# 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

#### Attached Files:

• ###### shiryaev_pg59.jpg
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2. Jun 14, 2008

### Staff: Mentor

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.