Local Limit Theorem: Explaining Little o Notation

[ehrenfest]

Homework Statement

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)|.

Homework Equations

The Attempt at a Solution

 

[Astronuc]

Does this seem to apply - http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation

See this thread - https://www.physicsforums.com/showthread.php?t=85146

 

[ehrenfest]

Does this seem to apply - http://en.wikipedia.org/wiki/Big_O_notation#Little-o_notation

See this thread - https://www.physicsforums.com/showthread.php?t=85146

Can you be more specific as to how either of those relate to what I asked?

 

[Hurkyl]

It makes no sense to me how they say that a FIXED real number is little o of something.

Where did they say that?

 

[ehrenfest]

Where did they say that?

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

 

[Hurkyl]

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

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

 

[ehrenfest]

That doesn't sound like a FIXED real number to me.

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.

 

[Hurkyl]

I don't know how else to interpret it.

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

 

[ehrenfest]

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

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

 

[Hurkyl]

How did they interpret it?

as $n \rightarrow \infty$,

$$\sup_{\left\{ x : |x| \leq \psi(n) \right\}}<br /> \left|<br /> \frac{P_n(np + x \sqrt{npq})}{<br /> \frac{1}{\sqrt{2\pi npq}} e^{-x^2 / 2}<br /> } - 1<br /> \right| \rightarrow 0,$$​

where $\psi(n) = o(npq)^{1/6}$.

 

[ehrenfest]

as $n \rightarrow \infty$,

$$\sup_{\left\{ x : |x| \leq \psi(n) \right\}}<br /> \left|<br /> \frac{P_n(np + x \sqrt{npq})}{<br /> \frac{1}{\sqrt{2\pi npq}} e^{-x^2 / 2}<br /> } - 1<br /> \right| \rightarrow 0,$$​

where $\psi(n) = o(npq)^{1/6}$.

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 $$o(npq)^{1/6}$$?

 

[ehrenfest]

anyone?

 

[ehrenfest]

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