Hello, I hope everyone is doing well. I am currently having a hard time seeing where I am supposed to arrive at for this particular question. I will try my best to explain:
I know, conceptually, that this is a binomial distribution, and I need to show that the the random variable in standard form converges in distribution to the standard normal random variable $Z$. I understand that if enough samples are taken, it will slowly converge into a normal distribution. What I do not understand is how I would go about this. What exactly do I "show"? How would I do that? I cannot just say that i.i.d. random sample implies convergence in distribution; I must show it.
I also would like an explanation of how to go about part b). My instructor told me that I have to use the central limit theorem to show it converges to $0$ mean and $p(1-p)$ variance. I found this explicitly for the mean by doing $$E[\sqrt{n}(\overline{X_n}-p)] = \sqrt{n}E[p-p] = 0 $$
He did not accept that as THE answer he wanted.
Any help would be great. I have been searching for examples for hours now. Here are some resources https://www.ma.imperial.ac.uk/~ayoung/m2s1/Convergencedistribution.PDF
https://www.youtube.com/watch?v=dRUm8U-r0IM
I cannot seem to wrap my head around doing this on paper. I know the answer is right in front of me, but I am stumped. I know this may seem like a trivial subject, but I am in a high paced statistics class. I do not have a lot of time to learn things thoroughly.
The central limit theorem says if $X_1,X_2,\dots$ are independent observations from any distribution with mean $\mu$ and variance $\sigma^2$ then $\frac{\bar X_n-\mu}{\sigma/\sqrt n}\rightarrow^d \mathcal N(0, 1)$.
If $X\sim\text{Bernoulli}(p)$ then $EX=p$ and $\sigma^2=p(1-p)$ and so the CLT says that $\sqrt n\frac{\bar X_n-p}{\sqrt{p(1-p)}}\rightarrow ^d\mathcal N(0, 1)$.
For part b, use part a. By Slutkey's lemma, $\frac{\sqrt n(\bar X_n-p)}{\sqrt{p(1-p)}}\cdot \sqrt {p(1-p)}\rightarrow \mathcal N(0, 1)\sqrt {p(1-p)}=\mathcal N(0, p(1-p))$.