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# HW4 - Probability Theory and Statistics(EE/TE 3341 Homework...

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Probability Theory and Statistics (EE/TE 3341) Homework 4 Solutions Problem Solutions : Yates and Goodman, 2.5.1 2.5.8 2.5.11 2.6.2 2.6.6 2.7.3 2.7.6 2.7.7 2.8.1 2.8.5 2.9.4 and 2.9.6 Problem 2.5.1 Solution For this problem, we just need to pay careful attention to the definitions of mode and median. (a) The mode must satisfy P X ( x mod ) P X ( x ) for all x . In the case of the uniform PMF, any integer x between 1 and 100 is a mode of the random variable X . Hence, the set of all modes is X mod = { 1 , 2 , . . . , 100 } (1) (b) The median must satisfy P [ X < x med ] = P [ X > x med ]. Since P [ X 50] = P [ X 51] = 1 / 2 (2) we observe that x med = 50 . 5 is a median since it satisfies P [ X < x med ] = P [ X > x med ] = 1 / 2 (3) In fact, for any x satisfying 50 < x < 51, P [ X < x ] = P [ X > x ] = 1 / 2. Thus, X med = { x | 50 < x < 51 } (4) Problem 2.5.8 Solution The following experiments are based on a common model of packet transmissions in data networks. In these networks, each data packet contains a cylic redundancy check (CRC) code that permits the receiver to determine whether the packet was decoded correctly. In the following, we assume that a packet is corrupted with probability ǫ = 0 . 001, independent of whether any other packet is corrupted. (a) Let X = 1 if a data packet is decoded correctly; otherwise X = 0. Random variable X is a Bernoulli random variable with PMF P X ( x ) = 0 . 001 x = 0 0 . 999 x = 1 0 otherwise (1) The parameter ǫ = 0 . 001 is the probability a packet is corrupted. The expected value of X is E [ X ] = 1 ǫ = 0 . 999 (2) 1

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(b) Let Y denote the number of packets received in error out of 100 packets transmitted. Y has the binomial PMF P Y ( y ) = braceleftbigg ( 100 y ) (0 . 001) y (0 . 999) 100 y y = 0 , 1 , . . . , 100 0 otherwise (3) The expected value of Y is E [ Y ] = 100 ǫ = 0 . 1 (4) (c) Let L equal the number of packets that must be received to decode 5 packets in error. L has the Pascal PMF P L ( l ) = braceleftbigg ( l 1 4 ) (0 . 001) 5 (0 . 999) l 5 l = 5 , 6 , . . . 0 otherwise (5) The expected value of L is E [ L ] = 5 ǫ = 5 0 . 001 = 5000 (6) (d) If packet arrivals obey a Poisson model with an average arrival rate of 1000 packets per second, then the number N of packets that arrive in 5 seconds has the Poisson PMF P N ( n ) = braceleftbigg 5000 n e 5000 /n ! n = 0 , 1 , . . . 0 otherwise (7) The expected value of N is E [ N ] = 5000. Problem 2.5.11 Solution We write the sum as a double sum in the following way: summationdisplay i =0 P [ X > i ] = summationdisplay i =0 summationdisplay j = i +1 P X ( j ) (1) At this point, the key step is to reverse the order of summation. You may need to make a sketch of the feasible values for i and j to see how this reversal occurs. In this case, summationdisplay i =0 P [ X > i ] = summationdisplay j =1 j 1 summationdisplay i =0 P X ( j ) = summationdisplay j =1 jP X ( j ) = E [ X ] (2) Problem 2.6.2 Solution From the solution to Problem 2.4.2, the PMF of X is P X ( x ) = 0 . 2 x = 1 0 . 5 x = 0 0 . 3 x = 1 0 otherwise (1) 2
(a) The PMF of V = | X | satisfies P V ( v ) = P [ | X | = v ] = P X ( v ) + P X ( v ) (2) In particular, P V (0) = P X (0) = 0 . 5 P V (1) = P X ( 1) + P X (1) = 0 . 5 (3) The complete expression for the PMF of V is P V ( v ) = 0 . 5 v = 0 0 . 5 v = 1 0 otherwise (4) (b) From the PMF, we can construct the staircase CDF of V .

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HW4 - Probability Theory and Statistics(EE/TE 3341 Homework...

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