Directional bivariate quantiles: a robust approach based on the cumulative distribution function

Abstract

The definition of multivariate quantiles has gained considerable attention in previous years as a tool for understanding the structure of a multivariate data cloud. Due to the lack of a natural ordering for multivariate data, many approaches have either considered geometric generalisations of univariate quantiles or data depths that measure centrality of data points. Both approaches provide a centre-outward ordering of data points but do no longer possess a relation to the cumulative distribution function of the data generating process and corresponding tail probabilities. We propose a new notion of bivariate quantiles that is based on inverting the bivariate cumulative distribution function and therefore provides a directional measure of extremeness as defined by the contour lines of the cumulative distribution function which define the quantile curves of interest. To determine unique solutions, we transform the bivariate data to the unit square. This allows us to introduce directions along which the quantiles are unique. Choosing a suitable transformation also ensures that the resulting quantiles are equivariant under monotonically increasing transformations. We study the resulting notion of bivariate quantiles in detail, with respect to computation based on linear programming and theoretical properties including asymptotic behaviour and robustness. It turns out that our approach is especially useful for data situations that deviate from the elliptical shape typical for ‘normal-like’  bivariate distributions. Moreover, the bivariate quantiles inherit the robustness of univariate quantiles even in case of extreme outliers.

A Further proofs

1.1 A.1 Proof of Theorem 2

Proof

For fixed $b\in \mathbb {R}$, we define the expected loss as a function of $v\in \mathbb {R}$ by

$$\begin{aligned} \mathbb {E}(\rho _{b,\tau }(\varvec{Y},\varvec{q}))=\mathbb {E}(\rho _{\tau }(\varvec{Y},(v,v+b)')). \end{aligned}$$

(12)

Clearly, the bivariate quantile curves from (1) can be obtained as

$$\begin{aligned} \mathcal {Q}_{\tau }=\bigcup _{b\in \mathbb {R}}\left\{ v\in \mathbb {R}|F(v,v+b)=\tau \right\} \end{aligned}$$

which intuitively means that we describe $\mathbb {R}^2$ by straight lines with slope one and intercepts b. With the definition

$$\begin{aligned} u(\varvec{y},v)=\max \left( y_1-v,y_2-v-b\right) , \end{aligned}$$

the expected loss for fixed $b\in \mathbb {R}$ is given by

$$\begin{aligned} \mathbb {E}\left( \rho _{b,\tau }\left( \varvec{Y},v\right) \right)&= (\tau -1)\displaystyle \int \limits _{-\infty }^{ v}\displaystyle \int \limits _{-\infty }^{v+b}u\left( \varvec{y},v\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1 \\&\quad + \tau \displaystyle \int \limits _{ v}^{\infty }\displaystyle \int \limits _{v+b}^{\infty }u(\varvec{y},v) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad + \, \tau \displaystyle \int \limits _{ v}^{\infty }\displaystyle \int \limits _{-\infty }^{v+b}u(\varvec{y},v) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1 \\&\quad + \tau \displaystyle \int \limits _{-\infty }^{ v}\displaystyle \int \limits _{v+b}^{\infty }u(\varvec{y},v) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&=(\tau -1)\displaystyle \int \limits _{-\infty }^{ v}\displaystyle \int \limits _{-\infty }^{y_1+b}\left( y_1-v\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\ {}&\quad +\,(\tau -1)\displaystyle \int \limits _{-\infty }^{ v}\displaystyle \int \limits _{y_1+b}^{v+b}\left( y_2-v-b\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1 \\ {}&\quad +\,\tau \displaystyle \int \limits _{ v}^{\infty }\displaystyle \int \limits _{v+b}^{y_1+b}\left( y_1-v\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad +\,\tau \displaystyle \int \limits _{ v}^{\infty }\displaystyle \int \limits _{y_1+b}^{\infty }\left( y_2-v-b\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad + \, \tau \displaystyle \int \limits _{ v}^{\infty }\displaystyle \int \limits _{-\infty }^{v+b}\left( y_1-v\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\ {}&\quad +\, \tau \displaystyle \int \limits _{-\infty }^{ v}\displaystyle \int \limits _{v+b}^{\infty }\left( y_2-v-b\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1. \end{aligned}$$

Our strategy is now to show that for all $b\in \mathbb {R}$ the expected loss $\mathbb {E}(\rho _{b,\tau }(\varvec{Y},v))$ is uniquely minimised at $q\in \mathbb {R}$ and fulfils the condition $\mathbb {P}(Y_1\le q,Y_2\le q+b)=\tau$. We therefore investigate the first derivative of $\mathbb {E}(\rho _{b,\tau }(\varvec{Y},v))$ with respect to v. The derivative is obtained by applying the Leibniz rule for parameter integrals twice.

