Probset6_sol_1

# Probset6_sol_1 - Probability and Stochastic Processes A...

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Unformatted text preview: Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Edition 2 Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman, 3.4.1 3.4.3 3.4.6 3.4.9 3.5.1 3.5.2 3.5.5 3.5.7 3.6.1 3.6.2 3.6.3 3.6.4 3.6.6 3.6.7 and 3.6.8 Problem 3.4.1 Solution The reflected power Y has an exponential ( λ = 1 /P ) PDF. From Theorem 3.8, E [ Y ] = P . The probability that an aircraft is correctly identified is P [ Y > P ] = integraldisplay ∞ P 1 P e − y/P dy = e − 1 . (1) Fortunately, real radar systems offer better performance. Problem 3.4.3 Solution From Appendix A, an Erlang random variable X with parameters λ > 0 and n has PDF f X ( x ) = braceleftbigg λ n x n − 1 e − λx / ( n- 1)! x ≥ otherwise (1) In addition, the mean and variance of X are E [ X ] = n λ Var[ X ] = n λ 2 (2) (a) Since λ = 1 / 3 and E [ X ] = n/λ = 15, we must have n = 5. (b) Substituting the parameters n = 5 and λ = 1 / 3 into the given PDF, we obtain f X ( x ) = braceleftbigg (1 / 3) 5 x 4 e − x/ 3 / 24 x ≥ otherwise (3) (c) From above, we know that Var[ X ] = n/λ 2 = 45. Problem 3.4.6 Solution We know that X has a uniform PDF over [ a,b ) and has mean μ X = 7 and variance Var[ X ] = 3. All that is left to do is determine the values of the constants a and b , to complete the model of the uniform PDF. E [ X ] = a + b 2 = 7 Var[ X ] = ( b- a ) 2 12 = 3 (1) Since we assume b > a , this implies a + b = 14 b- a = 6 (2) 1 Solving these two equations, we arrive at a = 4 b = 10 (3) And the resulting PDF of X is, f X ( x ) = braceleftbigg 1 / 6 4 ≤ x ≤ 10 otherwise (4) Problem 3.4.9 Solution Let X denote the holding time of a call. The PDF of X is f X ( x ) = braceleftbigg (1 /τ ) e − x/τ x ≥ otherwise (1) We will use C A ( X ) and C B ( X ) to denote the cost of a call under the two plans. From the problem statement, we note that C A ( X ) = 10 X so that E [ C A ( X )] = 10 E [ X ] = 10 τ . On the other hand C B ( X ) = 99 + 10( X- 20) + (2) where y + = y if y ≥ 0; otherwise y + = 0 for y < 0. Thus, E [ C B ( X )] = E bracketleftbig 99 + 10( X- 20) + bracketrightbig (3) = 99 + 10 E bracketleftbig ( X- 20) + bracketrightbig (4) = 99 + 10 E bracketleftbig ( X- 20) + | X ≤ 20 bracketrightbig P [ X ≤ 20] + 10 E bracketleftbig ( X- 20) + | X > 20 bracketrightbig P [ X > 20] (5) Given X ≤ 20, ( X- 20) + = 0. Thus E [( X- 20) + | X ≤ 20] = 0 and E [ C B ( X )] = 99 + 10 E [( X- 20) | X > 20] P [ X > 20] (6) Finally, we observe that P [ X > 20] = e − 20 /τ and that E [( X- 20) | X > 20] = τ (7) since given X ≥ 20, X- 20 has a PDF identical to X by the memoryless property of the exponential random variable. Thus, E [ C B ( X )] = 99 + 10 τe − 20 /τ (8) Some numeric comparisons show that E [ C B ( X )] ≤ E [ C A ( X )] if τ > 12 . 34 minutes. That is, the flat price for the first 20 minutes is a good deal only if your average phone call is sufficiently long....
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## This note was uploaded on 02/13/2011 for the course 332 226 taught by Professor Staff during the Spring '08 term at Rutgers.

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Probset6_sol_1 - Probability and Stochastic Processes A...

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