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    moment generating function problem




    Hi everyone, I'm stuck on a probability problem and was wondering if anyone could help me get started? I would really appreciate it!

    Let X,Y,Z be random variables, either continuous or discrete. The joint MGF of X,Y,Z is defined by

    Mxyz(t1,t2,t3) = Ee^(t1(X)+t2(Y)+t3(Z))

    Show that:



    Thanks in advance to anyone who can give me a hint as to what I should be looking up to try and do this on my own!

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    Re: moment generating function problem

    I guess just follow the definition is ok.

    \frac {\partial^{k_1 + k_2 + k_3}}
{\partial t_1^{k_1} \partial t_2^{k_2} \partial t_3^{k_3} }
M_{XYZ}(t_1, t_2, t_3)

    = \frac {\partial^{k_1 + k_2 + k_3}}
{\partial t_1^{k_1} \partial t_2^{k_2} \partial t_3^{k_3} }
E[e^{t_1 X + t_2 Y + t_3 Z}]

    = E\left[\frac {\partial^{k_1 + k_2 + k_3}}
{\partial t_1^{k_1} \partial t_2^{k_2} \partial t_3^{k_3} }
e^{t_1 X} e^{t_2 Y} e^{t_3 Z}\right]

    = E\left[ \frac {\partial^{k_1}} {\partial t_1^{k_1}} e^{t_1 X}
\frac {\partial^{k_2}} {\partial t_2^{k_2}} e^{t_2 Y}
\frac {\partial^{k_3}} {\partial t_3^{k_3}} e^{t_3 Z} \right]

    = E\left[ X^{k_1} e^{t_1 X} Y^{k_2} e^{t_2 Y} 
Z^{k_3} e^{t_3 Z} \right]

    Hence
    \left . \frac {\partial^{k_1 + k_2 + k_3}}
{\partial t_1^{k_1} \partial t_2^{k_2} \partial t_3^{k_3} }
M_{XYZ}(t_1, t_2, t_3) \right|_{t_1 = t_2 = t_3 = 0} 
= E\left[ X^{k_1} Y^{k_2} Z^{k_3} \right]

  3. #3
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    Re: moment generating function problem

    Quote Originally Posted by BGM View Post
    = \frac {\partial^{k_1 + k_2 + k_3}}{\partial t_1^{k_1} \partial t_2^{k_2} \partial t_3^{k_3} }E[e^{t_1 X + t_2 Y + t_3 Z}]

    = E\left[\frac {\partial^{k_1 + k_2 + k_3}}{\partial t_1^{k_1} \partial t_2^{k_2} \partial t_3^{k_3} }e^{t_1 X} e^{t_2 Y} e^{t_3 Z}\right]
    It looks like I am on the same problem.

    Question 1: Are you able to swap the Expected Value with the differentiation (quoted) because the transforms for expected value don't include any of the t's? It's not readily intuitive that

    \frac {d} {dt} E\left[h(t,x)\right] = E\left[\frac {d} {dt} h(t,x)\right].

    \frac {d} {dt} \int_{-\infty}^{\infty} \! h(t,x) f_{X}(x)\,dx = \int_{-\infty}^{\infty} \! \frac {d} {dt} \left[h(t,x)\right] f_{X}(x)\,dx.

    Question 2: This regards a previous part of the same problem. We are to show that

    M_{XYZ}(t, t, t) = M_{X+Y+Z}(t).

    I can easily do this if I assume that X, Y, and Z are also independent, but this wasn't given in the problem. I can see that

    M_{XYZ}(t, t, t) = E\left[e^{t(X+Y+Z)}\right]

    This seems like Mx+y+z, but how do I make sense of the actual integrals? It seems like

    M_{XYZ}(t, t, t) = E\left[e^{t(X+Y+Z)}\right] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \! e^{t(X+Y+Z)} f_{XYZ}(x,y,z)\,dx\,dy\,dz \,

    which doesn't intuitively seem to equal

    M_{S}(t) = E\left[e^{tS}\right] = \int_{-\infty}^{\infty} \! e^{tS} f_{S}(s)\,ds, \,\, where \, S = X + Y + Z.

    Am I over-thinking this?
    Last edited by kastchei2112; 11-21-2010 at 05:31 PM. Reason: Trimming quote.

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    Re: moment generating function problem


    http://en.wikipedia.org/wiki/Leibniz_integral_rule

    Yes I am quite lazy the check the technical condition.

    You need the Fubini Theorem/Dominated Convergence Theorem
    to exchange the order of limit.

    But I guess in most of the case, when you required to use the moment
    generating function to generate the moment, you need your mgf to satisfy
    such a technical condition; otherwise, differentiating the mgf may not
    yield the desired result.

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