Convergence in probability implies convergence in quantile of inverse quantile.

Question

I'm having some problems proving the following result.

Let $F_{n}, n=0,1,2, \ldots$, be c.d.f.'s such that $F_{n} \rightarrow{ }_{w} F_{0} .$ Let $G_{n}(U)=$ $\sup \left\{x: F_{n}(x) \leq U\right\}, n=0,1,2, \ldots$, where $U$ is a random variable having the uniform $U(0,1)$ distribution. Show that $G_{n}(U) \rightarrow{ }_{p} G_{0}(U)$.

Here I think G represents a sort of inverse quantile functions. If we assume that the r.v. are continuous, then G is just the random variables themselves. However, I have no idea how to prove this general case when it's not assumed that the inverse quantile exists. Any help is appreciated. Thanks.

Answer 1

We show here something stronger: convergence almost surely of cumulants.

Without loss of generality consider the probability space $((0,1),\mathscr{B}(0,1)),\lambda)$ where $\lambda$ is LEbesue measure restricted to the unit interval $(0,1)$. The function $U(t)=t$ is a uniform distributed random variable.

A few initial remarks:

• For any measure $\mu$ in the real line define $F_\mu(x)=\mu((-\infty,x])$ ($F_\mu$ is known as the cumulative probability distribution of measure $\mu$). Observe that (a) $F_\mu$ is monotone non decreasing, (b) right continuous with left limits, (c) and $\lim_{x\rightarrow-\infty}F_\mu(x)=0$, $\lim_{x\rightarrow\infty}F_\mu(x)=1$. Conversely, function $F$ that satisfies (a)-(c) yields a unique measure $\mu$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ such that $F_\mu=F$.

• Given a probability measure $\mu$ in $(\mathbb{R},\mathscr{B}(\mathbb{R}))$, the quantile function $Q_\mu:(0,1)\rightarrow\mathbb{R}$ is defined as $$ \begin{align} Q_\mu(t)=\inf\{x\in\mathbb{R}: F_\mu(x)\geq t\} \end{align}$$ By monotonicity and right-continuity of $F$, it is easy to check that $Q_\mu$ satisfies $$\begin{align} F_\mu(Q(t))\geq t,&\qquad t\in(0,1)\tag{0}\label{zero}\\ F_\mu(x)\geq t\quad &\text{iff}\qquad Q_\mu(t)\leq x \tag{1}\label{one} \end{align}$$ whence it follows that $$\lambda\big(\{t\in(0,1): Q_\mu(t)\leq x\}\big)=\lambda\big(t\in(0,1): F_\mu(x)\geq t\}\big)=F_\mu(x)$$ Hence $Q_\mu$ is a random variable on $(0,1)$ whose distribution is $\mu$. It is clear by definition of the quantile function that $Q_\mu$ is monotone nondecreasing, and left-continuous with right-limits.

Solution to OP:

Let $\mu_n$ the measure with cumulative distribution $F_n$ and $\mu$ the measure with cumulative distribution $F$.

1. Then, $Q_n:=Q_{\mu_n}$ and $Q:=Q_\mu$ are random variables with ditributions $\mu_n$ and $\mu$ respectively. Let $D$ be the set points in $(0,1)$ at which $Q$ is discontinuous. Since $Q$ is monotone, $D$ is at most countable and so, $\lambda(D)=0$, i.e., $Q$ is continuous almost surely (a.s.)

2. Recall that $\mu_n$ converges weakly to $\mu$ iff $F_n(x)\xrightarrow{n\rightarrow\infty}F(x)$ for $x$ where $F$ is continuous. Since $F$ is monotone, the set of discontinuities of $F$ is countable.

