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Math Help - Poisson Process Question

  1. #1
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    Poisson Process Question

    Assume that you perform and experiment where the outcomes occur according to a Poisson Process with rate 4 per minute, and that the experiment is complete once you observe 250 outcomes. Calculate the mean time and variance time of the experiment.

    for the mean would think that it's just \frac{250}{4}=62.5

    but for the variance I'm not too sure but would think that it's \frac{250}{4^2}=15.625

    are these correct?
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  2. #2
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    Quote Originally Posted by lllll View Post
    Assume that you perform and experiment where the outcomes occur according to a Poisson Process with rate 4 per minute, and that the experiment is complete once you observe 250 outcomes. Calculate the mean time and variance time of the experiment.

    for the mean would think that it's just \frac{250}{4}=62.5

    but for the variance I'm not too sure but would think that it's \frac{250}{4^2}=15.625

    are these correct?
    The interval between events has a exponential distribution with a mean of 1/4 min and a variance of (1/4)^2, so a SD of 1/4 min, also the intervals are independent RV's.

    Now you are looking at the sum of 250 such intervals so the mean is 250(1/4), and the SD is (1/4)*sqrt(250).

    CB
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    If I wanted to calculate the probability that the experiment finishes in less then 56.4 minutes? The would it follow:

    \lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!} = 4 e^{-4(56.4)} \frac{(4(56.4))^{250-1}}{(250-1)!}
    Last edited by lllll; December 10th 2008 at 11:21 PM.
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  4. #4
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    Quote Originally Posted by lllll View Post
    If I wanted to calculate the probability that the experiment finishes in less then 56.4 minutes? The would it follow:

    \lambda e^{-\lambda t} \frac{(\lambda t)^{n-1}}{(n-1)!} = 4 e^{-4(56.4)} \frac{(4(56.4))^{250-1}}{(250-1)!}
    No!

    You need to know the distribution of the sum of the 250 independent exponential random variables each with parameter \lambda = 4 (since the mean is 1/4).

    The following is a well known result (and not too difficult to prove):

    Let X_1, \, X_2, \, .... \, X_n be independent exponential random variables with parameter \lambda. Then Y = X_1 + X_2 + \, .... \, + X_n has a gamma distribution with pdf f(y) = \frac{y^{n-1} e^{-y/\lambda}}{\Gamma(n) \lambda^n} = \frac{y^{n-1} e^{-y/\lambda}}{(n-1)!\,  \lambda^n}.

    In your case \lambda = 4 and n = 250.

    Use the pdf to calculate \Pr(Y < 56.4).

    Note: The Central Limit Theorem is valid here.
    Last edited by mr fantastic; December 11th 2008 at 03:15 AM.
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  5. #5
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    Quote Originally Posted by lllll View Post
    If I wanted to calculate the probability that the experiment finishes in less then 56.4 minutes?
    If you prefer to use Poisson random variables, you can also say that what you're looking for is:

    P(X_1+\cdots+X_{250}\leq 56.4)=P(N_{56.4}\geq 250),

    where the X_i are those of mr fantastic, and N_t is the number of points in the Poisson process before t. This is just a dual way to express the probability.

    For any t, N_t is distributed according to a Poisson distribution of parameter 4t.

    So, the probability is P(N_{56.4}\geq 250)=1-P(N_{56.4}<250)=1-\sum_{k=0}^{249} e^{-\lambda} \frac{\lambda^k}{k!} where \lambda=4\times 56.4.

    However, this is not so easy to compute. Like mr fantastic pointed out, a CLT would provide a simpler yet approximate answer.
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