problem with mgf

lpd

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!

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Anonymous1

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

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

Anonymous1

Yes, it goes to 1...

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