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I am having difficulty with a Gamma Distribution problem

  1. Apr 28, 2015 #1
    1. The problem statement, all variables and given/known data
    Automobile accidents occur in the United States over a 72 hour holiday period like events in a Poisson process with parameter lambda=10/hr. V is the time until the 10th accident
    a) what is expected value of V or E[V] and standard deviation?
    b) What is the probability that V is larger than 1 hour?

    2. Relevant equations
    variance = n/(Lambda^2)
    fX(x)={λαxα−1e−λxΓ(α) x>0 otherwise

    ∫∞0λαxα−1e−λxΓ(α)dx=λαΓ(α)∫∞0xα−1e−λxdx

    I'm going to post to Wolfram, as I think I'm doing a terrible job of explaining the above...I hope that is okay http://mathworld.wolfram.com/GammaDistribution.html


    3. The attempt at a solution
    a)so since variance = n/(Lambda^2) or 10/100, we can take square root to get both
    b)...no idea
     
  2. jcsd
  3. Apr 28, 2015 #2
    sorry should clarify, for a, I took the square root and it matched up with the answer in the back of the book.
     
  4. Apr 29, 2015 #3

    Ray Vickson

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    So, do you now know how to do (b)?

    BTW: a random variable such as V--- which is a sum of a fixed, integer number of independent, identically-distributed exponential random variables---occurs so often it is given a name all on its own: it is an Erlang distribution. In this case, V has a 10-Erlang distribution (10 being the number of arrivals you are waiting for).

    Of course, Erlang is a special case of Gamma, but has many nice properties that a general Gamma may not have. In particular, there is a close relationship between things like P(V <= t) and Poisson random variables on the interval [0,t]. Google "Erlang distribution".
     
    Last edited: Apr 29, 2015
  5. May 2, 2015 #4
    Not really, eventually guy i was studying with figured out...or rather just tried...I guess the expression is brute forcing it. It's basically Poisson, but several of them summed, in this case, sum of i=0 to 10 to the i power 10 being equal to (lambda*x) lambda is 10 as it is mean (10/hr) and x is one as while normally right oriented, is greater than one.

    i might be kind of off on this....

    but we then multiply the entire thing by exp(LAMBDA*x) and it gives approx the correct answer within about 1000th of a decimal point. or .459....

    But I think I'm supposed to have some simpler way to sort it out.
     
  6. May 2, 2015 #5
    please forgive the lateness of this response.
     
  7. May 2, 2015 #6

    Ray Vickson

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    I presume that by "brute force" you mean you applied integration by parts 10 times and arrived at ##P(V > 10) = \sum_{i=0}^{10} p_i (10)##, where ##p_i(a) = e^{-a} a^i/i!## and ##a = \lambda t = 10 \times 1 = 10##. Is that what you did? Well: that is not correct; the summation should be from ##i = 0## to ##i = 9## only, and the numerical answer should be 0.4579. If the summation goes up to i = 10 the numerical value is .5830. Check your work again.

    Anyway, a much easier way to derive the formula is to note that ##V > 1## means that the 10th arrival occurs later than 1 hour, so equivalently, the number of arrivals within 1 hour is 9 or fewer; that is: ##P(V > 1) = \sum_{n=0}^9 p_n(10)##. Of course, you still need to do some work to compute the sum, but you can by-pass all the integrations by parts.

    In general, if ##T_n## is the time of the nth arrival in a rate-##\lambda## Poisson process, we have
    [tex] P(T_n > t) = \sum_{k = 0}^{n-1} p_k(\lambda t) [/tex]
    Here, ##p_k(a) = e^{-a} a^k/k!##.
     
  8. May 2, 2015 #7

    Okay, yeah, i got the right answer but did the wrong work!

    It looks like I miscounted and just didn't do it until ten.....or rather I counted up ten and included zero so i had a count of ten, but set the bounds incorrectly. I got 10086.57319 *exp(-10) or .457929714.

    Okay, I'll try it again the other way you showed.
     
  9. May 3, 2015 #8

    Ray Vickson

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    There is nothing to try; you will need to do exactly the same computation. The only difference is in the method of developing/justifying the formula; the formula itself is the same.
     
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