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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

$$ \sqrt{n}\left((\hat{\xi}_p-\xi_p)-\frac{p-F_n(\xi_p)}{f(\xi_p)}\right)=o_p(1)? $$

Hermi
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  • Do you need to use this approach to this proof? It may be easier express the sample quantile as an M-estimator to establish the limiting distribution of the p-th sample quantile. – Eli Mar 03 '22 at 14:22
  • @Eli Can you please give more detiails? Thanks! – Hermi Mar 06 '22 at 00:35

1 Answers1

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Stefanski and Boos use M-estimator theory in Example 8 to show asymptotic normality of the sample quantile function.

Eli
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  • That seems a lot weaker than what is requested. The asymptotic normality is established in our thread at https://stats.stackexchange.com/questions/45124, btw, along with conditions for it to hold (this result is not always true, especially not for non-continuous distributions). – whuber Mar 07 '22 at 03:37