Question
r.v. $X \sim B(n, \theta)$.
I want to show that $\displaystyle\frac{\sqrt{n}(\hat{\theta}-\theta)}{\sqrt{\theta(1-\theta)}}$ converges in distribution to the standard normal distribution when $n\to \infty$.
What I know
From the asymptotic normality of the maximum likelihood estimator, the following holds
\begin{align} \sqrt{n}(\hat\theta - \theta) \overset{L}{\to} N(0, 1/\mathbb{I}_\theta) \end{align}
where $\mathbb{I}_\theta$ is the Fisher information.
\begin{align} \mathbb{I}_\theta = \frac{n}{\theta(1-\theta)}. \end{align} \begin{align} \therefore \sqrt{n}(\hat\theta - \theta)\overset{L}{\to} N(0, \frac{\theta(1-\theta)}{n})\\ \therefore \frac{\sqrt{n}(\hat\theta - \theta)}{\sqrt{\frac{\theta(1-\theta)}{n}}} \overset{L}{\to} N(0,1)\ ??? \end{align}