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Moment generating functions

  1. Jun 16, 2011 #1
    Hi, when asked to show that the addition of 2 independent RV's with the same distribution is once again of the same distribution, eg showing in gaussian that if X has mean m and variance v and Y has mean n and variance w then if i want to show that X + Y has gaussian distribution with mean m+n and variance w+v I use the product of their m.g.fs which is fine , when i get the result of this product, what is a good way to basically say that this shows the stability under additivity and that the parameters are m+n and w+v ?

  2. jcsd
  3. Jun 16, 2011 #2


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    I don't fully understand your question. However, if X and Y are independent (not necessarily normal), as long as the means and variances exist, they add as you described. You just need to wotk with the first and second moments, where independence is used to give E(XY) = E(X)E(Y) in the variance calculation.

    The Gaussian assumption is used only to get the fact that the sum is also Gaussian.
  4. Jun 16, 2011 #3
    What I mean is, im asked to show that the sum of 2 independent random variables each with the same distribution has that distribution again, i do this using the product of the mgfs, but how do i phrase / conclude throughly in words how this proves that the sum of the variables has the same distribution as the original ones?
  5. Jun 16, 2011 #4

    Stephen Tashi

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    You don't mean that the sum of 2 random variables has "has that distribution again". You mean that the sum is in the same "family" of distributions as the random variables that were added. (Whether it is or not, will depend on how you define the "family". There is no requirement that a "family" of distributions be defined by parameters, although this is the usual way of doing it.)

    I don't think that, in general, that you can prove such a result using an argument based on the product of moment generating functions. Two random variables with different probability distributions can have the same moment generating function. You would need to show that for the "family" of distributions in question, the product of the moment generating functions could only be the moment generating function of a distribution in that family.

    Try using characteristic functions instead of moment generating functions.
  6. Jun 16, 2011 #5
    Really? I saw once that the mgf unique determines the distribution, provided that all the moments exist. Maybe I recall it wrong, I'll search for the reference...
  7. Jun 16, 2011 #6
    The statement in "Probability and measure" in Billingsley is

    If P is a probability distribution on the real line having finite moments [itex]\alpha_k[/itex]of all orders and if the power series
    has a positive radius of convergence, then P is the unique probability distribution with [itex]\alpha_k[/itex] as it's moments.

    In particular, the mgf determines the distribution uniquely. The only problem is that not every random variable has an mgf, that's why one uses the characteristic function.
  8. Jun 16, 2011 #7

    Stephen Tashi

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    You are correct. I was confusing the question of the uniqueness of the moment generating function (when it exists) with the fact that two distributions can have the same moments and not be the same distribution ( in a case where the moment generating function does not converge).
  9. Jun 17, 2011 #8


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    This is one reason why characteristic function is preferred to moment generating function. It always exists.
  10. Jun 18, 2011 #9
    I'll provide you with a framework using the gamma distribution.

    [tex]X ~ is ~ Gamma(\alpha_1, \beta); Y ~ is~ Gamma(\alpha_2, \beta) [/tex]

    E[e^{t(X+Y)}] = E[e^{tX}]*E[e^{tY}] = (\frac{\beta}{\beta - t})^{\alpha_1} (\frac{\beta}{\beta - t})^{\alpha_2} = (\frac{\beta}{\beta - t})^{\alpha_1 + \alpha_2}


    This matches the MGF of a gamma random variable with new alpha:
    [tex]\alpha ' = \alpha_1 + \alpha_2[/tex]

    Therefore X + Y ~ gamma(alpha1 + alpha2, beta). Basically just show that the resulting MGF is the same MGF as before with new parameters.
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