problem with mgf

lpd

Sep 2009
100
1
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 a moderator:
Nov 2009
517
130
Big Red, NY
I'm having a lot of trouble with this problem...
It is not clear what \(\displaystyle \bar Y_n\) is, but assuming your MGF is correct...


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

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

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

\(\displaystyle \implies Z_{n\to\infty} \sim ... \)
 
Last edited:

lpd

Sep 2009
100
1
doesn't the limit go to zero? or is it 1?