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Homework Help: Nasty Integrals

  1. Oct 27, 2007 #1
    I'm in a mathematical statistics class and it is spanking me. Please help.

    I have two questions that will really help me understand things if I get a nice explanation.

    1. A random variable Y has the following probability function p(y) = y[tex]^{2}[/tex]/15 for y = 1, 2, 3. Findthe moment generating function for Y.

    What this problem requires is the integration of m(t) = E[e^ty] = [tex]\int[/tex] e[tex]^{ty}[/tex]y[tex]^{2}[/tex]/15dy integrated from 1 to 3.

    I used integration by parts but succeeded in getting something very large and ugly.

    The second question is along the same lines:

    2. Let Y be a random variable with [tex]\mu[/tex][tex]^{'}_{k}[/tex]=[1 + 2[tex]^{k+1}[/tex] + 3[tex]^{k+1}[/tex]]/6

    I need to inegrate m(t) = E[e[tex]^{ty}[/tex]] = [tex]\int[/tex][tex]e^{ty}[/tex][1 + 2[tex]^{k+1}[/tex] + 3[tex]^{k+1}[/tex]]/6 dy from 0 to infinity finding the first four terms and indicating the sum continues.

    Anyone?
     
  2. jcsd
  3. Oct 27, 2007 #2

    HallsofIvy

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    Science Advisor

    Large and ugly? I don't see why you would! Any time you have a power of x times an integrable function, it should be reasonably easy to reduce. Let u= y2 and dv= etydy. Then du= 2y dy and v= (1/t)ety. You now have
    [tex](1/t)e^{ty}y^2\right|_{y=1}^{3} -(2/t)\int_{y= 1}^3 ye^{ty}dy[/tex]
    Use integration by parts again with u= y, dv= etydy.

    Are you sure that's what you have? [1+ 2k+1+ 3k+1]/6 is a constant! You only need to integrate ety.
     
  4. Oct 27, 2007 #3
    I see where I went wrong on the first problem. my dv and my v for the first integration by parts was switched. Thanks for clearing that up.
     
  5. Oct 27, 2007 #4
    On the second problem it isn't an integral. Its a sum. Here is how I should have written it.

    [tex]\sum[/tex]e[tex]^{ty}[/tex][(1+2[tex]^{k+1}[/tex]+3[tex]^{k+1}[/tex])/6]
     
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