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Homework Help: Independent poisson random variables

  1. Nov 2, 2009 #1
    1. The problem statement, all variables and given/known data
    There are two urns, A and B. Let v be a random number of balls. Each of these balls is put to urn A with probably p and to urn B with probability q = 1 - p. Let v_a and v_b denote the numbers of balls in A and B, respectively. Show that random variables v_a and v_b are independent if and only if v is a Poisson random variable.

    2. Relevant equations
    The probability mass function of a Poisson random variable is


    3. The attempt at a solution
    We showed in an earlier problem that if two variables X, Y are independent Poisson random variables then X + Y is also a Poisson random variable. Therefore, if v_A and v_B are Poisson random variables, then v must be as well. Noting that v_A + v_B = v, I think that proving that v must be a Poisson random variable might be sufficient, but I also don't know the level of rigor wanted here. It would be easy to show that v CAN be a Poisson random variable, but I don't see how to go about showing that it MUST be. I thought maybe I could show that the basis of "Poisson random variable space" is independent from the basis of "non-Poisson random variable space", but I don't know if those are even sensible ideas.

    Random thing I noted that may or may not be relevant:
    E(v) = E(v_A) + E(v_B). If v_A and v_B are Poisson random variables then their expectation value is their parameter \lambda.
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  3. Nov 2, 2009 #2


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    I think the following might be a good direction to start with, but I'm pressed for time at the moment so I can't promise that it will bear fruit.

    Suppose v is a Poisson random variable. I define random variables [itex]a_1, a_2, \ldots, a_v[/itex] such that

    [tex]a_j = \left\{\begin{array}{ll}1, & \textrm{if ball j goes into urn A} \\
    0, & \textrm{if ball j goes into urn B}\end{array}\right.[/tex]

    The [itex]a_j[/itex]'s are independent, and

    [tex]v_A = \sum_{j=1}^{v} a_j[/tex]

    Then I can obtain the marginal probability mass function for [itex]v_A[/itex] as follows:

    [tex]P(v_A = k) = \sum_{m=0}^{\infty} P(v_A = k \quad \cap \quad v = m) = \sum_{m=0}^{\infty}P(v_A = k \quad | \quad v = m) P(v = m)[/tex]

    and similarly for [itex]v_B[/itex]. The right-hand side should be straightforward from here.

    So that gets you the marginal distributions, and now you need the joint distribution, and to show that [itex]v_A[/itex] and [itex]v_B[/itex] are statistically independent, i.e.,

    [tex]P(v_A = k \quad \cap \quad v_B = n) = P(v_A = k) P(v_B = n)[/tex]

    If no one else chimes in, and you still need ideas, I'll try to write more later this evening.
  4. Nov 2, 2009 #3
    This helped a lot - but I still don't see how to prove that ONLY Poisson random variables will work.
  5. Nov 3, 2009 #4


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    I'm not sure how to prove that. I may be way off base, but my guess is that you could make some headway using characteristic functions. The argument would start as follows:

    Suppose [itex]v_A[/itex] and [itex]v_B[/itex] are independent. Then

    [tex]\phi_v(t) = \phi_{v_A}(t) \phi_{v_B}(t)[/tex]

    where the notation

    [tex]\phi_x(t)[/tex] denotes the characteristic function of the random variable [itex]x[/itex], defined as

    [tex]\phi_x(t) = E[e^{itx}] = \sum_k e^{itk} P(x = k)[/tex]

    You could then use (deriving if not presumed known) the characteristic functions for binomial and Poissan random variables, which are

    [tex](1 - p + pe^{it})^{n}[/tex] (binomial with n trials and probability p per trial)


    [tex]e^{\lambda(e^{it}-1)}[/tex] (Poisson with parameter [itex]\lambda[/itex])

    This isn't the whole story, because [itex]v_A[/itex] and [itex]v_B[/itex] are only conditionally binomial, given a fixed value of [itex]v[/itex], but it seems like it might be a promising avenue to try. [Of course, if this is for a class and you haven't covered characteristic functions yet, then scratch that idea...]
  6. Nov 3, 2009 #5
    Yeah, it is for class, and we haven't gotten to characteristic functions yet. I ended up just throwing in what I had and mentioning something about the space of non-Poisson random variables hopefully not being so large that it could be dense in the space of Poisson random variables, but I was definitely just talking out of my ***.
  7. Nov 3, 2009 #6


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    Hmm, I guess without characteristic functions, you would make the analogous argument using probability mass functions, except the math might be grungier:

    Suppose [itex]v_A[/itex] and [itex]v_B[/itex] are independent. Then the pmf of [itex]v[/itex] is the convolution of the pmfs of [itex]v_A[/itex] and [itex]v_B[/itex]:

    [tex]P(v = k) = \sum_n P(v_A = n) P(v_B = k - n)[/tex]

    and you would write

    [tex]P(v_A = n) = \sum_j P(v_A = n \quad | v = j) P(v = j)[/tex]

    and similarly for [itex]v_B[/itex]

    Then use the fact that [itex]v_A[/itex] and [itex]v_B[/itex] are conditionally binomial, given [itex]v = j[/itex]. Hopefully you could then do a bunch of simplification and solve for [itex]P(v = k)[/itex] and show it to be Poisson.

    If you get a chance, I'll be curious to know if that's the solution your instructor gives, or if there is a simpler way.
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