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Math Help - Poisson Proofs

  1. #1
    Junior Member BrooketheChook's Avatar
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    Poisson Proofs

    Let (N_t) be a Poisson process with rate \lambda and N_0 = 0.

    Prove the following:

    (a) for 0 \leq u \leq t and \ j = 0,1,2,...n, Pr(N_u = j | Nt_t = n) = {n \choose j} \left(\frac{u}{t}\right)^2 \left(1 - \frac{u}{t}\right)^{n-j}
    i.e. the conditional distribution of N_u given N_t = n is Bin (n,\frac{u}{t})

    (b) for s,t \geq 0, \mathbb{E}[N_t \ \ N_{t+s}] = \lambda t(1+\lambda(t+s))
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  2. #2
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    Quote Originally Posted by BrooketheChook View Post
    Let (N_t) be a Poisson process with rate \lambda and N_0 = 0.

    Prove the following:

    (a) for 0 \leq u \leq t and \ j = 0,1,2,...n, Pr(N_u = j | Nt_t = n) = {n \choose j} \left(\frac{u}{t}\right)^2 \left(1 - \frac{u}{t}\right)^{n-j}
    i.e. the conditional distribution of N_u given N_t = n is Bin (n,\frac{u}{t})

    (b) for s,t \geq 0, \mathbb{E}[N_t \ \ N_{t+s}] = \lambda t(1+\lambda(t+s))
    What did you try for (a)? If you write the definition of conditional probability, this should go easily.

    For (b), a hint (yes it is) is: N_{t+s}=N_t+(N_{t+s}-N_t).
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  3. #3
    Junior Member BrooketheChook's Avatar
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    For (a) I get

    Pr(N_u = j | N_t = n) = \frac{(\lambda u)^j e^{-\lambda u}(\lambda (t-u))^{n-j}e^{-\lambda(t-u)}n!}{j!(n-j)!(\lambda t)^n e^{-\lambda t}}

    Maybe its because im really tired, but I cant see how this leads to

    {n \choose j} \left(\frac{u}{t}\right)^2 \left(1 - \frac{u}{t}\right)^{n-j}
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  4. #4
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    Quote Originally Posted by BrooketheChook View Post
    For (a) I get

    Pr(N_u = j | N_t = n) = \frac{(\lambda u)^j e^{-\lambda u}(\lambda (t-u))^{n-j}e^{-\lambda(t-u)}n!}{j!(n-j)!(\lambda t)^n e^{-\lambda t}}

    Maybe its because im really tired, but I cant see how this leads to

    {n \choose j} \left(\frac{u}{t}\right)^{\color{red}j} \left(1 - \frac{u}{t}\right)^{n-j}
    Many a simplification happens here. The exponentials simplify, and so do the terms with lambda. It remains \frac{u^j (t-u)^{n-j}n!}{t^n j!(n-j)!}=\frac{u^j t^{n-j}(1-u/t)^{n-j}}{t^n} {n\choose j} =\left(\frac{u}{t}\right)^j \left(1-\frac ut\right)^{n-j}{n \choose j}.
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