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Thread: Poisson Proofs

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

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

    Prove the following:

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

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

    Prove the following:

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

    (b) for $\displaystyle s,t \geq 0, \mathbb{E}[N_t \ \ N_{t+s}] = \lambda t(1+\lambda(t+s))$
    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: $\displaystyle N_{t+s}=N_t+(N_{t+s}-N_t)$.
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  3. #3
    Junior Member BrooketheChook's Avatar
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    For (a) I get

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

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

    $\displaystyle {n \choose j} \left(\frac{u}{t}\right)^2 \left(1 - \frac{u}{t}\right)^{n-j}$
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  4. #4
    MHF Contributor

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    Quote Originally Posted by BrooketheChook View Post
    For (a) I get

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

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

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