Local limit theorem

1. Jun 14, 2008

ehrenfest

1. The problem statement, all variables and given/known data
This theorem is from Shiryaev. Can someone PLEASE explain how they are using the little o notation here. It makes no sense to me how they say that a FIXED real number is little o of something. I thought f(n) = o(g(n)) mean that for ever c>0 there exists an n_o such that when n => n_0, |f(n)|<c|g(n)|.

2. Relevant equations

3. The attempt at a solution

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• shiryaev_pg59.jpg
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2. Jun 14, 2008

Astronuc

Staff Emeritus
3. Jun 14, 2008

ehrenfest

4. Jun 14, 2008

Hurkyl

Staff Emeritus
Where did they say that?

5. Jun 14, 2008

ehrenfest

It says for all $$x \in R^1$$

6. Jun 14, 2008

Hurkyl

Staff Emeritus
That doesn't sound like a FIXED real number to me. And besides, they clarify what they mean immediately afterwards.

Last edited: Jun 14, 2008
7. Jun 14, 2008

ehrenfest

They say for all $$x \in R^1$$ such that $$x = o(npq)^{1/6}$$. To me that means that you can take any real number that you want and see if it is $$o(npq)^{1/6}$$. I don't know how else to interpret it.

8. Jun 14, 2008

Hurkyl

Staff Emeritus
How about exactly how they interpreted it in the attachment you posted?

9. Jun 14, 2008

ehrenfest

That is precisely my question: How did they interpret it?

10. Jun 14, 2008

Hurkyl

Staff Emeritus
as $n \rightarrow \infty$,
$$\sup_{\left\{ x : |x| \leq \psi(n) \right\}} \left| \frac{P_n(np + x \sqrt{npq})}{ \frac{1}{\sqrt{2\pi npq}} e^{-x^2 / 2} } - 1 \right| \rightarrow 0,$$​
where $\psi(n) = o(npq)^{1/6}$.

11. Jun 14, 2008

ehrenfest

OK. I will have to think about that. I am kind of confused about the sup above. Is \psi(n) fixed? If not, I don't understand how the sup is taken. Is it taken over all x AND over all \psi(n) that are $$o(npq)^{1/6}$$?

12. Jun 15, 2008

ehrenfest

anyone?

13. Jul 8, 2008

ehrenfest

I FINALLY figured this out! Hurky! was right that they clarify what they mean-that just wasn't clicking for me.