$$\begin{aligned} \frac{\partial }{\partial v}\mathbb {E}\left( \rho _{b,\tau }\left( \varvec{y},v\right) \right)&=(\tau -1)\displaystyle \int \limits _{-\infty }^{v}\frac{\partial }{\partial v}\displaystyle \int \limits _{-\infty }^{y_1+b}\left( y_1-v\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad +\,(\tau -1)\displaystyle \int \limits _{-\infty }^{v}\frac{\partial }{\partial v}\displaystyle \int \limits _{y_1+b}^{v+b}\left( y_2-v-b\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad +\,\tau \displaystyle \int \limits _{v}^{\infty }\frac{\partial }{\partial v}\displaystyle \int \limits _{v+b}^{y_1+b}\left( y_1-v\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad +\,\tau \displaystyle \int \limits _{v}^{\infty }\frac{\partial }{\partial v}\displaystyle \int \limits _{y_1+b}^{\infty }\left( y_2-v-b\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad -\tau \displaystyle \int \limits _{v+b}^{\infty }(y_2-v-b)f(v,y_2)\mathrm {d}y_2\\&\quad +\,\tau \displaystyle \int \limits _{v}^{\infty }\frac{\partial }{\partial v}\displaystyle \int \limits _{-\infty }^{v+b}\left( y_1-v\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad +\,\tau \displaystyle \int \limits _{-\infty }^{v}\frac{\partial }{\partial v}\displaystyle \int \limits _{v+b}^{\infty }\left( y_2-v-b\right) f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad +\,\tau \displaystyle \int \limits _{v+b}^{\infty }(y_2-v-b)f(v,y_2)\mathrm {d}y_2 \\&=-(\tau -1)\displaystyle \int \limits _{-\infty }^{v}\displaystyle \int \limits _{-\infty }^{y_1+b} f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad -(\tau -1)\displaystyle \int \limits _{-\infty }^{v}\displaystyle \int \limits _{y_1+b}^{v+b} f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad -\,\tau \displaystyle \int \limits _{v}^{\infty }\displaystyle \int \limits _{v+b}^{y_1+b} f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1-\tau \displaystyle \int \limits _{v}^{\infty }(y_1-v)f(y_1,v+b)\mathrm {d}y_1\\&\quad -\,\tau \displaystyle \int \limits _{v}^{\infty }\displaystyle \int \limits _{y_1+b}^{\infty } f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1-\tau \displaystyle \int \limits _{v+b}^{\infty }(y_2-v-b)f(v,y_2)\mathrm {d}y_2\\&\quad -\,\tau \displaystyle \int \limits _{v}^{\infty }\displaystyle \int \limits _{-\infty }^{v+b} f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1+\tau \displaystyle \int \limits _{v}^{\infty }(y_1-v)f(y_1,v+b)\mathrm {d}y_1\\&\quad -\,\tau \displaystyle \int \limits _{-\infty }^{v}\displaystyle \int \limits _{v+b}^{\infty } f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1+\tau \displaystyle \int \limits _{v+b}^{\infty }(y_2-v-b)f(v,y_2)\mathrm {d}y_2 \\&=-(\tau -1)\displaystyle \int \limits _{-\infty }^{v}\displaystyle \int \limits _{-\infty }^{v+b} f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1-\tau \displaystyle \int \limits _{v}^{\infty }\displaystyle \int \limits _{v+b}^{\infty } f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&\quad -\,\tau \displaystyle \int \limits _{v}^{\infty }\displaystyle \int \limits _{-\infty }^{v+b} f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1-\tau \displaystyle \int \limits _{-\infty }^{v}\displaystyle \int \limits _{v+b}^{\infty } f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1\\&=\displaystyle \int \limits _{-\infty }^{v}\displaystyle \int \limits _{-\infty }^{v+b}f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1-\tau . \end{aligned}$$

In summary, we have

$$\begin{aligned} \begin{aligned} \frac{\partial }{\partial v}\mathbb {E}\left( \rho _{b,\tau }\left( \varvec{y},v\right) \right) = \displaystyle \int \limits _{-\infty }^{v}\displaystyle \int \limits _{-\infty }^{v+b}f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1-\tau . \end{aligned} \end{aligned}$$

(13)

Let us first assume that $\mathbb {P}(Y_1\le q,Y_2\le q+b)=\tau$ holds. It then follows from Eq. (13) that

$$\begin{aligned} \frac{\partial }{\partial v}\mathbb {E}\left( \rho _{b,\tau }\left( \varvec{y},v\right) \right) \bigg |_{v=q}=\tau -\tau =0. \end{aligned}$$

In addition,

$$\begin{aligned} \frac{\partial ^2}{\partial v^2}\mathbb {E}\left( \rho _{b,\tau }\left( \varvec{y},v\right) \right) \bigg |_{v=q}=\displaystyle \int \limits _{-\infty }^{q}f(y_1,q+b)\mathrm {d}y_1+\displaystyle \int \limits _{-\infty }^{q+b}f(q,y_2)\mathrm {d}y_2>0 \end{aligned}$$

holds since we assumed $f(y_1,y_2)>0$. Consequently, $(q,q+b)'$ is a minimiser of $\mathbb {E}\left( \rho _{b,\tau }\left( \varvec{y},v\right) \right)$ and in particular of $\mathbb {E}\left( \rho _{\tau }\left( \varvec{y},\varvec{q}\right) \right)$.

