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Moment generating function to calculate the mean and variance

  1. Oct 27, 2012 #1
    I attached a pdf. The questions are not really what is stumping me. Its the wording of the question I dont understand. When it says, "Assume that 0 < λ < 1. Note that your answers will be in terms of the constant λ." and "Assume that λ > 0. Note that your answers will be in terms of the constant λ." I dont understand what they want me to really do here when it says that. To show I know how to do the questions before anyone helps me, I will briefly explain the solution to each one.

    1a. Multiply by e^tx. Its a geometric series so just use the formula for the geometric series.
    1b. To find the mean of X take the derivative of the answer to 1a.
    1c. To find the variance of X take the second derivative minus first derivative squared.

    2a. Multiply by e^tx then integrate from 0 to infinity.
    2b. Same idea as 1b.
    2c. Same idea as 1c.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2012 #2

    Ray Vickson

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    I do not understand what your difficulty is. You have a distribution; it contains a parameter λ. You do not know the value of λ, but you are told it is between 0 and 1. The question asks you to determine the mean and variance. Of course there will be a λ in the answers, since there is a λ in the input. What is unclear about that?

    RGV
     
  4. Oct 27, 2012 #3

    HallsofIvy

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    1a) asks you to find the moment generating function of the distribution [itex](1- \lambda)\lambda^x[/itex] and you are told "1a. Multiply by [itex]e^tx[/itex]." Okay, the moment generating function is, by definition, the expected value of [itex]e^{tx}[/itex] and the expected value of any function is the sum of function values for each x times the probabilty of that particular x. Here, that would be
    [tex]\sum_{x=0}^\infty (1- \lambda)\lambda^xe^{tx}= (1- \lambda)\sum_{n=0}^\infty (\lambda e^t)^x[/tex]
    which is, as they say, a geometric series with common multiple [itex]\lambda e^t[/itex]. Use the fact that the sum of a geometric series is given by
    [tex]\sum_{x=0}^\infty r^x= \frac{1}{1- r}[/tex].
     
  5. Oct 27, 2012 #4
    Thanks everyone. I dont know why I was letting it confuse me, but it seems clarified now. One last question though. What purpose did it serve to tell me that in the first question, 0 < λ < 1, and in the second question λ > 0? Couldnt they of just said that λ is just a constant?
     
  6. Oct 27, 2012 #5

    Ray Vickson

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    No. In question 1 you need lambda between 0 and 1 in order to avoid negative probabilities or probabilities > 1. Question 2 is completely different; there you could have ANY positive value for lambda (but, again, a negative value would give nonsense such as negative probability, etc).

    RGV
     
  7. Oct 28, 2012 #6
    Gotcha. Thanks. Last question. The sum of the geometric series in the first question is

    (1 - λ)/(1 - λe^t) right?

    Since the first term is (1 - λ) and the common ratio is λe^t.
     
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