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Exponential Distribution, Mean, and Lamda confusion

  1. Jun 30, 2017 #1
    1. The problem statement, all variables and given/known data

    Accidents at a busy intersection follow a Poisson distribution with three accidents expected in a week.
    What is the probability that at least 10 days pass between accidents?

    2. Relevant equations
    F(X) = 1- e-λx
    μ = 1/λ

    3. The attempt at a solution
    Let x = amount of time between accidents in days
    My r.v. is continuous so x~Exponential(λ=?)

    E(x) = 3/7 (in days)
    Since E(x) = μ = 1/λ = 3/7
    λ = 7/3

    Thus x~Exponential(λ=7/3)
    P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-70/3

    Answer in the back of the book says:
    λ = 3/7
    P(X≥10) = 1-P(X≤10) = 1- F(10) = 1- e-30/7

    I'm confused why λ = 3/7 and not 7/3 if my expected value is 3/7. Shouldn't lambda, by definition, be its reciprocal?
     
  2. jcsd
  3. Jun 30, 2017 #2

    Charles Link

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    I agree that ## \lambda=3/7 ##. ## \lambda ## is the rate at which things occur. I think your very last line is in error though. As I understand it, the exponential distribution comes from the differential equation ## \frac{dN}{dt}=-\lambda N ##. This would give ## P(X \geq x)=e^{- \lambda x} ## and ## P(X \leq x)=1-e^{- \lambda x} ##.
     
  4. Jun 30, 2017 #3

    Ray Vickson

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    For a Poisson random variable ##N## with mean ##m## the probability mass function is
    $$P(N = k) = \frac{m^k}{k!} e^{-m}, \; k = 0,1,2,\ldots$$
    In your case if you measure arrivals per day your ##m = 3/7##.

    On the other hand, the times between successive arrivals are iid exponentials with mean ##\mu = 1/m##, so ##\mu = 7/3## (days) in your case.

    That makes perfect sense: on average, fewer than one accident per day occurs, so it is reasonable that the mean time between accidents is greater than 1 day.

    I have used different names ##m## and ##\mu##, but you can relate them to some ##\lambda## if you want.
     
  5. Jun 30, 2017 #4

    Charles Link

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    I can add a hint to the OP to do the calculation in this manner. For a Poisson distribution, let's use the letter ## \nu ## for the mean, i.e. ## \nu=Np ##. They give you data for a 7 day period, and ## \nu=3 ##. You need to compute ## \nu ## (which is ## \lambda ## in my post #2 and ## m ## in post #3 above) for a 10 day period . Once you have that, the formula given by @Ray Vickson should be able to lead you to the answer. Additional comment=I believe formula as given by @Ray Vickson is the result for the probability of ## k ## accidents on a single day. (Note: In some ways this method by @Ray Vickson is a better solution than the route I presented in post #2, because this way also allows you to compute the probability that there is 1 collision in the 10 day period,the probability that there are 2 collisions, the probability that there are 3 collisions, the probability that there are 4 collisions in the 10 day period, etc.)
     
    Last edited: Jun 30, 2017
  6. Jun 30, 2017 #5
    Ahhh, I think I found out where I was confused. I was confusing the continuous exponential distribution with a discrete Poisson distribution (where mean = μ = m = lamda). Rather than viewing them as their own independent definitions, I got used to viewing them as synonyms. :confused:

    Thanks.
     
  7. Jun 30, 2017 #6

    Charles Link

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    It actually is advantageous to learn both methods. See my edited post #4. For the calculation at hand, using the @Ray Vickson approach, you would compute the probability that there are 0 collisions in the 10 day period.
     
  8. Jun 30, 2017 #7

    Charles Link

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    @Of Mike and Men Notice the Poisson distribution has ## e^{- \nu} ## built into it. This is certainly necessary because ## \sum\limits_{k=0}^{+\infty} \frac{\nu^k}{k!}=e^{\nu} ##. (The result follows because of the Taylor expansion for ## e^{\nu} ##.) Thereby the Poisson distribution is correctly normalized to 1... The Poisson distribution is a limiting case of the binomial distribution, which is in fact properly normalized, so that the Poisson distribution that is computed from it would of course be properly normalized. ## \\ ## Additional comment: This problem of the intersection with accidents can be treated similar to the problem of the life of a light bulb, where the probability that a given bulb survives without an event for a time ## X \geq x ## is ## P(X \geq x)=e^{- \lambda x} ## (Comes from ## \frac{dN}{dt}=-\lambda N ##) . ## \\ ## Alternatively, it can be solved by using the Poisson distribution which gives the probability of ## k ## events in a given time span as a function of integer ## k ##, and in this case setting ## k=0 ## for the ten day time span. ## \\ ## In setting up the Poisson distribution for the case at hand, you don't know ## N ## (although you know it is a large number ,which will be larger for a longer time span in whatever you are analyzing), and you don't know ## p ## (## p ## is small), but you do know the product ## \nu=Np ## (the letters ## \lambda ## and ## m ## are also used in place of ## \nu ##), which is the mean number of events that occurs with the scenario you are analyzing over the time span that is selected. ## \\ ## (In some cases you will know both ## N ## and ## p ##, but the only requirement is that ## N ## is large and ## p ## is small and that you know the product ## \nu=Np ##).
     
    Last edited: Jul 1, 2017
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