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Unformatted text preview: Stochastic Processes nctuee07f Homework 3 Solutions 1. Exercise 3.10 in Gallager’s note. Solutions are on page 8 (labelled as 3). 2. Exercise 3.11 in Gallager’s note. Solutions are on page 9 (labelled as 4). 3. (a) The maximum likelihood detection rules states L(H 1 ) H 1 ≷ H 2 L(H 2 ) , where L(H 1 ) and L(H 2 ) are the likelihood of the signal s associated with hypoth esis H 1 and H 2 , respectively. More specifically, by taking natural log of both sides, we have 1 2 ( y s ) H K 1 n ( y s ) H 1 ≷ H 2 1 2 y H K 1 n y . With some rearrangement, it follows that s H K 1 n  {z } = w H ML · y H 1 ≷ H 2 1 2 s H K 1 n s . That is, w ML = K 1 n s and η = 1 2 s H K 1 n s . (b) We can carry out SNR as SNR = w H ss H w w H k n w = w H K 1 2 n ¶ K 1 2 n ss H K 1 2 n K 1 2 n w ¶ w H k 1 2 n ¶ k 1 2 n w ¶ , which takes the form SNR = α H ββ H α α H α , where α = k 1 2 n w and β = K 1 2 n s . With CauchySchwartz inequality, it yields SNR = α H ββ H α α H α ≤  α  2  β  2  α  2 , with equality holds when α = c · β , with arbitrary constant c . Taking c = 1 , it follows k 1 2 n w = K 1 2 n s . We therefore have w MR = K 1 n s . 4. (a) The number X of Poisson events with parameter λ occurred in the time span [0 ,t ] has the probability mass function P [ X ( t ) = n ] = e λt ( λt ) n n ! . Let A be the event that the number of Poisson events occurred in the interval [0 ,y ] is greater than k . The probability distribution function is F Y k ( y ) = P £ Y k ≤ y / = P [ A ] = P h X ( y ) ≥ k i = 1 P h X ( y ) < k i = 1 k 1 X n =0 e λy ( λy ) n n ! . (b) The probability density function of Y k is f Y k ( y ) = dF Y k ( y ) /dy = ( λy ) k 1 λe λy ( k 1)!...
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 Winter '10
 GFung

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