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Unformatted text preview: EE 278 October 16, 2009 Statistical Signal Processing Handout #7 Homework #3 Solutions 1. Family planning. Alice and Bob choose a number X at random with equal probability from the set { 2 , 3 , 4 } . If the outcome is X = x , they decide to have children until they have a girl or x children, whichever comes first. Assume that each child is a girl with probability 1 / 2 (independent of the number of children and gender of other children). Let Y be the number of children they will have. a. Find the conditional pmf p Y  X ( y  x ) for all possible values of x and y . b. Find the pmf of Y . Solution (10 points) a. Note that Y { 1 , 2 , 3 , 4 } . The conditional pmf is as follows p Y  X (1  2) = 1 2 , p Y  X (2  2) = 1 2 , p Y  X (1  3) = 1 2 , p Y  X (2  3) = 1 4 , p Y  X (3  3) = 1 4 , p Y  X (1  4) = 1 2 , p Y  X (2  4) = 1 4 , p Y  X (3  4) = 1 8 , p Y  X (4  4) = 1 8 . b. The pmf of Y is: p Y (1) = 4 summationdisplay x =2 p X ( x ) p Y  X (1  x ) = 1 / 2 , p Y (2) = 4 summationdisplay x =2 p X ( x ) p Y  X (2  x ) = 1 / 3 , p Y (3) = 4 summationdisplay x =3 p X ( x ) p Y  X (3  x ) = 1 / 8 , p Y (4) = p X (4) p Y  X (4  4) = 1 / 24 . 2. First available teller. A bank has two tellers. The service times for tellers 1 and 2 are indepen dent exponential random variables X 1 Exp( 1 ) and X 2 Exp( 2 ). You arrive at the bank and find that both tellers are busy but nobody else is waiting to be served. You are served by the first available teller who becomes free. What is the probability that you are served by the teller 1? Solution (5 points) The tellers service times are exponentially distributed, hence memoryless. Thus the service time distribution does not depend on my arrival time. The probability that I will be served by the first teller is P { X 1 < X 2 } = integraldisplay integraldisplay x 1 1 e 1 x 1 2 e 2 x 2 dx 2 dx 1 = integraldisplay 1 e ( 1 + 2 ) x 1 dx 1 = 1 1 + 2 . In other words, the probability of being served first by teller i is proportional to the tellers service rate i . 3. Optical communication channel. The input signal to an optical channel is X = braceleftBigg 1 with probability 1 2 10 with probability 1 2 . The output of the channel is Y , and the conditional pmf of Y given X = a is Poisson with intensity a ; i.e., Y { X = 1 } Poisson(1) and Y { X = 10 } Poisson(10). Show that the MAP rule reduces to D ( y ) = braceleftBigg 1 if y < y * 10 otherwise. Find y * and the corresponding probability of error. Solution (10 points) The probability of decoding error is minimized by the MAP rule: g ( y ) = braceleftBigg 1 p X  Y (1  y ) > p X  Y (10  y ) 10 otherwise Since the a priori probabilities for the two X values are equal, by Bayes rule the decision rule is equivalent to the maximumlikelihood decision rule: g ( y ) = braceleftBigg 1 p Y  X ( y  1) > p Y  X ( y  10) 10 otherwise For the optical communication channel of this problem...
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This note was uploaded on 11/28/2009 for the course EE 278 taught by Professor Balajiprabhakar during the Fall '09 term at Stanford.
 Fall '09
 BalajiPrabhakar
 Signal Processing

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