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
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 $\epsilon$.
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.
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?
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.
