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Moment generating function of a continuous variable

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the moment generating of:

    [tex]f(x)=.15e^{-.15x}[/tex]

    2. Relevant equations

    [tex]M_x(t)= \int_{-\infty}^{\infty}{e^{tx}f(x)dx}[/tex]

    3. The attempt at a solution

    I get down to the point (if i've done my calculus correctly) and gotten:

    [tex]\frac{.15e^{(t-.15)x}}{t-.15} \Bigr| _{0}^{\infty}[/tex]

    and I don't know how to evaulate this. Am I right this far? And if I am, how does this equate to the answer in the back of the book:

    [tex]\frac{.15}{.15-t}[/tex]
     
  2. jcsd
  3. Mar 23, 2010 #2

    vela

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    Remember that x is your variable of integration, not t. Assume t<0.15, otherwise the integral won't converge, and just plug in the limits like you would normally do.
     
  4. Mar 23, 2010 #3
    That's what I'm having trouble with. How do I evaluate it at infinity?
     
  5. Mar 23, 2010 #4

    vela

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    What's the limit of e-kx, k>0, as x goes to infinity?
     
  6. Mar 23, 2010 #5
    -inf?
     
  7. Mar 23, 2010 #6

    vela

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    Instead of guessing, try plotting the function or plugging numbers into your calculator to see if you can see what the limit is.
     
  8. Mar 23, 2010 #7
    I'm not guessing. I plotted it out, and as x goes to -inf, e goes to inf.

    Oh. typo above.
     
  9. Mar 23, 2010 #8

    vela

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    I think you're screwing up the signs. You have a function of the form e-kx where k>0 (this is why you need t<0.15). What does it do as x goes to positive infinity, which is the upper limit of the integral?
     
  10. Mar 23, 2010 #9
    Oh. I see. My head wouldn't release the reading error.

    [tex]\frac{-.15e^{(t-.15)\infty}}{t-.15} - \frac{-.15e^{(t-.15)0}}{t-.15}[/tex]

    [tex]0-\frac{.15}{t-.15} = \frac{-.15}{t-.15}=\frac{.15}{.15-t}[/tex]

    Do I have it yet? I'm way too tired to think clearly.
     
  11. Mar 23, 2010 #10

    vela

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    Yup, you got it!
     
  12. Mar 23, 2010 #11
    Okay. I have to take the second derivative of that to find the Variance of the pdf.

    First derivative = [tex]\frac{.15}{(.15-t)^2}[/tex]

    Second derivative = [tex]\frac{.3}{(.15-t)^3}[/tex]

    I've even confirmed the second derviative with Wolfram Alpha. However, when I plug 0 into it (to find the second moment, or the variance), I get 88.88. But the book says 44.44. Any idea what I'm doing wrong?
     
  13. Mar 23, 2010 #12

    vela

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    I think the book is wrong. I multiplied f(x) by x2 and integrated to calculate E(x2) and got the same answer you did.
     
  14. Mar 23, 2010 #13

    vela

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    Oh, wait. Were you looking for E(x2) or the variance? If it's the latter, you need to subtract off the square of the mean.
     
  15. Mar 23, 2010 #14
    Well, V(X) = E(x2)-[E(X)]2 but it also says that if I take the 2nd derivative of the moment generating function, I get V(X) or [itex]\sigma ^2[/itex]
     
  16. Mar 23, 2010 #15

    vela

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    That's not correct. The moment generating function only allows you to calculate E(xn) by taking the n-th derivative.
     
  17. Mar 23, 2010 #16
    So if E(x2) is the second derivative, I have to subtract off [E(X)]^2, or 44.44, which makes it 44.44 just like the book says.

    I think I understand that now. Thanks.
     
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