# Convergence in probability implies convergence in quantile of inverse quantile.

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.

• Could you clarify these $\to_w$ and $\to_p$ pieces of notation? Regarding your question I think a first step would be to show that $G_n(u)\to G_0(u)$ for every fixed $u\in(0,1)$ such that $G_0$ is continuous at $u$. Sep 26 at 12:08
• The OP is offering to wet convergence of distributions (w) and convergence in probability of random variables (p). Sep 26 at 12:09
• @OliverDiaz thanks. But I think the almost sure convergence also holds (and is as easy to establish, isn't it)? Sep 26 at 13:22
• @nejimban: Yes, it does. It is not difficult to established a.s convergence, but one needs a little analysis of the properties of the quantile function as well as the monotonicity properties of both the cumulative distribution function and the quantile function. Sep 26 at 14:10
• @OliverDiaz, you're right… I might have not emphasized too much on that in my answer, so I upvoted yours 😅. Sep 26 at 14:25

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 $$y, we have from \eqref{one} that $$F(y). Since $$F_n(y)\xrightarrow{n\rightarrow\infty}F(y)$$, there is $$N\in\mathbb{N}$$ such that $$F_n(y) for all $$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, $$Q(t)\leq Q(t'). Fix $$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'). 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 $$\lim_nQ_n(t)=Q(t),\qquad t\in(0,1)\setminus D$$

• Many thanks for your detailed solution! I appreciate your help. Sep 26 at 18:15

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 $$0<\varepsilon. Let $$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).$$

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.

• I should add this gives also a proof of Skorokhod's representation theorem for real-valued random variables. Sep 26 at 13:34
• In fact, the proofs presented here (which are basically the same) prove Skorokhod's imbedding theorem of $\mathbb{R}$. For Polish spaces one can also use this result by first mapping isometrically the Polish space to $\mathbb{R}$. Sep 26 at 15:06