If $X_1,\dots, X_n$ iid samples from CDF $F(x)$ and assume that $F$ is first order differentiable at $\xi_p$ with $f(\xi_p)>0$. Let $F_n$ be the empirical CDF of $F(x)$. Then $\hat{\xi}_p=F^{-1}_n(p)$ is the sample p-th quantile. Ghosh (1971) obtained a weaker version of Bahadur’s (1966) representation:
$$
\hat{\xi}_p-\xi_p=\frac{p-F_n(\xi_p)}{f(\xi_p)}+o_p(n^{-1/2}).
$$
I try to prove this result as follows.
Note that $F(\xi_p)=p$ and
$$
\{\sqrt{n}(\hat{\xi}_p-\xi_p)\le t\}=\{p\le F_n(\xi_p+\frac{t}{\sqrt{n}})\}=\{
\frac{F(\xi_p+\frac{t}{\sqrt{n}})-F_n(\xi_p+\frac{t}{\sqrt{n}})}{f(\xi_p)}\le \frac{F(\xi_p+\frac{t}{\sqrt{n}})-F(\xi_p)}{f(\xi_p)}
\} \, (*)
$$
where
$$
F(\xi_p+\frac{t}{\sqrt{n}})-F(\xi_p)=f(\xi_p)(\frac{t}{\sqrt{n}})+o(n^{-1/2})
$$
then the RHS on (*) $$\sqrt{n}\times \frac{F(\xi_p+\frac{t}{\sqrt{n}})-F(\xi_p)}{f(\xi_p)}\to t$$
Also, the LHS on (*) is
$$
\sqrt{n}\times\left(\frac{F(\xi_p+\frac{t}{\sqrt{n}})-F_n(\xi_p+\frac{t}{\sqrt{n}})}{f(\xi_p)}-\frac{F(\xi_p)-F_n(\xi_p)}{f(\xi_p)}\right)\to 0
$$
in probability.
There is an asymptotic distribution of $\sqrt{n}\frac{F(\xi_p)-F_n(\xi_p)}{f(\xi_p)}$ to a Normal distribution from CLT.
But how to prove that
$$
\sqrt{n}\left((\hat{\xi}_p-\xi_p)-\frac{p-F_n(\xi_p)}{f(\xi_p)}\right)=o_p(1)?
$$
Best Answer
Stefanski and Boos use M-estimator theory in Example 8 to show asymptotic normality of the sample quantile function.