Find the limiting distribution of $\displaystyle \frac{\sqrt{n}(\arcsin(\overline{X_n}) – \arcsin(p))}{\overline{X_n}}$ where $\{X_i\}$ is an iid sequence of Bernoulli$(p), 0 < p < 1$ random variables.
If this were $\sqrt{n}(\arcsin(\overline{X_n}) – \arcsin(p))$ we could use $\Delta$-Method to get the limiting distribution as $N(0, (p(1-p))/(1-p^2))$ since by the CLT $\sqrt{n}(\overline{X_n} – p) $ converges in distribution to $N(0, p(1-p)$ thus by $\Delta$-method states we will have a limiting distribution of $N(0,(g'(p))^2p(1-p))$ where $g(p) = \arcsin(p)$.
I don't know how to apply $\Delta$-method with this quotient by the sample mean though.
Best Answer
By applying Slutsky's theorem as pointed out by openspace above,
Let $Z_n = \sqrt{n}(\arcsin(\overline{X_n}) - \arcsin(p))$ and $\overline{X_n} \to p$
And, $Z_n$ has limiting distribution of $N(0,p(1-p)/(1-p^2)$. So then $\frac{Z_n}{\overline{X_n}} \to \frac{Z}{p}$ where $Z\sim N(0,p(1-p)/(1-p^2))$
Thus, we have that $\frac{Z_n}{\overline{X_n}}$ has limiting distribution $N(0, p^3(1-p)/(1-p^2))=N(0,p^3/(1+p))$