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Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 4: Solutions Counting Random Variables, MaximumLikelihood Estimation 1. [Graphical study of binomial pmfs] (a) Work out the numbers yourself. (b) The graphs are as shown below. 0.2 0.4 0 1 2 3 4 5 6 7 8 9 10 p = 0.1 p = 0.9 0.1 0.2 0.3 0 1 2 3 4 5 6 7 8 910 p = 0.25 p = 0.75 0.1 0.2 0.3 0 1 2 3 4 5 6 7 8 9 10 p = 0.4 p = 0.6 p = 0.5 0.2 0.4 1 2 3 4 5 6 7 8 9 10 (c) The pmfs of X p and X 1 p are “reverses” of each other. For each choice of k , k = 0 , 1 ,..., 10, and any value of p , P { X p = k } = P { X 1 p = 10 k } . Note that X p counts the number of occurrences on 10 independent trials of an event A of probability p . But, if X p = k , then the complementary event A c (of probability 1 p ), must have occurred 10 k times, and the number of occurrences of A c is counted by X 1 p . 2. [The probability of an even number of successes] P { X is even } = N (1 p ) N + N 2 (1 p ) N 2 p 2 + N 4 (1 p ) N 4 p 4 + ··· Now, ( x + y ) N + ( x y ) N = 2 N x N + N 2 x N 2 y 2 + ··· , and so, setting x = 1 p , y = p , we get P { X is even } = 1 2 (1 p + p ) N + (1 p p ) N = 1 2 1 + (1 2 p ) N . Notice that the probability is 1/2 for p = 1 / 2. 3. [The pricing of airline tickets] (a) X takes on values i = 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 with probabilities p X ( i ) = 1 256 , 8 256 , 28 256 , 56 256 , 70 256 , 56 256 , 28 256 , 8 256 , 1 256 respectively. E [ X ] = 8 X i =0 i · p X ( i ) = 4 by tedious calculation, or, more simply, E [ X ] = np = 4. (b) Y = max { X 5 , } takes on values 1 , 2 , 3 and 0 with probabilities 28 256 , 8 256 , 1 256 , and 1 28+8+1 256 = 219 256 respectively....
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 Spring '08
 Milenkovic,O
 Poisson Distribution, Probability theory, Maximum likelihood, Estimation theory, NML, maximumlikelihood estimate

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