Cupitor

Have a good night. And my name is Naji.

Or we can find some other way to chat. I am not sure if this will be available tomorrow, but we can try this as well.

Yeah sure. I am on [email protected]

Really!

Thank you so much for taking the time!

Sounds good!

Yes in general it can have elemtns and that's why it is called measure zero. I understand. Now I additionally know (based on my reasoning) that 𝜎 = 𝜎' all the time, i.e. for all the omega. Therefore that particular measure zero set we are talking about needs to be empty.

No you can't because the set that 𝜎 = 𝜎' + 1 is of measure zero. The only thing that proves is that such a set, i.e. where 𝜎 = 𝜎' + 1 holds is exactly the empty set.

Or better said \Omega

Holding for a single omega is enough to ensure it holds all the time, because sigma and sigma' are not relevant to omega.

No you cant is in that case that set would be of measure 1! because then it should hold for any omega with the same reasoning but we know the other set is of measure zero!

Does it really matter? Write the error term. I am saying that there is one such omega where the equality holds. It should therefore hold for any omega because sigma and sigma' do not depend on omega any way!

𝜎 and 𝜎' are cosntant independent of the sample space

But 𝜎 and 𝜎' are constant. So if it holds for one omega it needs to hodl all the time.

right?

So we agree that there is a 𝜔 where this holds

what does this mean "𝜎 = 𝜎' a.s."?

Then it follows that 𝜎 = 𝜎'

well 𝜎 = 𝜎' a.s. as you say. Take one 𝜔 that this holds. It is measure 1 right?

It is in my case.

S-S' is not constant in what you're implying

Exactly because they're constant I am only willing to take one such 𝜔. Nothing at this point depends on 𝜔.

It is not but it is actually equal to \log(\sigma)-\log(sigma') which is still fine

Yes, but I am summing those sequences basically to cancel out S and S' because I can!

𝑃{lim𝑋𝑛=𝜎2}=1 and 𝑃{lim𝑌𝑛=𝜎′2}=1. But additionally I know 𝑋𝑛=𝑌𝑛 and that's where I can do everything before the final value of the limit is realized to free up myself from dependence on 𝜔.

lol ok

how are you writing in latex?

No no, hold on

Yes the way I wrote it is meaningless, and quite handwavy, but I think I can formalize it. I know your phd is in math and my bachelors is in math, so I have a lot to catch up on formalism. But either way may argument all happens before the final limit is approached.

It is where I say Writing above limits in 𝜖/𝛿 form (sum lhs of both and rhs of both) we can get

lol, I just copied an pasted from the original comment

Does that make sense to you or should I explain how I get that?

No no, I am talking about how I get from your equation above to 𝑃{𝜔:|𝜎−𝜎′|>𝜖}→0

That's not the part that I am applying things. When I have things in paranthesis and I am calculating limits, I use triangle inequality to get rid of anything dependant on $N$ does it make sense until that part?

(Why LaTeX is not parsed here?? :|)

As you said yourself!

So I understand your example. I just don't think my reasoning suffers from this counterexample. It drops N at some point of reasoning, as there is no $N$ involved.

And in that sense I can get rid of $N$ in the final limiting relationship I have.

Yes but the point is that in my case I know for every $n$ and for any $\omega$ the said equality holds!

Let me process what you wrote.

No problem! Thanks for being so active in open source community and helping people!

I'll let you write until you say you're done. Btw we can also zoom or something. I am the same person who wrote about your CV and the website HTTPS issue.

And I clarified this with the last proof by contradiction I implied.

I just disagree in this case one cannot make such a conclusion

Believe it or not I know the definitions of convergence in probability, in distribution and a.s.

I wrote $\sigma-\sigma'$ not $\omega-\omega'$. If I did then I made a typo.

Hello! I wasn't expecting you to be awake! Thanks for taking the time to address this!

That's my bad and I agree with you, but I don't get what you mean by " If the the 𝑌𝑖(𝜔) "do something normal" that leads to that sample variance converging to $\sigma'^2$, then we have a value of 𝜔 where the sample variances are equal even in the limit, and don't cancel out. ". Either way, I stil think the reasoning I made 6 comments ago stands. The summation I presented there literally implies $$P\{\omega:|\sum(X_n(\omega)-\mu)^2 - \sum(X_n'(\omega)-\mu')^2+\sigma-\sigma'|>\epsilon\}=P\{\omega:|\sigma-\sigma'|>\epsilon\}\to 0 .$$ In fact let's assume that's not the case...