Reversely, if $(q,q+b)'$ is a minimiser of $\mathbb {E}\left( \rho _{b,\tau }\left( \varvec{y},v\right) \right)$, a zero first derivative

$$\begin{aligned} \frac{\partial }{\partial v}\mathbb {E}\left( \rho _{b,\tau }\left( \varvec{y},v\right) \right) \bigg |_{v=q}=\displaystyle \int \limits _{-\infty }^{q}\displaystyle \int \limits _{-\infty }^{q+b}f(y_1,y_2)\mathrm {d}y_2\mathrm {d}y_1-\tau =0 \end{aligned}$$

is required which is equivalent to

$$\begin{aligned} \mathbb {P}(Y_1\le q,Y_2\le q+b)=\tau . \end{aligned}$$

$\square$

1.2 A.2 Proof of Theorem 6

Proof

Recall first that ${\tilde{\varvec{Y}}}=({\tilde{Y}}_1,{\tilde{Y}}_2)'=(F_{1}(Y_1),F_{2}(Y_2))'$, and let $\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)>0$ be the density of ${\tilde{\varvec{Y}}}$. From Sect. 3.1, we furthermore have that $\rho _{\alpha ,\tau }({\tilde{\varvec{y}}},\tilde{r})$ can be decomposed into

$$\begin{aligned} \begin{aligned} {\left\{ \begin{array}{ll} (1-\tau )\left( \frac{1-{\tilde{y}}_{1}}{\cos (\alpha )} -\tilde{r}\right) &{}\quad \text{ if } \frac{1-{\tilde{y}}_{1}}{\cos (\alpha )} -\tilde{r}\le \frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}<0\\ (1-\tau )\left( \frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}\right) &{}\quad \text{ if } \frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}<\frac{1-{\tilde{y}}_{1}}{\cos (\alpha )} -\tilde{r}<0\\ -\,\tau \quad \;\;\,\left( \frac{1-{\tilde{y}}_{1}}{\cos (\alpha )} -\tilde{r}\right) &{}\quad \text{ if } \min \left( \frac{1-{\tilde{y}}_{1}}{\cos (\alpha )} -\tilde{r},\frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}\right) \ge 0 \text{ and } \frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}\ge \frac{1-{\tilde{y}}_{1}}{\cos (\alpha )}-\tilde{r}\\ -\,\tau \quad \;\;\,\left( \frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}\right) &{}\quad \text{ if } \min \left( \frac{1-{\tilde{y}}_{1}}{\cos (\alpha )}-\tilde{r},\frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}\right) \ge 0 \text{ and } \frac{1-{\tilde{y}}_{2}}{\sin (\alpha )} -\tilde{r}< \frac{1-{\tilde{y}}_{1}}{\cos (\alpha )}-\tilde{r}.\\ \end{array}\right. } \end{aligned} \end{aligned}$$

Accordingly, the expected loss is

$$\begin{aligned} \mathbb {E}(\rho _{\alpha ,\tau }({\tilde{\varvec{y}}},\tilde{r}))&=(1-\tau )\displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\displaystyle \int _{0}^{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}\frac{1-\tilde{r}\cos (\alpha )-{\tilde{y}}_1}{\cos (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,(1-\tau )\displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\displaystyle \int _{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}^{1-\tilde{r}\sin (\alpha )}\frac{1-\tilde{r}\sin (\alpha )-{\tilde{y}}_2}{\sin (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\displaystyle \int _{1-\tilde{r}\sin (\alpha )}^{1}\frac{{\tilde{y}}_2-1+\tilde{r}\sin (\alpha )}{\sin (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{1-\tilde{r}\cos (\alpha )}^{1}\displaystyle \int _{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}^{1}\frac{{\tilde{y}}_2-1+\tilde{r}\sin (\alpha )}{\sin (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{1-\tilde{r}\cos (\alpha )}^{1}\displaystyle \int _{1-\tilde{r}\sin (\alpha )}^{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}\frac{{\tilde{y}}_1-1+\tilde{r}\cos (\alpha )}{\cos (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{1-\tilde{r}\cos (\alpha )}^{1}\displaystyle \int _{0}^{1-\tilde{r}\sin (\alpha )}\frac{{\tilde{y}}_1-1+\tilde{r}\cos (\alpha )}{\cos (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1 \end{aligned}$$

