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Probability Mass Function/Moment Generating Function

  1. Oct 13, 2011 #1
    1. The problem statement, all variables and given/known data

    The pmf of a random variable X is given by f(x) = π(1 − π)x for x = 0, 1, ..., ∞, and 0 ≤ π ≤ 1.
    a) Show that this function actually is a pmf.
    b) Find E(X).
    c) Find the moment generating function of X, MX(t) = E(etX).

    2. The attempt at a solution
    My solution was done numerically in MATLAB, but I suppose that there is probably an analytical solution as well. My biggest issue is the interpretation of π.

    [PLAIN]http://img401.imageshack.us/img401/2015/proofox.png [Broken]

    For the second part, I also did this numerically, by solving the series:

    [tex]\sum_{x}x \cdot p(x) = \sum_{x}x \cdot f(X=x)[/tex]
    Which evaluates, also through MATLAB, to be: E(X) = 1, but I suspect there is probably an analytical method for this as well?

    The part I am struggling the most with is the last bit. I can get my MGF down to:

    [tex]\sum_{\forall x} e^{tx} \pi (1-\pi)^{x}[/tex]
    But I am not sure how to get rid of the infinite summation. I tried an infinite geometric series, but it only holds true for:

    [tex]|e^{t}(1-\pi)| < 1[/tex]
    Which means that E(X) cannot be found with the MGF.

    Any ideas?
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 13, 2011 #2

    Ray Vickson

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    You should not need Matlab to do part (a); in fact, relying on software instead of thinking means you will _never_ learn the material. You need to show that sum{(1-p)^k, k=0..infinity} is 1/p. Are you really sure you have never, ever, seen this type of material before? It is just elementary algebra.

    Part (b) is just a little bit trickier, but again, involves only standard results that are widely available If q = 1-p, you want to evaluate S = 0 + 1*q + 2*q^2 + 3*q^3 + ... . You can write this as q + q^2 + q^3 + ... + [q^2 + 2*q^3 + 3*q^4 + ... ], and [...] = q*(q + 2*q^2 + 3*q^3 +...) = q*S. The first summation is q*(1 + q + q^2 + ...) = q/(1-q), so we have S = q/(1-q) + q*S. This is an equation for S that you can solve. (There are many other ways of getting the result, but the above is the most "elementary".)

    You can get part (c) similarly: just write out the first few terms of what you are supposed to evaluate, then look for a pattern.

    RGV
     
    Last edited by a moderator: May 5, 2017
  4. Oct 13, 2011 #3
    To be honest, I cannot recall if I've seen the material or not. It seems like I have, but it's been so long that I cannot remember.

    To be honest, this really confuses me. Also, when solving your final equation, I get:

    [tex]S = \frac{q}{1-q} + qS[/tex]
    [tex]S = \frac{q}{(q-1)^2}[/tex]

    Which should return 1 for any value of q such that 0 <= q <= 1, no? For q = 0.5, it evaluates to 2, for q = 0.75, it evaluates to 12, etc.

    Regardless, even if I have learned about series (however long ago), I've never done series manipulation like this, as far as I am aware.
     
  5. Oct 14, 2011 #4

    Ray Vickson

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    Sorry: I should have said S = p*(0 + q + 2*q^2 + 3*q^3+...), which is p times what I wrote before, so should have gotten S = p*q/(1-q)^2 = p*q/p^2 = q/p. Here, p = 1-q. There is no reason to expect S to be 1, because S is NOT the sum of the probabilities (which would be sum p(x)); it is the expected value of X, which is sum x*p(x).

    RGV
     
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