Let $X^1,\dots,X^n$ be a sample of (not necessarily iid) random variables and denote $$F^n(x)=\frac{1}{n}\sum_{i=1}^n \mathbf 1_{X^i\leq x}$$ the empirical distribution function. Suppose that we know that $F^n(x)\to F(x)$ in probability for all continuity points $x$ of $F$, where $F$ is a given cdf. Denote $$Q^n(p)=\inf\{x\in\mathbb R: F^n(x)\geq p\}, \quad Q(p)=\inf\{x\in\mathbb R: F(x)\geq p\}.$$ Can we say that $Q^n(p)\to Q(p)$ in probability for all continuity points $p$ of $Q$?
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2$\begingroup$ Is the convergence in probability important, as compared to in distribution? If the latter suffices then I am quite sure that the treatment in van der Vaart & Wellner will give you what you need. They prove Hadamard differentiability of the generalized inverse functional and then the functional delta method provides the rest. However, if it is crucial that the convergence be in probability I would have to take a closer look at the arguments (hence, for now a comment rather than an answer). $\endgroup$– PierreCommented Jun 16, 2014 at 20:58
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Yes, this is true. Note that $Q^n \to Q$ in probability iff every subsequence of $Q^n$ has a subsubsequence that converges almost surely to $Q$. Now take any subsequence of $Q^n$, and let us again denote it by $Q^{n}$ for convenience. Then the corresponding subsequence $F^{n}$ admits a subsubsequence, denoted by $F^{n_k}$, that converges almost surely to $F$. Since convergence in distribution implies convergence of quantile functions, we find that $Q^{n_k}(p) \to Q(p)$ almost surely in every continuity point of $p$.