Now, in analogy to the strategy of the proof of Theorem 2, we apply the Leibniz rule for integrals twice, add or subtract terms with identical limits of integration afterwards and after some further basic calculations obtain

$$\begin{aligned} \frac{\partial }{\partial \tilde{r}}\mathbb {E}(\rho _{\alpha ,\tau }({\tilde{\varvec{y}}},\tilde{r}))&=(\tau -1)\displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\displaystyle \int _{0}^{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,(\tau -1)\displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\displaystyle \int _{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}^{1-\tilde{r}\sin (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\displaystyle \int _{1-\tilde{r}\sin (\alpha )}^{1}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{1-\tilde{r}\cos (\alpha )}^{1}\displaystyle \int _{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}^{1}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{1-\tilde{r}\cos (\alpha )}^{1}\displaystyle \int _{1-\tilde{r}\sin (\alpha )}^{1-(1-{\tilde{y}}_1)\tfrac{\sin (\alpha )}{\cos (\alpha )}}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1\\&\quad +\,\tau \displaystyle \int _{1-\tilde{r}\cos (\alpha )}^{1}\displaystyle \int _{0}^{1-\tilde{r}\sin (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1 \end{aligned}$$

Adding together the different integrals yields

$$\begin{aligned} \frac{\partial }{\partial \tilde{r}}\mathbb {E}(\rho _{\alpha ,\tau }({\tilde{\varvec{y}}},\tilde{r}))=\tau -\displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\displaystyle \int _{0}^{1-\tilde{r}\sin (\alpha )}\tilde{f}({\tilde{y}}_1,{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2\mathrm {d}{\tilde{y}}_1 \end{aligned}$$

(14)

A necessary condition for $\tilde{r}$ giving a minimum of the expected loss is that (14) is zero, i.e. that $\tfrac{\partial }{\partial \tilde{r}}\mathbb {E}(\rho _{\alpha ,\tau }({\tilde{\varvec{y}}},\tilde{r}))|_{\tilde{r}=\tilde{r}_{\alpha ,\tau }}=0$. This implies that the quantile condition $\mathbb {P}({\tilde{Y}}_1\le {\tilde{q}}_1,{\tilde{Y}}_2\le {\tilde{q}}_2)=\tau$ is fulfilled if and only if the first derivative of the expected loss at $\tilde{r}$ is zero. What remains to show is that $\tfrac{\partial ^2}{\partial \tilde{r}^2}\mathbb {E}(\rho _{\alpha ,\tau }({\tilde{\varvec{y}}},\tilde{r}))|_{\tilde{r}=\tilde{r}_{\alpha ,\tau }}>0$ holds.

• For $\alpha \in (0,\pi /2)$, this follows from

  $$\begin{aligned} \frac{\partial ^2}{\partial \tilde{r}^2}\mathbb {E}(\rho _{\alpha ,\tau }({\tilde{\varvec{y}}},{\tilde{\varvec{q}}}))= & {} \displaystyle \int _{0}^{1-\tilde{r}\cos (\alpha )}\sin (\alpha )\tilde{f}({\tilde{y}}_1,1-\tilde{r}\sin (\alpha ))\mathrm {d}{\tilde{y}}_1\nonumber \\&+\,\displaystyle \int _{0}^{1-\tilde{r}\sin (\alpha )}\cos (\alpha )\tilde{f}(1-\tilde{r}\cos (\alpha ),{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2 \end{aligned}$$

  (15)

  since $\tilde{f}(\cdot ,\cdot )>0$ and $\cos (\alpha )>0$, $\sin (\alpha )>0$.

• In case of $\alpha =0$, we have $\sin (\alpha )=0$, $\cos (\alpha )=1$ such that the second integral in (15) is zero while the first one is $\int _0^1 \tilde{f}(1-\tilde{r},{\tilde{y}}_2)\mathrm {d}{\tilde{y}}_2=\tilde{f}_1(1-\tilde{r})>0$.

• In case of $\alpha =\pi /2$, we have $\cos (\alpha )=0$, $\sin (\alpha )=1$ such that the first integral in (15) is zero while the second one is $\int _0^1 \tilde{f}({\tilde{y}}_1,1-\tilde{r})\mathrm {d}{\tilde{y}}_1=\tilde{f}_2(1-\tilde{r})>0$. $\square$

1.3 A.3 Proof of Lemma 10

In the following and in order to prove asymptotic results of Sect. 4, we treat the observed data as i.i.d. replications of $\varvec{Y}$ defined on the probability space $(\Omega ,\mathcal F, \mathbb {P})=(\mathbb {R}^2,\mathcal B(\mathbb {R}^2),F)$. Consequently, the transformed data are i.i.d. replicates of $\tilde{\varvec{Y}}=({\tilde{Y}}_1,{\tilde{Y}}_2)'=(F_1(Y_1), F_2(Y_2))$ supplemented with the probability space $(\tilde{\Omega }, \tilde{\mathcal F}, \tilde{\mathbb {P}})=([0,1]^2,\mathcal B([0,1]^2),\tilde{F})$ and CDF $\tilde{F}({\tilde{y}}_1, {\tilde{y}}_2)=\mathbb {P}(Y_1\le F_1^{-1}({\tilde{y}}_1),Y_2\le F_2^{-1}({\tilde{y}}_2))$. In addition, we introduce

$$\begin{aligned} {\tilde{R}}=\min \left( \tfrac{1-{\tilde{Y}}_{1}}{\cos (\alpha )},\tfrac{1-{\tilde{Y}}_{2}}{\sin (\alpha )}\right) ,\;\alpha \in D(\alpha ) \end{aligned}$$

as a random variable on the probability space $(\tilde{\Omega }_{\alpha }, \tilde{\mathcal F}_{\alpha }, \tilde{\mathbb {P}}_{\alpha })=(D_{\tilde{r}}(\alpha ),\mathcal B(D_{\tilde{r}}(\alpha )),\tilde{F}_{\alpha })$.

