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Question about poisson distributed variables

  1. Nov 14, 2003 #1

    I'm trying to prove if X~Po(m) => 2X~Po(2m)
    But I'm not sure how to prove or disprove it.
    I'm thinking about using the addition formula, but is this the right approach?

    n=m => X_1=X_2 => 2X_1~Po(2m)

    Any help is appreciate.

  2. jcsd
  3. Nov 14, 2003 #2


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    I might be able to help you if you could explain your terminology. Specifically, what is the definition of "~", what is X, what is Po(m)?
  4. Nov 14, 2003 #3


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    Adding two random variables is definitely not the same as multiplying a random variable by 2.

    Not convinced? Suppose your random variable is rolling a 6 sided die. The sum of two of these random variables could be anything from 2 through 12, with an uneven distribution. Multiplying the result of a roll by 2 can only be even numbers from 2 through 12 with an even distribution.
  5. Nov 15, 2003 #4
    Okay, Hurkyl, I can understand your reasoning there.

    So my idea of the proof is no good. Any other ideas how I should be able to prove/disprove this implication?


  6. Nov 15, 2003 #5


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    It would be nice to know what your notation means.

    Anyways, I imagine you want to use the fact:

    P(2X < &alpha;) = P(X < &alpha; / 2)

    to prove that 2X has the right cumulative distribution.
  7. Nov 16, 2003 #6
    Okay, lets see if I can explain my notation...

    X is my random variable with a poisson distribution

    I'm trying to verify if the statement
    X in Po([lamb]) implies that 2X in Po(2[lamb])

    I hope this is clear enough

  8. Nov 16, 2003 #7


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    For starters, X can only assume non-negative integer values. 2X is then restricted to EVEN non-negative integer values and cannot have a Poisson distribution.
  9. Nov 20, 2003 #8
    Ok, I'm not really sure if I follow your reasoning there.
    The beginning is ok, due to the fact that X is a discrete random variable.

    2X will only give us even non-negative values; I'm still with you. But in the next step, I'm lost.
    Why can't 2X be Poisson distributed ([tex]2\lambda[/tex]) when X is Poisson distributed ([tex]\lambda[/tex])?
    ie [tex] X\in Po(\lambda)\Rightarrow 2X\in Po(2\lambda) [/tex]

    Thank you [tex]\LaTeX[/tex]

    /farbror - feels silly that he can't grasp this
    Last edited: Nov 20, 2003
  10. Nov 20, 2003 #9


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    Poisson distribution is very specific. One feature is that random variables have non-zero probabilities for ALL non-negative integer values. 2X will have probability 0 for all odd integers.
  11. Oct 9, 2011 #10
    My suggestion is that you should try the moment generating function technique,ofcos u will need the theorem on limiting mgfs,bt it is the easiet way,or do a transformation
  12. Oct 9, 2011 #11
    The sum of 2 poissons is obviously poisson just like the sum of 2 binomials is binomial,ur reasoning is flawed,if X ε Poi(λ) and Y=X+X,THEN Y ε Poi(2λ),it follows quite simply evn by general reasoning.
  13. Oct 9, 2011 #12
    Are you trying to show that X + X ~ Po(2m) or that 2X ~ Po(2m)?

    And do you understand why these are not the same? (The earlier dice nicely shows you that they aren't so understand that)

    So if Z = X + Y, then P(Z=1) is the probability of the events {X=1} and {Y=0} or {X=0} and {Y=1}.

    Where as if Z = 2X then P(Z=1) is the probability of the event that {X=1/2} which can't occur.
  14. Oct 10, 2011 #13
    Maybe a goodness-of-fit test could help.
  15. Oct 10, 2011 #14
    I'm sorry, I misread the question; please ignore my previous post.
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