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Suppose real-valued random variables $\{X_{n}\} $ converges to $X$ in distribution. Then, will the quantile of the distribution of $\{X_n\}$ converge to the quantile of $X$? .

Srivatsan
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webster
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2 Answers2

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Yes. If $X$ is a random variable with distribution function $F$, then for $0<p<1$ define the quantile function as $Q(p)=\inf(x: F(x)\geq p)$. Then $X_n\to X$ in distribution if and only if $Q_n(p)\to Q(p)$ at all continuity points $p$ of $Q$.

Added: It's a nice exercise to prove this result from the definition. On the other hand, it is Proposition 5 (page 250) in A Modern Approach to Probability Theory by Bert Fristedt and Lawrence Gray, and is also proved in Chapter 21 of Asymptotic Statistics by A. W. van der Vaart.

  • But what if $Q$ is not continuous at $p$? – webster Dec 25 '11 at 11:09
  • Suppose we know that $F$ is strictly increasing. Will this additional condition help? – webster Dec 25 '11 at 14:08
  • The discontinuity points of $Q$ correspond to the "flat spots" on the graph of $F$ (Draw a picture!). So if $F$ is strictly increasing, then $Q$ is continuous at all $0<p<1$. –  Dec 25 '11 at 15:26
  • Yes. But what will be the formal proof? – webster Dec 25 '11 at 16:25
  • By the way, if $F$ is strictly increasing on the support of $X$, and the supports of all $X_n$ are contained in the support of $X$, then $Q$ is continuous on $[0,1]$, and the convergence of the quantile functions is uniform. This simple observation has applications in the theory of Toeplitz matrices. http://dx.doi.org/10.1007/s40590-016-0105-y – Egor Maximenko Oct 23 '16 at 03:04
  • I don't think the cited Proposition 5 in Fristedt & Gray has anything to do with quantiles, although it does have the letter Q; it starts, "Let Q and Qn, n=1,2,..., be probability measures..." and says nothing about quantiles. (The point of the proposition is conditions for sequence of rv Xn to converge a.s. to rv X, instead of just convergence in distribution.) – David M Kaplan Jul 04 '22 at 17:13
  • The references are not convincing. – Snoop Dec 14 '22 at 08:26
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Since the provided references seem not appropriate I will provide a proof. First, some notation: If $X$ is a random variable with distribution function $F$, then for $0<p<1$ define the quantile function as $Q(p)=\inf(x: F(x)\geq p)$. Then $X_n\to X$ in distribution if and only if $Q_n(p)\to Q(p)$ at all continuity points $p$ of $Q$.

Proof: $[\Rightarrow]$ Assume $X_n \to X$ in distribution and let $p$ be a continuity point of $Q$. Then, there is a sequence $\epsilon_k$ with $\epsilon_k \to 0$ such that $Q(p) - \epsilon_k$ and $Q(p) + \epsilon_k$ are continuity points of $F$ for every $k$.

Now fix $k > 0$. Since $p$ is a continuity point of $Q$, it is impossible for $F$ to be constant on $[Q(p)-\epsilon_k, Q(p)]$ or $[Q(p), Q(p) + \epsilon_k]$, so we find $$F(Q(p) -\epsilon_{k}) < F(Q(p)) < F(Q(p) + \epsilon_k). $$ Since $p$ is a continuity point of $Q$, we have $F(Q(p)) = p$. Furthermore, by convergence in distribution and since $Q(p) - \epsilon_{k}$ and $Q(p) + \epsilon_k$ are continuity points of $F$, there is an $N \in \mathbb{N}$ such that for all $n \geq N$ $$ F_n(Q(p)-\epsilon_{k}) < p < F_n(Q(p) + \epsilon_{k}).$$ By definition of $Q_n$, this implies $$ Q(p) - \epsilon_{k} \leq Q_n(p) \leq Q(p) + \epsilon_{k}.$$ It follows that $[\liminf_{n\to\infty} Q_n (p), \limsup_{n\to\infty} Q_n (p)] \subseteq [Q(p)-\epsilon_k, Q(p) + \epsilon_k]$, and letting $\epsilon_{k} \to 0$ proves the claim.

$[\Leftarrow]$ Assume $Q_n(p) \to Q(p)$ in all continuity points $p$ of $Q$. Let $(\Omega, \mathbb{P})$ be a probability space and $U \sim (0,1)$ be a uniform random variable on this space. It is well-known that $Q_n(U) \sim X_n$ and $Q(U) \sim X$ (cf. Inversion sampling), and since $Q$ is increasing and therefore only has countably many discontinuities, it holds that $Q_n(U) \to Q(U)$ almost surely with respect to $\mathbb{P}$, since the discontinuity set of $Q$ has Lebesgue measure $0$. As almost sure convergence implies convergence in distribution, this concludes the proof.

user159517
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