Proof

• On 1. The claim follows directly due to the i.i.d. property of $\varvec{Y}_1,\ldots ,\varvec{Y}_n$.

• On 2. We introduce the random variables $Z_i=\mathbb {1}_{\lbrace ({\tilde{Y}}_{i1}\le 1-\tilde{r}\cos (\alpha ),{\tilde{Y}}_{i2}\le \tilde{1}-\tilde{r}\sin (\alpha ))\rbrace }$ which are i.i.d. since $\tilde{\varvec{Y}}_1,\tilde{\varvec{Y}}_2,\ldots$ are assumed to be i.i.d. We then have

  $$\begin{aligned} \mathbb {P}(Z_i=1)= & {} \mathbb {P}({\tilde{Y}}_{i1}\le 1-\tilde{r}\cos (\alpha ),{\tilde{Y}}_{i2}\le 1-\tilde{r}\sin (\alpha ))=\tilde{S}_{\alpha }(\tilde{r})\\ \mathbb {P}(Z_i=0)= & {} 1-\tilde{S}_{\alpha }(\tilde{r}) \end{aligned}$$

  and hence $\mathbb {E}(Z_i)=\tilde{S}_{\alpha }(\tilde{r})$. With the strong law of large numbers, we immediately find

  $$\begin{aligned} \tilde{S}_{n,\alpha }(\tilde{r})=\frac{1}{n}\sum _{i=1}^n Z_i\xrightarrow {a.s.}\mathbb {E}(Z_i)=\mathbb {E}(Z_1)=\tilde{S}_{\alpha }(\tilde{r}). \end{aligned}$$

• On 3. From 2., we have that $\mathbb {E}(Z_i)=\tilde{S}_{\alpha }(\tilde{r})$ and ${{\,\mathrm{Var}\,}}(Z_i)=\tilde{S}_{\alpha }(\tilde{r})(1-\tilde{S}_{\alpha }(\tilde{r}))$. Applying the central limit theorem implies

  $$\begin{aligned} \sqrt{n}\frac{\tilde{S}_{n,\alpha }(\tilde{r})-\tilde{S}_{\alpha }(\tilde{r})}{\sqrt{\tilde{S}_{\alpha }(\tilde{r})(1-\tilde{S}_{\alpha }(\tilde{r}))}}{=}\frac{\tfrac{1}{n}\sum _{i=1}^n Z_i-\mathbb {E}(Z_1)}{\sqrt{{{\,\mathrm{Var}\,}}(Z_1)}}{=}\frac{\sum _{i=1}^n Z_i-n\mathbb {E}(Z_1)}{\sqrt{n{{\,\mathrm{Var}\,}}(Z_1)}}\xrightarrow {d}{{\,\mathrm{N}\,}}(0,1). \end{aligned}$$

• On 4. Define

  $$\begin{aligned} \tilde{D}_n:=\sup _{\tilde{r}\in D_{\tilde{r}}(\alpha )} |\tilde{S}_{n,\alpha }(\tilde{r})-\tilde{S}_{\alpha }(\tilde{r})|. \end{aligned}$$

  • (i) $\tilde{S}_{\alpha }$ is continuous and monotonically decreasing in $\tilde{r}$. Hence, we can find a decomposition $\tilde{r}_{\min }=z_0<z_1<z_2<\cdots<z_{m-1}<z_m=\tilde{r}_{\max }$ such that $\tilde{S}_{\alpha }(z_0)=1,\tilde{S}_{\alpha }(z_1)=\tfrac{m-1}{m},\tilde{D}_{\alpha }(z_2)=\tfrac{m-2}{m},\ldots ,\tilde{S}_{\alpha }(z_{m-1})=\tfrac{1}{m},\tilde{S}_{\alpha }(z_m)=0$ and where $\tilde{r}_{\min }$ is the smallest $r\in D_{\tilde{r}}(\alpha )$ and similar $\tilde{r}_{\max }$ the largest $r\in D_{\tilde{r}}(\alpha )$.

