TITLE: Sample proportion and the Central Limit Theorem
QUESTION [5 upvotes]: Suppose that $ (\Omega,\Sigma,\mathsf{P}) $ is a probability space and that $ (X_{k})_{k \in \mathbb{N}} $ is a sequence of i.i.d. Bernoulli trials on $ (\Omega,\Sigma,\mathsf{P}) $, each with probability of success $ p \in (0,1) $. If we define another sequence $ (\hat{P}_{n})_{n \in \mathbb{N}} $ of random variables on $ (\Omega,\Sigma,\mathsf{P}) $ by
$$
\forall n \in \mathbb{N}: \qquad
\hat{P}_{n} \stackrel{\text{df}}{=} \frac{1}{n} \sum_{k = 1}^{n} X_{k},
$$
then according to the Central Limit Theorem, we have
$$
\forall z \in \mathbb{R}: \qquad
  \lim_{n \to \infty}
  \mathsf{P} \!
  \left( \frac{\hat{P}_{n} - p}{\sqrt{p (1 - p) / n}} \leq z \right)
= \Phi(z),
$$
where $ \Phi $ denotes the standard normal c.d.f. For each $ n \in \mathbb{N} $, we call $ \hat{P}_{n} $ a sample proportion for a sample of size $ n $.
When most statistics textbooks discuss confidence intervals for a sample proportion, they implicitly claim that
$$
\frac{\hat{P}_{n} - p}{\sqrt{\hat{P}_{n} (1 - \hat{P}_{n}) / n}}
\stackrel{\text{d}}{\longrightarrow}
\operatorname{N}(0,1),
$$
which is the same as saying that
$$
\forall z \in \mathbb{R}: \qquad
  \lim_{n \to \infty}
  \mathsf{P} \!
  \left(
  \frac{\hat{P}_{n} - p}{\sqrt{\hat{P}_{n} (1 - \hat{P}_{n}) / n}} \leq z
  \right)
= \Phi(z).
$$
However, I was unable to rigorously establish this claim using the Central Limit Theorem.
Could anyone kindly provide references? Thanks!

REPLY [5 votes]: The most straightforward proof of this result requires knowledge of Slutsky's theorem, which in turn requires the concept of convergence in probability. Write
$$\frac{\hat{P}_{n} - p}{\sqrt{\hat{P}_{n} (1 - \hat{P}_{n}) / n}}=
\frac{\hat{P}_{n} - p}{\sqrt{p(1-p) / n}}
\cdot
\sqrt{
\frac{p(1-p)}{\hat P_n(1-\hat P_n)}},
$$
a product of two factors. The first factor converges in distribution to the standard normal, by the central limit theorem. The second factor converges almost surely to the constant value $1$, by the law of large numbers. Now apply Slutsky's theorem, since convergence a.s. implies convergence in probability.