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 ?

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!

2. Originally Posted by lpd
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 $\displaystyle e^{\frac{(\frac{3}{2}\sqrt{n}t)}{n}} \propto e^{\frac{t}{n}} \text{ as } n\to \infty$

$\displaystyle \implies Z_{n\to\infty} \sim ...$

3. doesn't the limit go to zero? or is it 1?

4. Yes, it goes to 1...