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Math Help - random sample from a geometric distribution

  1. #1
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    random sample from a geometric distribution

    Hello everyone!

    I've got this problem on multivariable distributions I'm stuck on...

    Let X_1;X_2;X_3denote a random sample of size n = 3 from a distribution with the geometric pdf
    P(X=k)=(\frac{3}{4}) ( \frac{1}{4} )^{k-1} and given the distribution Z = X_1 + X_2 + X_3
    Question find the pdf of Z then find P(Z=15).
    Now I considered using the moment generating function for 3 reasons: (a) X_1, X_2, \text{and} X_3 are a random sample size \rightarrow \ X_1, X_2, \text{and } X_3 are independent (b) they have the same pdf thus mgf, and (c) well, mgf can be used for this.
    So i started out like this:
    M_Z (t) = E(e^{tZ})=E(e^{(X_1+X_2+X_3 )\cdot t})=E(e^{X_1 t} \cdot e^{X_2 t} \cdot e^{X_3 t})= ( E(e^{X_i t}) )^3 = (\frac{p e^t}{1-(1-p) e^t})^3

    Then I got stuck there... I presume I don't know how to "transform" variables in descrete distributions unlike continuous ones.
    So any help is much appreciated!
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  2. #2
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    Quote Originally Posted by rebghb View Post
    Hello everyone!

    I've got this problem on multivariable distributions I'm stuck on...

    Let X_1;X_2;X_3denote a random sample of size n = 3 from a distribution with the geometric pdf
    P(X=k)=(\frac{3}{4}) ( \frac{1}{4} )^{k-1} and given the distribution Z = X_1 + X_2 + X_3
    Question find the pdf of Z then find P(Z=15).
    Now I considered using the moment generating function for 3 reasons: (a) X_1, X_2, \text{and} X_3 are a random sample size \rightarrow \ X_1, X_2, \text{and } X_3 are independent (b) they have the same pdf thus mgf, and (c) well, mgf can be used for this.
    So i started out like this:
    M_Z (t) = E(e^{tZ})=E(e^{(X_1+X_2+X_3 )\cdot t})=E(e^{X_1 t} \cdot e^{X_2 t} \cdot e^{X_3 t})= ( E(e^{X_i t}) )^3 = (\frac{p e^t}{1-(1-p) e^t})^3

    Then I got stuck there... I presume I don't know how to "transform" variables in descrete distributions unlike continuous ones.
    So any help is much appreciated!
    Try comparing your moment generating function with the MGF of a negative binomial distribution.
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  3. #3
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    I tried but there's no exponential in the denominator, I looked up the binomial mgf too... Still, no success
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  4. #4
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    Ok, is it right to say:
    \frac{0.75^3e^{3t}}{(1-0.25e^{3t})^3} =27\cdot (\frac{0.25}{1-0.25e^{3t}})^3 \ e^{3t} =27 C_{k+3+3-1}^{3-1}(0.25)^3(0.75)^k =27 C_{k+5}^{4}(0.25)^3(0.75)^k
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  5. #5
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    Quote Originally Posted by rebghb View Post
    I tried but there's no exponential in the denominator, I looked up the binomial mgf too... Still, no success
    Unfortunately, there is no agreement on how to define the geometrical and negative binomial distributions. You have to use a consistent set of definitions if you want your MGFs to match.

    Try defining the geometrical distribution as
    f(x) = (1-p)^x p for x=0,1,2, \dots.
    That should get rid of the exponential in the denominator of the MGF. Then I think you will have a match with your definition of the negative binomial.

    Another possible source of confusion is whether p is the probability of success or the probability of failure. Again, you need a consistent set of definitions in order for everything to work out.
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