1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding distribution by using mgf(moment generating function)

  1. Sep 29, 2005 #1
    i have X_1,X_2,...X_n independant poisson-distributed variables with parameters: alfa_i and i=1,...k(unsure about this. however says so in the excercise)

    i am supposed to find the distribution of
    Y= SUM(from 1 to n) a_i*X_i where a_i>0

    maybe one could use the "poisson paradigm" by thinking of each variable as a trial with p_i as the chance for success. so that

    E[e^tX_i]=1+p_i(e^t - 1)

    and

    E[e^tX] is approximately
    (product from i=1 to n) EXP{p_i(e^t - 1)

    the problem is the a_i part. how do i find the mfg of Y?
     
  2. jcsd
  3. Sep 29, 2005 #2

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    I'll go with MathWorld notation.

    [tex]P_\nu(n)=\nu^n e^{-\nu}/n![/tex]

    MGF of Y is defined as [itex]m(t)=\sum_y e^{ty}f_Y(y)[/itex] where fY is the pdf of Y. It seems to me that you first need to derive fY with brute force then substitute it in the MGF formula.

    In general, fY will not be a Poisson pdf.

    P.S. For large [itex]\nu[/itex] (that's your alfa BTW), [itex]P_\nu[/itex] can be approximated as a normal pdf with mean = standard dev. = [itex]\nu[/itex]. If you use that approximation, then Y itself will be normal.
     
    Last edited: Sep 30, 2005
  4. Sep 30, 2005 #3
    got some help and this is what i have so far.

    m(t; X_i) = exp[alpha_i*(exp(t)-1)].

    The mgf of a_i*X_i is

    m(t; a_i*X_i) = m(t*a_i; X_i) = exp[alpha_i*{exp(a_i*t)-1}].

    The mfg of Y is

    m(t; Y) = PROD[m(t; a_i*X_i)] = exp[SUM{alpha_i*(exp(a_i*t)-1)}].


    the problem is now to say what distribution Y is....
     
  5. Sep 30, 2005 #4

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    Excuse me, why isn't m(t; Y) = PROD[m(t; a_i*X_i)] = exp[SUM{alpha_i*(exp(a_i*t)-1)}] the answer to the problem?
     
  6. Sep 30, 2005 #5
    i don't know. is the mgf also the distribution...?

    the exercise asked us to find the distribution of Y, by finding the mgf of Y.
     
  7. Sep 30, 2005 #6
    [tex]E(e^{t\sum X_i})=\prod_{i}E(e^{t X_i})=e^{(e^t-1)(\lambda_1+..\lambda_n)}[/tex]
    Hence the resultant distribution is poisson with poisson parameter
    [tex]\alpha_r[/tex]
    which is given by
    [tex]\alpha_r=\sum_{i=1}^{n}\alpha_i[/tex]
     
    Last edited: Sep 30, 2005
  8. Sep 30, 2005 #7

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    You left the weights [itex]\alpha_i[/itex] out of the definition of Y.

    P.S. Balakrishnan, you have labeled the Poisson parameter once [itex]\lambda[/itex] and once [itex]\alpha[/itex]. The [itex]\alpha[/itex] labels are prone to confusion as the OP used [itex]\alpha[/itex] for sum weights.
     
    Last edited: Oct 1, 2005
  9. Sep 30, 2005 #8

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    Excuse me, you're right. MGF is definitely not the pdf.
     
  10. Oct 1, 2005 #9
    [tex] $E(e^{tY})=\prod_{i}E(e^{ta_{i}X_{i}})=e^{\sum_{i}\left( e^{a_{i}t}-1\right) \left( \lambda _{i}\right) }$ [/tex]

    this is what i found the the MGF of Y to be. how do i know what the distribution of Y is?
     
  11. Oct 1, 2005 #10

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    As I have posted above, I would use "brute force" (in Arabic, al jabr) to derive Y's pdf. Then see whether or how it also can be obtained from the MGF by comparing the MGF and the pdf formulas.
     
  12. Oct 2, 2005 #11
    ok, but how do i do that then? how do i find the pdf of x_i*a_i ?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?