  • (ii) We use this decomposition to obtain approximations of $\tilde{S}_{n,\alpha }(z)-\tilde{S}_{\alpha }( z)$ for arbitrary $z\in D_{\tilde{r}}(\alpha )$. Let k be such that $z\in [z_{k},z_{k+1})$. Then,

    $$\begin{aligned} \tilde{S}_{n,\alpha }( z)-\tilde{S}_{\alpha }( z)&\le \tilde{S}_{n,\alpha }( z_{k})-\tilde{S}_{\alpha }( z_{k+1})=\tilde{S}_{n,\alpha }( z_{k})-\left( \tilde{S}_{n,\alpha }( z_{k})-\frac{1}{m}\right) \\ \tilde{S}_{n,\alpha }( z)-\tilde{S}_{\alpha }( z)&\ge \tilde{S}_{n,\alpha }( z_{k+1})-\tilde{S}_{\alpha }( z_{k})=\tilde{S}_{n,\alpha }( z_{k+1})-\left( \tilde{S}_{n,\alpha }( z_{k+1})+\frac{1}{m}\right) \end{aligned}$$

    due to the monotonicity of $\tilde{S}_{\alpha }$.

  • (iii) For $m\in \mathbb {N}$, $k=0,1,\ldots ,m$, define

    $$\begin{aligned} A_{m,k}:=\left\{ \tilde{\omega }_{\alpha }\in \tilde{\Omega }_{\alpha }{:}\,\lim _{n\rightarrow \infty }\tilde{S}_{n,\alpha }( z_{k};\tilde{\omega }_{\alpha })=\tilde{S}_{\alpha }( z_{k})\right\} . \end{aligned}$$

    Due to the almost sure convergence of $\tilde{S}_{n,\alpha }$ from 2., we have

    $$\begin{aligned} \mathbb {P}[A_{m,k}]=1\quad \forall m\in \mathbb {N},\quad k=0,1,\ldots ,m. \end{aligned}$$

  • (iv) Define $A_{m}=\cap _{k=0}^m A_{m,k}.$ This is a finite intersection of sets such that $\mathbb {P}[A_{m}]=1$ for all $m\in \mathbb {N}.$ Define $A=\cap _{m\in \mathbb {N}} A_{m}.$ This is a countable intersection of sets such that $\mathbb {P}[A]=1.$

  • (v) Consider now $\tilde{\omega }_{\alpha }\in A_{m}$. By definition of $A_{m,k}$, there exists an $n(\tilde{\omega }_{\alpha },m)\in \mathbb {N}$ such that

    $$\begin{aligned} |\tilde{S}_{n,\alpha }( z_{k};\tilde{\omega }_{\alpha })-\tilde{S}_{\alpha }( z_{k})|<\frac{1}{m}\quad \forall n>n(\tilde{\omega }_{\alpha },m),\, k=1,\ldots ,m.\text { Hence,} \end{aligned}$$
    $$\begin{aligned} |\tilde{S}_{n,\alpha }(z)-\tilde{S}_{\alpha }(z)|<\frac{1}{m}\quad \forall \tilde{\omega }_{\alpha }\in A_{m},\, n>n(\tilde{\omega }_{\alpha },m),\, z\in D_{\tilde{r}}(\alpha ). \end{aligned}$$

    From (ii), it follows

    $$\begin{aligned} \tilde{D}_n(\tilde{\omega }_{\alpha }):=\sup _{\tilde{r}\in D_{\tilde{r}}(\alpha )} |\tilde{S}_{n,\alpha }(\tilde{r};\tilde{\omega }_{\alpha })-\tilde{S}_{\alpha }(\tilde{r})|<\frac{2}{m}. \end{aligned}$$

    Furthermore, due to the definition of A , $\tilde{\omega }_{\alpha }\in A$ is element of all $A_m$, $m\in \mathbb {N}.$ Hence, $\forall m\in \mathbb {N}$ there exists an $n(\tilde{\omega }_{\alpha },m)\in \mathbb {N}$ such that $\forall n>n(\tilde{\omega }_{\alpha },m)$

    $$\begin{aligned} 0\le \tilde{D}_n(\tilde{\omega }_{\alpha })<\frac{2}{m} \text{ and } \text{ in } \text{ consequence } \lim _{n\rightarrow \infty }\tilde{D}_n(\tilde{\omega }_{\alpha })=0\quad \forall \tilde{\omega }_{\alpha }\in A. \end{aligned}$$

    Finally, we have $\lbrace \tilde{\omega }_{\alpha }\in \tilde{\Omega }_{\alpha }{:}\,\lim _{n\rightarrow \infty }\tilde{D}_n(\tilde{\omega }_{\alpha })=0\rbrace \supseteq A$ and from 4. that $\mathbb {P}[A]=1$ holds such that

    $$\begin{aligned} \mathbb {P}[\left\{ \tilde{\omega }_{\alpha }\in \tilde{\Omega }_{\alpha }{:}\,\lim _{n\rightarrow \infty }\tilde{D}_n(\tilde{\omega }_{\alpha })=0\right\} ]\ge \mathbb {P}[A]=1. \end{aligned}$$

