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Math Help - Limiting Poisson distribution

  1. #1
    Super Member Anonymous1's Avatar
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    Limiting Poisson distribution

    Starting at some fixed time, 0, satellites are launched at times of a Poisson process with rate \lambda. After an independent amount of time having distribution F and mean \mu, the satellite stops working. Let X(t) be the number of satellites working at time t.

    (a) Find the distribution of X(t).
    (b) Let t \rightarrow \infty in (a) to show the limiting distribution is Poisson(\lambda\mu).
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  2. #2
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    Quote Originally Posted by Anonymous1 View Post
    Starting at some fixed time, 0, satellites are launched at times of a Poisson process with rate \lambda. After an independent amount of time having distribution F and mean \mu, the satellite stops working. Let X(t) be the number of satellites working at time t.

    (a) Find the distribution of X(t).
    (b) Let t \rightarrow \infty in (a) to show the limiting distribution is Poisson(\lambda\mu).
    Hi,
    the number N of satellites launched before t is Poisson with parameter \lambda t and, given N, the times of the launches are (without order) i.i.d. random variables X_1,\ldots,X_N uniformly distributed on [0,t], ok?
    Let \tau_1,\ldots,\tau_N be their lifespan: i.i.d. with distribution F.
    If you prove that P(X_1+\tau_1>t)=\frac{1}{t}E[\min(\tau_1,t)], then you see that the number of satellites that are still working at time t is obtained by picking each of the launched satellites with this same probability, hence X(t) is Poisson with parameter (\lambda t)(\frac{1}{t}E[\min(\tau_1,t)])=\lambda E[\min(\tau_1,t)]. (It's the usual "picking each point of a rate \lambda Poisson process with probability p yields a rate \lambda p Poisson process"). You can also check this conclusion by computing P(X(t)=n) explicitly: P(X(t)=n)=E[P(X(t)=n|N)]=E[{N\choose n}P(X_1+\tau_1>t)^nP(X_1+\tau_1<t)^{N-n}], etc. I let you fill in the dots... If you need further help, please tell us extendedly what you tried and so on.
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  3. #3
    Super Member Anonymous1's Avatar
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    Not many dots left to fill in . So for part (b)...

    \lambda E[\min(\tau_1,t)] \rightarrow \lambda E[\tau_1] as n \rightarrow \infty.

    Also, E[\tau_1]=\mu.

    Thanks!
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  4. #4
    Super Member Anonymous1's Avatar
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    E[{N\choose n} P(X_1+\tau_1>t)^n P(X_1+\tau_1)^{N-n}] = N \times P(X_1+\tau_1>t) Since we have a Bin(N,P(X_1+\tau_1>t))

    Therefore, X(t) ~ Poi(\lambda\times E[\min(\tau_1, t)]).
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