12:21 PM
@DeltaIV Hi

1:08 PM
 @S.Catterall Hi! Thanks for joining
Are you online?

Yes, I'm online - haven't use chat here before

1:33 PM
:-)
so, first of all thanks for the answer!
It's really nice
should be in a book, or a blog
so, going straight to questions
let's write $P(\frac{Y_n-n\mu}{\sigma\sqrt n}\leq \alpha)=F_{Z_n}(\alpha)$ for simplicity
if that's unreadable, you can cut&paste it here without the dollar signs
we know that $F_{Z_n}(\alpha)\to\Phi(\alpha)$ as $n\to\infty$
now, you say: "the continuity of the limiting cdf implies that we also have $$P\Big(\frac{Y_n-n\mu}{\sigma\sqrt n}\leq \alpha+\frac{\alpha^2 \sigma^2}{4\mu\sigma\sqrt n}\Big)\to\Phi(\alpha)$$ because the additional term on the right hand side of the inequality tends to zero"
I don't understand
shouldn't $F_{Z_n}(\alpha)$ the function that has to be continuous, in order for what you state to be true? And we don't know if $F_{Z_n}(\alpha)$ is continuous: its limit for $n\to\infty$ is, but we don't know about $F_{Z_n}(\alpha)$
The way I understand your statement, if $F_{Z_n}(\alpha)$ were continuous, then the composition of two continuous functions $F_{Z_n}(\alpha)$ and $\alpha+\frac{\alpha^2 \sigma^2}{4\mu\sigma\sqrt n}$, i.e., $F_{Z_n}(\alpha+\frac{\alpha^2 \sigma^2}{4\mu\sigma\sqrt n})$, would be continuous
and then, for $n\to\infty$, its limit would be $\Phi(\alpha)$
but we don't know if $F_{Z_n}(\alpha)$ is continuous
This is my first doubt

2:05 PM
Thanks for your comments/questions. I can't find this material in any book - if you do happen to come across anything then it would be great if you could let me know.
I left out a few steps at this point in the proof. Sorry about that. Also, the continuity requirement is a 'red herring' really. We only need continuity from the right - and this is automatically satisfied by all cdf's.
I'll try to explain in more detail why it is true that $P\Big(\frac{Y_n-n\mu}{\sigma\sqrt n}\leq \alpha+\frac{\alpha^2 \sigma^2}{4\mu\sigma\sqrt n}\Big)\to\Phi(\alpha)$. Let's define $Z_n=\frac{Y_n-n\mu}{\sigma\sqrt n}$ and let $Z$ be standard normal.
Let $\epsilon_n=\frac{\alpha^2 \sigma^2}{4\mu\sigma\sqrt n}$ and let $\epsilon>0$
Then $P(Z_n\leq\alpha)\leq P(Z_n\leq\alpha+\epsilon_n)\leq P(Z_n\leq\alpha+\epsilon)$ is true for all $n$ beyond the point where $\epsilon_n$ becomes less than $\epsilon$.

2:22 PM
ok....

Take lower limits in this equation to get $\Phi(\alpha)\leq \liminf_{n\to\infty} P(Z_n\leq\alpha+\epsilon_n)\leq \Phi(\alpha+\epsilon)$. Similarly, take upper limits. What we find is that the upper and lower limits of $P(Z_n\leq\alpha+\epsilon_n)$ are 'squeezed' between $\Phi(\alpha)$ and $\Phi(\alpha+\epsilon)$.
This is true for any $\epsilon$ so (by continuity of $\Phi$) we can conclude that the upper and lower limits are equal (to $\Phi(\alpha)$), which implies that $P(Z_n\leq\alpha+\epsilon_n)$ converges.

How did you substitute $P(Z_n\leq\alpha+\epsilon)$ with $\Phi(\alpha+\epsilon)$? Because in the limit as $n\to\infty$, $P(Z_n\leq\alpha+\epsilon)\to\Phi(\alpha+\epsilon)$, right?

Yes, I'm basically taking the limit as $n$ goes to infinity in the equation. However, I don't yet know if the term in the middle even converges as $n$ goes to infinity, which is why I'm taking lower limits (as these exist for any sequence)
Sorry, I meant 'inequality' rather than 'equation' in my previous comment.

which is "the term in the middle"? $P(Z_n\leq\alpha+\epsilon_n)$ right?

Yes that's right

2:38 PM
ok....and we know that $\Phi(\alpha)\leq \liminf_{n\to\infty} P(Z_n\leq\alpha+\epsilon_n)$ because $P(Z_n\leq\alpha)\leq P(Z_n\leq\alpha+\epsilon_n)$ and I took liminf of left and right side of the inequality, using the fact that when lim exists finite, it's equal to liminf and limsup. Right?

Yes, that all looks right to me

man, your brain has a huge compression ratio! You're either a genius or a calculus prof, or both :-)
ps when I said "this stuff should be in a book/blog", I meant you should put it in a book or blog. I agree that it's not covered in any elementary probability book, but it should because it's great!

Thanks. I chose not to include these details in my answer - not because it was 100% obvious to me, but so that the technical details did not obscure the main thrust of the argument in my answer. I think that the statement we're trying to prove here seems intuitively plausible, regardless of the associated technicalities.
Maybe I should start a blog - I don't currently have one. A book would be too much effort. Anyway, thanks for your question, which in my view was very reasonable.

3:19 PM
hey

1 hour later…
4:28 PM
@S.Catterall you're welcome
bye!