    $\square$

1.4 A.4 Proof of Lemma 11

Proof

The uniqueness of $\tilde{r}_{\alpha ,\tau }$ yields $\tilde{S}_{\alpha }(r_{\alpha ,\tau }+\varepsilon )<\tau <\tilde{S}_{\alpha }(r_{\alpha ,\tau }-\varepsilon )$ for any $\varepsilon >0$. The strong consistency of $\tilde{S}_{n,\alpha }(\tilde{r})$ furthermore ensures

$$\begin{aligned} \begin{aligned} \tilde{S}_{n,\alpha }(\tilde{r}_{\alpha ,\tau }-\varepsilon )&\xrightarrow {a.s.}\tilde{S}_{\alpha }(\tilde{r}_{\alpha ,\tau }-\varepsilon )\\ \tilde{S}_{n,\alpha }(\tilde{r}_{\alpha ,\tau }+\varepsilon )&\xrightarrow {a.s.}\tilde{S}_{\alpha }(\tilde{r}_{\alpha ,\tau }+\varepsilon )\end{aligned} \end{aligned}$$

which is equivalent to

$$\begin{aligned} \begin{aligned}&\mathbb {P}\left( \lim _{n\rightarrow \infty }\tilde{S}_{n,\alpha }(\tilde{r}_{\alpha ,\tau }-\varepsilon )=\tilde{S}_{\alpha }(\tilde{r}_{\alpha ,\tau }-\varepsilon )>\tau \right) =1\\&\mathbb {P}\left( \lim _{n\rightarrow \infty }\tilde{S}_{n,\alpha }(\tilde{r}_{\alpha ,\tau }+\varepsilon )=\tilde{S}_{\alpha }(\tilde{r}_{\alpha ,\tau }+\varepsilon )<\tau \right) =1. \end{aligned} \end{aligned}$$

Using that almost sure convergence $\mathbb {P}(\lim _{n\rightarrow \infty }X_n=X)=1$ is equivalent to $\lim _{n\rightarrow \infty }\mathbb {P}(|X_m-X|<\varepsilon \;\;\forall m\ge n)$ in combination with $\mathbb {P}(A\cap B)=1-\mathbb {P}(A^{\mathsf {c}}\cup B^{\mathsf {c}})\ge 1-\mathbb {P}(A^{\mathsf {c}})-\mathbb {P}(B^{\mathsf {c}})$ implies

$$\begin{aligned} \lim _{n\rightarrow \infty }\mathbb {P}\left( \tilde{S}_{m,\alpha }(\tilde{r}_{\alpha ,\tau }+\varepsilon )<\tau <\tilde{S}_{m,\alpha }(\tilde{r}_{\alpha ,\tau }-\varepsilon )\;\;\forall m\ge n\right) =1. \end{aligned}$$

Due to the monotonicity of $\tilde{S}_{\alpha }$, we have $\tilde{S}_{\alpha }(\tilde{r})\le \tau \Leftrightarrow \tilde{r}\le \tilde{S}_{\alpha }^{-1}(\tau )$ and therefore

$$\begin{aligned} \begin{aligned}&\lim _{n\rightarrow \infty }\mathbb {P}\left( \tilde{r}_{\alpha ,\tau }+\varepsilon<\tilde{S}_{m,\alpha }^{-1}(\tau )=\tilde{r}_{m,\alpha ,\tau }<\tilde{r}_{\alpha ,\tau }-\varepsilon \;\;\forall m\ge n\right) =1\\&\quad \Leftrightarrow \lim _{n\rightarrow \infty }\mathbb {P}\left( |\tilde{r}_{m,\alpha ,\tau }-\tilde{r}_{\alpha ,\tau }|<\varepsilon \;\;\forall m\ge n\right) =1. \end{aligned} \end{aligned}$$

Finally, ${\tilde{q}}_{j,n,\alpha ,\tau }\xrightarrow {a.s.}{\tilde{q}}_{j,\alpha ,\tau }$, $j=1,2$, is a direct consequence of the continuous mapping theorem which in turn implies ${\tilde{\varvec{q}}}_{n,\alpha ,\tau }=({\tilde{q}}_{1,n,\alpha ,\tau },{\tilde{q}}_{2,n,\alpha ,\tau })'\xrightarrow {a.s.}{\tilde{\varvec{q}}}_{\alpha ,\tau }=({\tilde{q}}_{1,\alpha ,\tau },{\tilde{q}}_{2,\alpha ,\tau })'$, compare  Serfling (1980, 1.P, 2.b on page 52). $\square$

1.5 A.5 Proof of Theorem 13

Proof

For the proof of Theorem 13, we will use Lemma 10 together with the following Lemma 14.

