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

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

View Full Document Right Arrow Icon
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 de±nitions 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 satis±es 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
Background image of page 1

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

View Full DocumentRight Arrow Icon
(b) Let Y denote the number of packets received in error out of 100 packets transmitted. Y has the binomial PMF P Y ( y ) = b ( 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 ) = b ( 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 ) = b 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: s i =0 P [ X > i ] = s i =0 s 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, s i =0 P [ X > i ] = s j =1 j 1 s i =0 P X ( j ) = s 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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 02/22/2012.

Page1 / 8

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

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online