Let $X_i$, $i=1,2...,$ be independent Bernoulli($p$) random variables and let $Y_n=\frac{1}{n}\sum\limits_{i-1}^{n} X_i$. Show that as $n\rightarrow\infty$, $\sqrt{n}(Y_n-p)\rightarrow N(0,p(1-p))$ in distribution.
I have seen this question answered on another post, but It did not explain the overall underlying concepts. What is this question actually requiring us to do (perhaps a trivial question)?
To converge in distribution, we show that $$\lim_{n\rightarrow\infty} F_{X_n}(x)=F_x(x) \ \ \ \forall\ x \ \text{where} \ F_X \ \text{is continuous}$$ So how do we use this result to derive the answer? Can the Central Limit Theorem or the Delta Method be used here?
Well, in general CLT is always a good approach.
It says that $\frac{nS_n-n\mathbb{E}(X_1)}{\sqrt{n\cdot \text{ var }(X_1)}}\longrightarrow N(0,1)$ in distribution as $n \rightarrow \infty$
where $S_n=\frac{1}{n}\sum_{i=1}^nX_i$
In this particular case, we get
$$\frac{nY_n-np}{\sqrt{np(1-p)}}\longrightarrow N(0,1)$$
which is equivalent to
$$\sqrt{n}(Y_n-p)\longrightarrow N(0,p(1-p))$$