Lemma 14

(Jump heights of $\tilde{S}_{n,\alpha })$ Given the general assumptions from Sect. 2.1, the ordered sample ${\tilde{R}}_{(1)}<{\tilde{R}}_{(2)}<\cdots<{\tilde{R}}_{(n-1)}<{\tilde{R}}_{(n)}$ of distances ${\tilde{R}}_i=\min (\tfrac{1-{\tilde{Y}}_{i1}}{\cos (\alpha )},\tfrac{1-{\tilde{Y}}_{i2}}{\sin (\alpha )})$ will almost surely have no ties and therefore

$$\begin{aligned} |\tilde{S}_{n,\alpha }(\tilde{S}_{n,\alpha }^{-1}(\tau ))-\tau |\le \frac{1}{n}\; a.s. \end{aligned}$$

From Lemma 10.3, we have that for any $\tilde{r}\in D_{\alpha }(\tilde{r})$ with survivor function $\tilde{S}_{\alpha }$

$$\begin{aligned} \sqrt{n}(\tilde{S}_{n,\alpha }(\tilde{r})- \tilde{S}_{\alpha }(\tilde{r}))\xrightarrow {d}{{\,\mathrm{N}\,}}(0,\tilde{S}_{\alpha }(\tilde{r})(1-\tilde{S}_{\alpha }(\tilde{r}))) \end{aligned}$$

holds. Let $\tilde{r}=\tilde{r}_{\alpha ,\tau }=\tilde{S}_{\alpha }^{-1}(\tau )$. Then, we know that

$$\begin{aligned} \sqrt{n}(\tilde{S}_{n,\alpha }(\tilde{r}_{\alpha ,\tau })- \tilde{S}_{\alpha }(\tilde{r}_{\alpha ,\tau }))\xrightarrow {d}{{\,\mathrm{N}\,}}(0, \tau (1-\tau )). \end{aligned}$$

Using the property of stochastic equicontinuity for $\tilde{S}_{n,\alpha }$ interpreted as an empirical process (for an introduction and definition of stochastic equicontinuity, see Andrews 1994), we can replace $\tilde{r}_{\alpha ,\tau }$ by a consistent estimator $\tilde{r}_{n,\alpha ,\tau }$ such that

$$\begin{aligned} \sqrt{n}(\tilde{S}_{n,\alpha }(\tilde{r}_{n,\alpha ,\tau })- \tilde{S}_{\alpha }(\tilde{r}_{n,\alpha ,\tau }))\xrightarrow {d}{{\,\mathrm{N}\,}}(0, \tau (1-\tau )) \end{aligned}$$

holds. From Lemma (ii) in Serfling (1980, Sec. 1.1.4, p. 3) it now follows that

$$\begin{aligned} \sqrt{n}(\tilde{S}_{n,\alpha }(\tilde{r}_{n,\alpha ,\tau })- \tilde{S}_{\alpha }(\tilde{r}_{n,\alpha ,\tau }))\ge \sqrt{n} (\tau - \tilde{S}_{\alpha }(\tilde{r}_{n,\alpha ,\tau })). \end{aligned}$$

Since $\tilde{f}_{\alpha }$ is continuous, the probability of observing duplicates of ${\tilde{R}}_i$ is zero. Hence, using Lemma 14

$$\begin{aligned} \sqrt{n}(\tilde{S}_{n,\alpha }(\tilde{r}_{n,\alpha ,\tau })- \tilde{S}_{\alpha }(\tilde{r}_{n,\alpha ,\tau }))= \sqrt{n}(\tau -\tilde{S}_{\alpha }(\tilde{r}_{n,\alpha ,\tau }))+\mathcal {O}_p(1/\sqrt{n}) \end{aligned}$$

holds with probability one which (using Lemma 10.2) implies

$$\begin{aligned} \sqrt{n}(\tau -\tilde{S}_{\alpha }(\tilde{r}_{n,\alpha ,\tau }))\xrightarrow {d}{{\,\mathrm{N}\,}}(\tau (1-\tau )). \end{aligned}$$

Applying the Delta-method, i.e. Taylor expansion around of $\tilde{S}_{\alpha }$, $\tilde{r}_{\alpha ,\tau }$ yields

$$\begin{aligned} \tilde{S}_{\alpha }(\tilde{r}_{n,\alpha ,\tau })\approx \tilde{S}_{\alpha }(\tilde{r}_{\alpha ,\tau })-\tilde{f}_{\alpha }(\bar{r}_{\alpha ,\tau })(\tilde{r}_{n,\alpha ,\tau }-\tilde{r}_{\alpha ,\tau }), \end{aligned}$$

for $\bar{r}_{\alpha ,\tau }$ on the line segment between $\tilde{r}_{n,\alpha ,\tau }$ and $\tilde{r}_{\alpha ,\tau }$. The last step is to apply Slutsky’s theorem and the fact that $\bar{r}_{\alpha ,\tau }\rightarrow \tilde{r}_{\alpha ,\tau }$ since $\tilde{r}_{n,\alpha ,\tau }\rightarrow \tilde{r}_{\alpha ,\tau }$, such that we obtain

$$\begin{aligned} \sqrt{n}( \tilde{r}_{n,\alpha ,\tau }- \tilde{r}_{\alpha ,\tau })=\sqrt{n}\frac{\tau -S_{\alpha }(\tilde{r}_{n,\alpha ,\tau })}{\tilde{f}_{\alpha }(\tilde{r}_{\alpha ,\tau })}\xrightarrow {d}{{\,\mathrm{N}\,}}\left( 0,\frac{\tau (1-\tau )}{(\tilde{f}_{\alpha }(\tilde{r}_{\alpha ,\tau }))^2}\right) . \end{aligned}$$

$\square$