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# HW11 - University of Illinois Fall 2009 ECE 313 Problem Set...

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University of Illinois Fall 2009 ECE 313: Problem Set 11: Solutions Poisson Processes; Function of a Random Variable; Decision-Making 1. [Conditional Probabilities in a Poisson Process] From the definition of conditional probability, P { N (0 , τ ] = k | N (0 , t ] = M } = P { N (0 , τ ] = k N (0 , t ] = M } P { N (0 , t ] = M } . (1) Now, { N (0 , τ ] = k } ∩ { N (0 , t ] = M } = { N (0 , τ ] = k } ∩ { N ( τ, t ] = M - k } . and observe that the intervals (0 , τ ] and ( τ, t ] are disjoint time intervals. So by the basic property of Poisson processes, we have that N (0 , τ ] and N ( τ, t ] are independent random variables. This means that P { N (0 , τ ] = k N (0 , t ] = M } = P { N (0 , τ ] = k } · P { N ( τ, t ] = M - k } . Since the number of packets arriving in any interval is Poisson (with mean proportional to the time interval length) we have P { N (0 , t ] = M } = e - λt ( λt ) M M ! P { N (0 , τ ] = k } = e - λτ ( λτ ) k k ! P { N ( τ, t ] = M - k } = e - λ ( t - τ ) ( λ ( t - τ )) M - k M - k ! . Substituting these values in Equation (1) and after some straightforward simplification, we get P { N (0 , τ ] = k | N (0 , t ] = M } = M k τ t k t - τ t M - k . In other words, the conditional probability being asked is the same as that of a binomial random variable with parameters ( M, τ t ) taking on the value k . The intuition is that any arrival in the time interval (0 , t ] occurs uniformly in that interval. 2. [Random chords] (a) X is uniformly distributed on [0 , 2 π ). From the diagram below, it should be obvious that the

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