{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

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

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern