L¹ Properties of the Nadaraya Quantile Estimator

Abstract

Let X be a real random variable having f as density function. Let F be its cumulative distribution function and Q its quantile function. For h > 0, let F_h and Q_h denote respectively the Nadaraya kernel estimator of F and Q. In the first part of this paper the almost sure convergence of the conventional L¹ distance between Q_h and Q is established. In the second part, the L¹ right inversion distance is introduced. The representation of this L¹ right inversion distance in terms of F_h and F is given. This representation allows us to suggest ways to choose a global bandwidth for the estimator Q_h.

Appendices

Appendix A: Proofs

A.1.1 Proof of Theorem 1

—Proof of Theorem 1 (i)

Let us assume that all the random variables X,X₁,…,X_n are defined on the probability space $({\Omega },\ {\mathscr{A}},\ \mathrm {P}).$ For ω ∈Ω, set

$$ \lVert F_{h}(.,\omega)-F(.,\omega)\rVert = \sup_{x\in{\mathbb{R}}} | F_{h}(x,\omega)-F(x,\omega)|. $$

By Theorem 2 of Chacón and Rodríguez-Casal (2010) the set

$$ {\Omega}_{1}=\{\omega\in {\Omega}:\quad \lVert F_{h}(.,\omega)-F(.,\omega)\rVert\xrightarrow{\quad a.s.\quad} 0 \} $$

is of probability 1. For ω ∈Ω, set

$$ T_{n}(\omega)=\frac{| X_{1}(\omega)|+\cdots+| X_{n}(\omega)|}{n} $$

and

$$ {\Omega}_{2}=\{\omega\in {\Omega}:\quad T_{n}(\omega) \xrightarrow{\quad a.s.\quad} \mathrm{E}| X| \}. $$

By the strong law of large numbers P(Ω₂) = 1. Let ω ∈Ω₁ ∩Ω₂. By Lemma 8.3 of Bickel and Freedman (1981), to prove Theorem 1 (i), it is enough to show that

$$ \begin{array}{@{}rcl@{}} (a)\quad &&F_{h}(x,\omega)\xrightarrow{\quad \quad} F(x),\quad \forall x\in {\mathbb{R}} \\ (b)\quad && \int | x| \mathrm{d}F_{h}(x,\omega)\xrightarrow{\quad \quad} \int | x| \mathrm{d}F(x)=\mathrm{E}| X|. \end{array} $$

(a) is true because ω ∈Ω₁. Let us prove (b). Using the substitution u = (x − X_i)/h we get

$$ \int | x| K\left( \frac{x-X_{i}}{h}\right) \mathrm{d}x=h \int | X_{i}+uh| K(u)\mathrm{d}u. $$

We have

$$ \int | x| \mathrm{d}F_{h}(x,\omega)=T_{n}(\omega)+\int | x| \mathrm{d}F_{h}(x,\omega) -T_{n}(\omega). $$

Setting

$$ {\Gamma}_{1}=\int | x| \mathrm{d}F_{h}(x,\omega) -T_{n}(\omega), $$

we have

$$ {\Gamma}_{1}=\frac{1}{n}\sum\limits_{i=1}^{n}\int\left( | X_{i}+uh| -| X_{i}|\right) K(u)\mathrm{d}u, $$

it follows that

$$ \begin{array}{@{}rcl@{}} | {\Gamma}_{1}| &\leq & \frac{1}{n}\sum\limits_{i=1}^{n}\int | uh| K(u)\mathrm{d}u \\ &= & h\int | u| K(u)\mathrm{d}u. \end{array} $$

From this last inequality we see that ${\Gamma }_{1} \xrightarrow {\quad \quad } 0.$ Hence, (b) is true because ω ∈Ω₂.

—Proof of Theorem 1 (ii)

We are going to use the following lemmas in the proof.

Lemma 1.

If (A.2) is true then

$$ \int | x | f_{h}(x)\mathrm{d} x<+ \infty. $$

Lemma 2.

Let Y be a random variable with cumulative distribution function G. If G is continuous, then $\mathrm {E}| Y|=+\infty$ if and only if

$$ {\int}_{-\infty }^{0} G(x)\mathrm{d} x=+\infty\quad \text{or}\quad {\int}_{0}^{+\infty }(1- G(x))\mathrm{d} x=+\infty. $$

(Let us continue with the proof of Theorem 1(ii)) It follows from Lemma 1 and Lemma 2 that

$$ {\int}_{-\infty }^{0} F_{h}(x)\mathrm{d} x<+\infty\quad \text{and}\quad {\int}_{0}^{+\infty }(1- F_{h}(x))\mathrm{d} x<+\infty. $$

Since $\mathrm {E}| X|=+\infty ,$ by Lemma 2 we have

$$ {\int}_{-\infty }^{0} F(x)\mathrm{d} x=+\infty\quad \text{or}\quad {\int}_{0}^{+\infty }(1- F(x))\mathrm{d} x=+\infty. $$

We have:

$$ \begin{array}{@{}rcl@{}} \int | F_{h}(x)-F(x)| \mathrm{d}x& = &{\int}_{-\infty }^{0} | F(x)-F_{h}(x)| \mathrm{d}x+ {\int}_{0}^{+\infty } | (1-F(x))-(1- F_{h}(x))| \mathrm{d}x \\ &\ge & {\int}_{-\infty }^{0} \left( F(x)-F_{h}(x) \right)\mathrm{d}x+ {\int}_{0}^{+\infty } \left( (1-F(x))-(1- F_{h}(x)) \right) \mathrm{d}x \\ &= & +\infty. \end{array} $$

A.1.1.1 Proof of Lemma 1

We have

$$ \int | x| K\left( \frac{x-X_{i}}{h}\right) \mathrm{d}x=h \int | X_{i}+uh| K(u)\mathrm{d}u. $$

It follows that

$$ \int | x| K\left( \frac{x-X_{i}}{h}\right) \mathrm{d}x\leq h | X_{i}|\int K(u)\mathrm{d}u+ h^{2} \int | u| K(u)\mathrm{d}u. $$

This inequality permits to get

$$ \int | x| f_{h}(x) \mathrm{d}x\leq \frac{1}{n} \sum\limits_{i=1}^{n}| X_{i}| +h\int | u| K(u)\mathrm{d}u<+\infty. $$

A.1.1.2 Proof of Lemma 2

It is well-known that

$$ \begin{array}{@{}rcl@{}} \mathrm{E} | Y| &= &{\int}_{0}^{+\infty} \mathrm{P}\left( | Y|>t\right)\mathrm{d}t \\ &= & {\int}_{0}^{+\infty} \mathrm{P}\left( Y<-t\right)\mathrm{d}t + {\int}_{0}^{+\infty} \mathrm{P}\left( Y>t\right)\mathrm{d}t\\ &= &{\int}_{-\infty}^{0}G(t) \mathrm{d}t+ {\int}_{0}^{+\infty} (1-G(t))\mathrm{d}t. \end{array} $$

A.2.1 Proof of Proposition 1

First we prove some useful equalities. Let (x_n) be a bounded sequence. Then we have:

$$ \begin{array}{@{}rcl@{}} F(\underset{n}{\sup} x_{n}) &= & \underset{n}{\sup} F(x_{n}) \end{array} $$

(A.3)

$$ \begin{array}{@{}rcl@{}} F(\underset{n}{\inf} x_{n})&= & \underset{n}{\inf} F(x_{n}). \end{array} $$

(A.4)

Proof of Eq. A.3. We have

$$ \begin{array}{@{}rcl@{}} x_{n}\leq \sup_{n} x_{n} &\implies & F(x_{n})\leq F(\underset{n}{\sup} x_{n}) \\ &\implies & \underset{n}{\sup} F(x_{n})\leq F(\underset{n}{\sup} x_{n}). \end{array} $$

Set $u_{0}=\sup _{n} x_{n}.$ For $\varepsilon =\frac {1}{k}\ (k\in {\mathbb {N}}^{*}),\ \exists n_{k}\in {\mathbb {N}}$ such that

$$ \begin{array}{@{}rcl@{}} && u_{0}-\frac{1}{k} < x_{n_{k}}\leq u_{0}\\ &\implies & F(u_{0}-\frac{1}{k})\leq F(x_{n_{k}})\\ &\implies & F(u_{0}-\frac{1}{k})\leq \sup_{n} F(x_{n}),\ \forall k\in{\mathbb{N}}^{*}\\ &\implies & F(u_{0})\leq \underset{n}{\sup} F(x_{n}). \end{array} $$

This last inequality completes the proof of Eq. A.3. The proof of Eq. A.4 is similar. Let α₀ ∈ (0, 1) and assume that Q(α₀) is the unique solution of the equation F(y) = α₀. To prove Proposition 1, it is enough to show that, if (s_n) is a sequence then

$$ F(s_{n})\xrightarrow{\quad \quad} \alpha_{0} \implies s_{n}\xrightarrow{\quad \quad} Q(\alpha_{0}). $$

(A.5)

Proof of the implication (A.5)

–(s_n) is necessarily bounded. Suppose for instance that $s_{n_{k}}\xrightarrow {\quad \quad } -\infty ,$ then $F(s_{n_{k}})\xrightarrow {\quad \quad } 0 \neq \alpha _{0}.$ This contradicts the fact that $F(s_{n_{k}})\xrightarrow {\quad \quad } \alpha _{0}.$ Next, set

$$ l_{n}=\inf \{s_{k},\quad k\geq n \},\quad u_{n}=\sup \{s_{k},\quad k\geq n \}. $$

Using Eq. A.4 we have

$$ \begin{array}{@{}rcl@{}} && F(l_{n})=\inf_{k\geq n} F(s_{k})\\ &\implies & \lim F(l_{n})=\underline{\lim} F(s_{n})\\ &\implies & F(\lim l_{n}) =\alpha_{0} \\ &\implies & F(\underline{\lim} s_{n}) =\alpha_{0}. \end{array} $$

Using Eq. A.3 we can prove similarly that

$$ F(\overline{\lim} s_{n}) =\alpha_{0}. $$

The uniqueness of Q(α₀) implies that

$$ \underline{\lim} s_{n}=Q(\alpha_{0})=\overline{\lim} s_{n} $$

and these equalities show that

$$ s_{n}\xrightarrow{\quad \quad} Q(\alpha_{0}). $$

A.3.1 Proof of Theorem 2

Theorem 2 follows from certain properties of cumulative distribution functions. Lemma 3 below is its restatement in terms of general cumulative distribution functions.

Lemma 3.

Let G and H be continuous cumulative distribution functions. Then we have:

$$ \begin{array}{@{}rcl@{}} \text{(i)}\ \ \ \ && {{\int}_{0}^{1}} \left| H\left( G^{-1}(\alpha)\right) -\alpha\right|\mathrm{d}\alpha = \int| G(x)-H(x)| \mathrm{d}H(x)\\ \text{(ii-a)}\ \ \ && {{\int}_{0}^{1}} \left| G\left( H^{-1}(\alpha)\right) -\alpha\right|\mathrm{d}\alpha = \int| G(x)-H(x)| \mathrm{d}H(x)\\ \text{(ii-b)}\ \ \ && \int| G(x)-H(x)| \mathrm{d}G(x)=\int| G(x)-H(x)| \mathrm{d}H(x)\\ \text{(iii)}\ \ \ \ && \int| H\left( G^{-1}(\alpha)\right) -G\left( H^{-1}(\alpha)\right) | \mathrm{d}\alpha = 2\int| G(x)-H(x)| \mathrm{d}H(x).\\ \end{array} $$

Note that (ii-b) is a restatement of Corollary 1. All the identities in Lemma 3 are presented in Schmid (1993). However, the author states and proves his results assuming that G and H are continuous and bijective. We are going to use the following lemma in our proof.

Lemma 4.

(Lemma 1 in Youndjé (2018)) Let G, H and L be three continuous cumulative distribution functions. Then we have:

$$ \int| G(x)-H(x)| \mathrm{d} L(x)= {{\int}_{0}^{1}} \left| L\left( G^{-1}(\alpha)\right) - L\left( H^{-1}(\alpha)\right)\right|\mathrm{d}\alpha. $$

Proof of Lemma 3

By Lemma 4 we have:

$$ \begin{array}{@{}rcl@{}} \int| G(x)-H(x)| \mathrm{d} H(x) & = & {{\int}_{0}^{1}} \left| H\left( G^{-1}(\alpha)\right) - H\left( H^{-1}(\alpha)\right)\right|\mathrm{d}\alpha \end{array} $$

(A.6)

$$ \begin{array}{@{}rcl@{}} & = & {{\int}_{0}^{1}} \left| H\left( G^{-1}(\alpha)\right) -\alpha\right|\mathrm{d}\alpha \end{array} $$

(A.7)

therefore (i) is established. Since H is supposed continuous, using the substitution α = H(x) (see Apostol (1974)) we have:

$$ \begin{array}{@{}rcl@{}} \int| G(x)-H(x)| \mathrm{d} H(x)&= & {{\int}_{0}^{1}} \left| G\left( H^{-1}(\alpha)\right) - H\left( H^{-1}(\alpha)\right)\right|\mathrm{d}\alpha\\ &= & {{\int}_{0}^{1}} \left| G\left( H^{-1}(\alpha)\right) -\alpha\right|\mathrm{d}\alpha. \end{array} $$

These equalities show that (ii-a) is true. Using the substitution α = G(x) we have:

$$ \begin{array}{@{}rcl@{}} \int| G(x)-H(x)| \mathrm{d} G(x)&= & {{\int}_{0}^{1}} \left| G\left( G^{-1}(\alpha)\right) - H\left( G^{-1}(\alpha)\right)\right|\mathrm{d}\alpha\\ &= & {{\int}_{0}^{1}} \left| H\left( G^{-1}(\alpha)\right) -\alpha\right|\mathrm{d}\alpha\\ &= & \int| G(x)-H(x)| \mathrm{d} H(x). \end{array} $$

This last equality follows from Eqs. A.6 and A.7. The proof of (ii-b) is obtained from this chain of equalities. Let us set L = (G + H)/2. On one hand we have

$$ \begin{array}{@{}rcl@{}} \int| G(x)-H(x)| \mathrm{d} L(x)&= &\frac{1}{2}\int| G(x)-H(x)| \mathrm{d} G(x)+ \frac{1}{2}\int| G(x)-H(x)| \mathrm{d} H(x)\\ &=&\int| G(x)-H(x)| \mathrm{d} H(x) \end{array} $$

by Lemma 3 (ii-b). On the other hand, using Lemma 4, we have

$$ \begin{array}{@{}rcl@{}} \int| G(x)-H(x)| \mathrm{d} L(x)&= & {{\int}_{0}^{1}} \left| L\left( G^{-1}(\alpha)\right) - L\left( H^{-1}(\alpha)\right)\right|\mathrm{d}\alpha\\ &= & {{\int}_{0}^{1}} \left|\left( \frac{G+H}{2}\right) \left( G^{-1}(\alpha)\right) - \left( \frac{G+H}{2}\right)\left( H^{-1}(\alpha)\right)\right|\mathrm{d}\alpha\\ &= & {{\int}_{0}^{1}} \left| \left( \frac{\alpha}{2}+ \frac{H\left( G^{-1}(\alpha)\right)}{2}\right) -\left( \frac{G\left( H^{-1}(\alpha)\right)}{2}+ \frac{\alpha}{2} \right)\right|\mathrm{d}\alpha\\ &= & \frac{1}{2}\int| H\left( G^{-1}(\alpha)\right) -G\left( H^{-1}(\alpha)\right) |\mathrm{d}\alpha. \end{array} $$

It follows from the two chains of equalities above that

$$ \int| G(x)-H(x)| \mathrm{d} H(x)= \frac{1}{2}\int| H\left( G^{-1}(\alpha)\right) -G\left( H^{-1}(\alpha)\right) |\mathrm{d}\alpha $$

and this identity is exactly Lemma 3 (iii).

A.4.1 Proof of Proposition 2

—Proof of Proposition 2 (i)

We give the proof for the Cauchy cumulative distribution function. The one for the Pareto cdf follows the same steps. We have

$$ \int \left( F_{h}(x)-F(x)\right)^{2} \mathrm{d}x= {\int}_{-\infty }^{0} \left( F(x)-F_{h}(x)\right)^{2} \mathrm{d}x+ {\int}_{0}^{+\infty } \left( (1-F(x))-(1- F_{h}(x))\right)^{2} \mathrm{d}x, $$

thus, to prove Proposition 2 it is enough to show that

$$ (i)\ {\int}_{-\infty }^{0} \left( F(x)-F_{h}(x)\right)^{2} \mathrm{d}x<+\infty\quad \text{and} \quad (ii)\ {\int}_{0}^{+\infty } \left( (1-F(x))-(1- F_{h}(x))\right)^{2} \mathrm{d}x<+\infty. $$

Proof of (ii): To prove (ii) it is enough to show that

$$ (a)\ {\int}_{0}^{+\infty } \left( 1-F(x)\right)^{2} \mathrm{d}x<+\infty\quad \text{and} \quad (b)\ {\int}_{0}^{+\infty } \left( 1-F_{h}(x)\right)^{2} \mathrm{d}x<+\infty. $$

We have

$$ {\int}_{0}^{+\infty } \left( 1-F_{h}(x)\right)^{2} \mathrm{d}x\leq {\int}_{0}^{+\infty } \left( 1-F_{h}(x)\right) \mathrm{d}x<+\infty $$

(A.8)

because of Lemma 1 and Lemma 2. Hence (b) is true. To prove (a), it is enough to show that if F is the Cauchy cdf, we have:

$$ 1-F(x)=\frac{1}{\pi x} +o\left( \frac{1}{x}\right)\quad \text{as}\quad x\xrightarrow{\quad \quad} +\infty. $$

(A.9)

We have

$$ F(x)=\frac{1}{\pi} \arctan(x)+\frac{1}{2}. $$

For x > 0, it is widely known that

$$ \arctan(x) +\arctan\left( \frac{1}{x}\right)=\frac{\pi}{2}. $$

It follows that

$$ \begin{array}{@{}rcl@{}} && \arctan(x)=\frac{\pi}{2}-\arctan\left( \frac{1}{x}\right)\\ &\implies & F(x)= \frac{1}{2} -\frac{1}{\pi}\arctan\left( \frac{1}{x}\right) +\frac{1}{2}\\ &\implies & 1- F(x)=\frac{1}{\pi}\arctan\left( \frac{1}{x}\right) \end{array} $$

hence, Eq. A.9 is obtained via Taylor expansion of order 1. Using the fact that

$$ \arctan(x) +\arctan\left( \frac{1}{x}\right)=-\frac{\pi}{2}\quad\text{for}\quad x<0, $$

the proof of (i) follows the same steps as that of (ii).

—Proof of Proposition 2 (ii) We have:

$$ \begin{array}{@{}rcl@{}} ISE(F_{h},F)&= &\int (F_{h}(x)-F(x))^{2} \mathrm{d}x\\ &\ge & {\int}^{+\infty }_{1} (F_{h}(x)-F(x))^{2} \mathrm{d}x \\ &= & {\int}^{+\infty }_{1} \left( (1-F(x))-(1- F_{h}(x)) \right)^{2} \mathrm{d}x. \end{array} $$

Thus the proof of Proposition 2 (ii) will be complete if we prove that:

$$ {\int}^{+\infty }_{1} \left( (1-F(x))-(1- F_{h}(x)) \right)^{2} \mathrm{d}x=\ +\infty. $$

(A.10)

Next, set $J=[1,\ +\infty )$ and

$$ L^{2}(J)=\{ h:J\longrightarrow {\mathbb{R}}, \quad \mid \ {\int}^{+\infty }_{1} h(x)^{2}\mathrm{d}x<+\infty \}. $$

It is well-known that L²(J) is a vector space, hence to prove (A.10) it is enough to show that

$$ 1-F_{h}\in L^{2}(J)\quad \text{and} \quad 1-F \notin L^{2}(J). $$

Equation (A.8) shows that 1 − F_h ∈ L²(J). When $F(x)=1 -\frac {1}{ax^{a}}$ for x > 1 (the Pareto cdf) we have:

$$ {\int}^{+\infty }_{1} \left( (1-F(x) \right)^{2} \mathrm{d}x=\frac{1}{a^{2}} {\int}^{+\infty }_{1} \frac{1}{x^{2a}} \mathrm{d}x=+\infty\quad \text{if}\quad 0<a\leq \frac{1}{2} $$

and this ends the proof of Proposition 2 (ii).