In our lecture, we learned about the CLT and we were shown the following application:
Let $X_1, X_2, \dots$ ~ $B(p)$, meaning that $P(X_i = 1) = p \in (0,1)$. Additionally, let $X_1, X_2, \dots$ be independent and identically distributed. Let $S_n = \sum_{i=1}^nX_i$, then (knowing the moment-conditions are fulfilled), the following convergence holds:
$\frac{S_n - np}{\sqrt{np(1-p)}} \longrightarrow N(0,1)$
Since $E[S_n] = np$ and $Var(S_n)=np(1-p)$ with $N(0,1)$ denoting a standard-normal.
Since also the following equation holds:
$\frac{S_n - np}{\sqrt{n}} \longrightarrow N(0,p(1-p))$
I was wondering if (or why?) $\frac{S_n}{\sqrt{n}} \longrightarrow N(p,p(1-p))$ is also a right expression.
Having $E[\frac{S_n}{\sqrt{n}}] = \sqrt{n}p$, I do not understand why the expectation of the asymptotic normal distribution should be simply p (which is what my lecturer told us).
Can anyone please help out?
Thanks.
Since $\frac{S_n-np}{\sqrt{n}} =\frac{S_n}{\sqrt{n}} - \sqrt{n}p$ converges in distribution to $\mathcal N\left(0,p(1-p)\right)$ as $n$ increases,
and $\sqrt{n}p$ increases without bounds as $n$ increases,
you will have $\frac{S_n}{\sqrt{n}}$ diverging to $+\infty$.
What your lecturer may have said is that $\frac{S_n}{n} \to p$ in probability and almost surely. This is the law of large numbers; note there is no square root in the denominator.