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It is known that $$\frac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$$ converges to $N(0,1)$ in distribution. However, I'm not able to prove that $$\frac{\hat{p}-p}{\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}}$$ converges as well to $N(0,1)$ in distribution.

My attempt: we know that $\sqrt{n}(\hat{p}-p)\to N(0,p(1-p))$. I tried to find a function $g$ to apply the Delta Method, that is $$\sqrt{n}\big(g(\hat{p})-g(p)\big)\to N\big(0,g'(p)^2p(1-p)\big).$$ However, it is not clear to me if it is possible to find such function $g$. On the other hand, the fact that $\text{E}\hat{p}=p$ seems to helpful, but I don't know how to apply it.

Any help would be appreciated!

Little Jonny
  • 825

1 Answers1

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I came up with the answer just after writing the question! Since $\hat{p}\to p$ in distribution, it follows that $$\sqrt{\hat{p}(1-\hat{p})}\to\sqrt{p(1-p)}$$ in distribution. Finally, by Slutsky's Theorem we have that $$\frac{\sqrt{p(1-p)}}{\sqrt{\hat{p}(1-\hat{p})}}\frac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\to \frac{\sqrt{p(1-p)}}{\sqrt{p(1-p)}}N(0,1)=N(0,1).$$

Little Jonny
  • 825