3. We claim that $Q_n(t)\xrightarrow{n\rightarrow\infty}Q(t)$ for all $t\in(0,1)\setminus D$ and thus, $Q_n\xrightarrow{n\rightarrow\infty}Q$ almost surely. Let $t\in(0,1)\setminus D$. If $y$ is a point of continuity of $F$ with $y<Q(t)$, we have from \eqref{one} that $F(y)<t$. Since $F_n(y)\xrightarrow{n\rightarrow\infty}F(y)$, there is $N\in\mathbb{N}$ such that $F_n(y)<t$ for all $n\leq N$. Again, by \eqref{one}, $Q_n(t)>y$ for all $n\geq N$. Hence $\liminf_nQ_n(t)\geq y$. Letting $y\nearrow Q(t)$ along points of continuity of $F$ yields $$\begin{align} Q(t)\leq\liminf_nQ_n(t)\tag{2}\label{two} \end{align}$$ Now, since $t\in (0,1)\setminus D$, given $\varepsilon>0$, there is $\delta>0$ such that if $t<t'<t+\delta$, $Q(t)\leq Q(t')<Q(t)+\varepsilon$. Fix $t'\in(t,t+\delta)$, and let $z\in(Q(t'),Q(t)+\varepsilon)$ at which $F$ is continuous. Then, by monotonicity of $F$ and \eqref{zero}, $F(z)\geq F(Q(t'))\geq t'>t$. Since $F_n(z)\xrightarrow{n\rightarrow\infty}F(z)$, there is $N'\in\mathbb{N}$ such that $F_n(z)>t$ for all $n\geq N'$. By \eqref{one}, $Q_n(t)\leq z$ for all $n\geq N$. Consequently $\limsup_nQ_n(t)\leq z$. Letting $z\searrow Q(t')$ along points of continuity of $F$ yields $\limsup_nQ(t)\leq Q(t')<Q(t)+\varepsilon$. As $\varepsilon>0$ can be taken to be arbitrarily small, we obtain $$\begin{align} \limsup_nQ_n(t)\leq Q(t)\tag{3}\label{three}\end{align}$$ Combining \eqref{two} and \eqref{three} gives $$\lim_nQ_n(t)=Q(t),\qquad t\in(0,1)\setminus D$$

Answer 2

This is a standard fact about the inverse method.

For $n=0,1,\ldots$ and $u\in(0,1)$, let $$G_n(u):=\sup\:\{x\in\mathbb R:F_n(x)\le u\}.$$ It is easy to check that $G_n$ is non-decreasing (and right-continuous*). In particular $G_0$ (just like $F_0$) has at most countably many point of discontinuities.

We now show that $G_n(u)\to G(u)$ for all $u\in(0,1)$ such that $G_0$ is continuous at $u$.

1. Let $0<\varepsilon<u$. Let $x\in\mathbb R$ be such that $F_0(x)\le u-\varepsilon$. Up to choosing an arbitrarily closer $x$, we can assume that $F_0$ is continuous at $x$. Then the weak convergence of distributions gives $F_n(x)\to F_0(x)$ as $n\to\infty$. In particular $F_n(x)\le F_0(x)+\varepsilon$ for all $n$ large enough. This implies that $x\le G_n(F_0(x)+\varepsilon)$ for such $n$, and then (because $F_0(x)\le u-\varepsilon$), $x\le G_n(u)$. Thus $$\liminf_{n\to\infty}G_n(u)\ge x$$ for any chosen $x$ with $F_0(x)\le u-\varepsilon$, so $$\liminf_{n\to\infty}G_n(u)\ge G_0(u-\varepsilon)$$ by definition of $G_0(u-\varepsilon)$. As $\varepsilon$ was arbitrary and $G_0$ is assumed to be continuous at $u$, we deduce that $$\liminf_{n\to\infty}G_n(u)\ge G_0(u).$$
2. For the other direction, let $x>G_0(u)$. Then $F_0(x)>u$ and again, the convergence in distribution gives $F_n(x)>u$ for all $n$ large enough, so $G_n(u)\le x$. Hence $$\limsup_{n\to\infty}G_n(u)\le x.$$ This holds for all $x>G_0(u)$ so $$\limsup_{n\to\infty}G_n(u)\le G_0(u).$$

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Therefore $G_n(u)\to G_0(u)$ for all point $u\in(0,1)$ of continuity of $G_0$. As a uniform variable $U\in(0,1)$ is almost surely a continuity point of $G_0$, we get that $G_n(U)\to G_0(U)$ almost surely (and obviously also in probability).

*: We also encounter (perhaps more often) the left-continuous inverses $\inf\:\{x\in\mathbb R:F_n(x)\ge u\}$ for which the above fact can be proved similarly.