Results 1 to 4 of 4

Math Help - problem with mgf

  1. #1
    lpd
    lpd is offline
    Member
    Joined
    Sep 2009
    Posts
    100
    I'm having a lot of trouble with this problem.

    Let be a random sample of sizze n from a geometric distribution with pmf ,

    Let .Find, the mgf of . Then Find the limiting mgf lim as n goes to infitinity. What is the limiting distribution of

    I'm a little lost.

    I can find the mgf of which is

    so... if there was a factor of of ... does that mean the mfg for is ?

    or... instead of i'll have

    How about the bit of ... ?

    and then the limit just equal to zero... if term exists....

    distribution just 0? doesn't make sense...

    I'm very confused!
    Last edited by mr fantastic; May 25th 2010 at 08:08 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    Quote Originally Posted by lpd View Post
    I'm having a lot of trouble with this problem...
    It is not clear what \bar Y_n is, but assuming your MGF is correct...


    M_{Y_n} = \left(\frac{.75e^{\frac{t}{n}}}{1-.25e^{\frac{t}{n}}}\right)^n

    \implies M_{Z_n} = e^{-2\sqrt{n}t}\cdot M_{Y_n}(\frac{3}{2}\sqrt{n}t)=e^{-2\sqrt{n}t}\cdot\left(\frac{.75e^{\frac{(\frac{3}{  2}\sqrt{n}t)}{n}}}{1-.25e^{\frac{(\frac{3}{2}\sqrt{n}t)}{n}}}\right)^n

    Since e^{-2\sqrt{n}t} \to 1 and e^{\frac{(\frac{3}{2}\sqrt{n}t)}{n}} \propto e^{\frac{t}{n}} \text{ as } n\to \infty

    \implies Z_{n\to\infty} \sim ...
    Last edited by Anonymous1; May 25th 2010 at 10:54 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    lpd
    lpd is offline
    Member
    Joined
    Sep 2009
    Posts
    100
    doesn't the limit go to zero? or is it 1?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Anonymous1's Avatar
    Joined
    Nov 2009
    From
    Big Red, NY
    Posts
    517
    Thanks
    1
    Yes, it goes to 1...
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum