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Finding moment generating functions

  1. Mar 20, 2013 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations
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    3. The attempt at a solution
    Well I believe it to be Binomial as the probabilities never change. So I used it to solve for the moment generating function, I do howvere want to know if what I have done is correct and also how I went from sigma((Pe^t)^y(n y)q^(n-y)) as the book did this?
     

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  3. Mar 20, 2013 #2

    haruspex

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    Isn't p the probability of one head in one toss? How is it 0.75?
     
  4. Mar 20, 2013 #3
    Ok so p should be .50 but does my work look ok other than that and the obvious mathematical error due to that on the second part
     
  5. Mar 20, 2013 #4
    After the final sigma part like I asked above how do I arrive at the moment generating function like I did?
     
  6. Mar 20, 2013 #5

    haruspex

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    In the third line of part 2, you've written E[Y] = M'(t) by mistake for M'(0).
    In the next line, you've left something out when applying the product rule to get M''.
     
  7. Mar 20, 2013 #6

    haruspex

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    It's just the binomial expansion (run in reverse).
     
  8. Mar 20, 2013 #7
    Ok a bit lost on what was wrong with m" it looks right to me hmmm where should I look
     
  9. Mar 20, 2013 #8

    haruspex

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    Looks like I need to take a break. You're quite right, and the rest of your work looks good too.
     
  10. Mar 20, 2013 #9
    Ok so should I change p to .5 and q to .5 and I wind up with 1.5
     
  11. Mar 20, 2013 #10
    I changed p from .75 to .5 as you said
     
  12. Mar 20, 2013 #11
    Also for the first part I now have (.50e^t+.50)^2
     
  13. Mar 20, 2013 #12

    haruspex

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    All good.
     
  14. Mar 20, 2013 #13
    Wait why is p what you said should I not also include the probability that you flip a head and head thus making it .75
     
  15. Mar 20, 2013 #14
    When you say all good was it all good before or all good now?
     
  16. Mar 20, 2013 #15

    haruspex

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    When you wrote the first two lines of your solution, what were you defining p to be? What do you get if you set n=2 and write out the sum in the second line?
    You showed E[Y] = np. If n=2 and p=.75, would you really expect an average of 1.5 heads in two tosses?
     
  17. Mar 20, 2013 #16
    I was defining p to be the possible heads observed by the 2 flips and to the final question no I would not so I guess what I have changed it to is correct
     
  18. Mar 20, 2013 #17

    haruspex

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    But p is a probability, not a number of heads. If you wanted to approach it that way you'd need three probabilities: 0, 1 or 2 heads.
     
  19. Mar 20, 2013 #18
    Ah ok got it now thanks so the way I have it now is correct
     
  20. Mar 20, 2013 #19

    haruspex

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    Yes.
     
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