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Thread: Poisson Process Exercise

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    Poisson Process Exercise

    Hi everyone i'm new in this forum and i wanna ask if there someone that can help me with this kind of exercises in which i have an arrival process according to a Poisson Process and life times according to an exponentially distributed random variable.

    Customers arrive at a certain facility according to a Poisson process of rate /lambda. Suppose that it is known that five customers arrived in the first
    hour. Each customer spends a time in the store that is a random variable, exponentially distributed with parameter /alpha and independent of the other customer times, and then departs. What is the probability that the store is empty at the end of this first hour?

  2. #2
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    Re: Poisson Process Exercise

    This is a classical example involving the expected value of a symmetric function of arrivial times of a Poisson process, conditional on the Poisson process. See e.g. Theorem 1.2 in


    Let Y_i be the i-th customer arrival time and
    X_i be the corresponding time spended in the store, i = 1, 2, ..., N(1)

    Now the required probability is

    \Pr\left\{\bigcap_{i=1}^{N(1)} X_i + Y_i \leq 1 \Bigg|N(1) = 5\right\}

    = \Pr\left\{\bigcap_{i=1}^5 X_i + U_i \leq 1 \right\} given by the theorem, in which U_i \stackrel {\text{iid}} {\sim} \text{Uniform}(0, 1)

    = \Pr\{X_1 + U_1 \leq 1\}^5 by iid assumption

    = \left(\int_0^1 \Pr\{X_1  \leq 1 - u\} du \right)^5 by Law of Total Probability

    = \left(\int_0^1 1 - e^{-\alpha(1 - u)} du\right)^5 by the exponential CDF

    = \left(\left. u - \frac {1} {\alpha} e^{-\alpha(1 - u)}\right|_0^1\right)^5

    = \left(1 - \frac {1} {\alpha} - 0 + \frac {1} {\alpha}e^{-\alpha}  \right)^5

    = \left[1 - \frac {1} {\alpha}\left(1 - e^{-\alpha}\right)  \right]^5

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