ C = 20 . (b) Find E [ X ]. Solution. By definition, for a discrete random variable, we have: E [ X ] = summationdisplay x x · p X ( x ) = 4 summationdisplay x =- 4 x | x | 20 = 1 20 [- 4 · 4 + (- 3) · 3 + (- 2) · 2 + (- 1) · 1 + 0 · 0 + 1 · 1 + 2 · 2 + 3 · 3 + 4 · 4] = 0 . In general, any random variable X with an even 1 PMF must have E [ X ] = 0. (c) Find the PMF of the random variable Z = ( X- E [ X ]) 2 . Solution. Since E [ X ]=0, we have Z = X 2 . It follows from definition that: p Z ( z ) = P ( { Z = z } ) = P ( { X 2 = z } ) . This is the probability that X 2 equals z . Since X can only be ± 1 , ± 2 , ± 3 , ± 4 (with non-zero probability), X 2 can only be 1, 4, 9, or 16. Therefore for z not equal to these four values, we must have p Z ( z ) = 0, and p Z ( z ) = P ( { X 2 = z } ) = p X (- √ z ) + p X ( √ z ) = | - √ z | 20 + | √ z | 20 = √ z 10 for z = 1 , 4 , 9 , 16 . (d) Using Part (c), compute the variance of X . Solution. By definition, the variance of X is: var( X ) = E { ( X- E [ X ]) 2 } = E [ Z ] = summationdisplay z =1 , 4 , 9 , 16 z · p Z ( z ) = 1 · √ 1 10 + 4 · √ 4 10 + 9 · √ 9 10 + 16 · √ 16 10 = 10 . 1 A function f is called even if f ( x ) = f (- x ). 1 (e) Compute the variance of X using the identity var( X ) = summationdisplay x ( x- E [ X ]) 2 p X ( x ). Solution. We only need to sum over x ’s which have non-zero probability: var( X ) = summationdisplay x = ± 1 , ± 2 , ± 3 , ± 4 ( x- E [ X ]) 2 p X ( x ) = (- 4) 2 p X (- 4) + (- 3) 2 p X (- 3) + (- 2) 2 p X (- 2) + (- 1) 2 p X (- 1) +(1) 2 p X (1) + (2) 2 p X (2) + (3) 2 p X (3) + (4) 2 p X (4) = 16 · 4 20 + 9 · 3 20 + 4 · 2 20 + 1 · 1 20 1 · 1 20 + 4 · 2 20 + 9 · 3 20 + 16 · 4 20 = (64 + 27 + 8 + 1) · 2 20 = 10 . Problem 2. The Curious Case of a 1992 Virginia Lottery. Before 1992, the state of Virginia ran a lottery where a winning six-number combination was selected as six distinct numbers out of 44 numbers. Anybody could buy any number of tickets and, on each ticket, select any set of six numbers. The price of one ticket was $1. Thus, in some drawings no one selected the winning combination and therefore no one won the jackpot. On the other hand, there were a few drawings where more than one person happened to select the winning combination. In this case, the jackpot was divided equally among the winners. Whenever there was a drawing with no jackpot winners, the jackpot for the next drawing would increase. This led to a $27M jackpot in the February 15, 1992 drawing, payable in 20 yearly installments of $1.35M.February 15, 1992 drawing, payable in 20 yearly installments of $1....
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This note was uploaded on 11/01/2009 for the course PSY 120 taught by Professor Donnely during the Spring '08 term at Purdue.

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hw4_sol - ECE 302 Fall 2009 Division 2 Homework 4 Solutions...

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