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Statistics: Poisson Distribution

  1. Jan 25, 2013 #1
    1. A Poisson random variable is such that it assumes the values 0 and 1 with equal probability. Find the value of the Poisson parameter, ρ ,for this variable.



    2. Poisson equation: f(x) = e-λs(λs)/x!



    3. I assumed the probability would be 0.5 because it can be either 0 or 1.
    0.5 = e-λs(λs)/x! But now I'm wondering where in the equation would ρ be?
     
  2. jcsd
  3. Jan 26, 2013 #2

    Ray Vickson

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    No, no, no. A Poisson random variables has nonzero probabilities at 0, 1, 2, 3, ..., and all these probabilities together sum to 1. Therefore, it is impossible to have probability 1/2 at both 0 and 1; all you have is that P(0) = P(1), and these must both be < 1/2.

    Anyway, your formula for the Poisson f(x) is incorrect.
     
  4. Jan 26, 2013 #3
    Ah why do they both have to be < 1/2?

    Oh I forgot the x! f(x) = (e-λs(λs)x)/(x!)
     
  5. Jan 26, 2013 #4

    haruspex

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    Could they add up to 1? What would that mean for the prob of 2?
     
  6. Jan 26, 2013 #5

    Ray Vickson

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    You were worried about "where ρ would be". Well, what do you think ρ stands for? How is it related to λ and s? BTW: λ and s have nothing to do with the problem!
     
  7. Jan 26, 2013 #6

    Curious3141

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    What's ##s## and what's ##\rho##? Never heard of these symbols used in the context of the Poisson distribution.

    The probability ##P(k)## of observing a value ##k## in a Poisson process with parameter ##\lambda## is given by:

    $$P(k) = \frac{\lambda^k e^{-\lambda}}{k!}$$

    (##k## is the usual symbol, and it's equivalent to the way ##x## has been used in this thread by the thread starter).

    It's as simple as determining ##P(0)## and ##P(1)##, both very simple expressions, then setting them equal to each other and solving for ##\lambda##. Even the solution of this equation is trivial, giving a very simple value for the parameter. Yes, you will find that the (equal) probabilities are less than half, but you don't even need to recognise that before you solve the equation.
     
  8. Jan 26, 2013 #7
    Hmm my book has the formula that way. I think ρ =λs.
    It says that:
    λ = the average number of occurrences of the event per unit.
    s = the length or size of the observation period.

    So I got,
    P(0) = P(1)
    e-λsλs = e-λs
    Substitute λs to ρ
    ρ = 1?
     
  9. Jan 26, 2013 #8

    haruspex

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  10. Jan 27, 2013 #9
    Thank you everyone!

    Could you also help me on this problem?

    An IT department uses the following probability distribution for the number X of computer system crashes occurring during a week:

    Number of crashes, x Probability, p(x)
    0 0.60
    1 0.30
    2 0.07
    3 0.03

    a. What is the probability that there would be at least 2 crashes in a given week?
    b. Find the expected number of crashes in a week, E(X).
    c. Find the variance of X, V(X).

    This is what I got but I'm not sure if it's correct.

    a. 1 - P[X=0] - P[X=1] = 0.1
    b. Using the E(X) formula, I got 0.53.
    c. V(x) formula = 0.5691
     
  11. Jan 27, 2013 #10

    Curious3141

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    All look right. For the first, you could also have added P(X=2) and P(X=3).
     
  12. Jan 27, 2013 #11
    Thank You!!!
     
  13. Jan 27, 2013 #12

    Ray Vickson

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    In principle, a Poisson random variable need not have anything to do with counting "occurrences" over time; that is, you can have a ρ without having a λ and an s. A Poisson random variable with mean ρ has probability mass function [tex] \Pr(k) = \frac{\rho^k e^{-\rho}}{k!}, \: k = 0, 1, 2, \ldots [/tex]
    period. Whether or not ρ happens to be related to some "arrival" process has no bearing on the Poisson distribution itself. I would have hoped your textbook made that clear.
     
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