On the Strong Consistency of the Kernel Estimator of Extreme Conditional Quantiles

Abstract

Nonparametric regression quantiles can be obtained by inverting a kernel estimator of the conditional distribution. The asymptotic properties of this estimator are well known in the case of ordinary quantiles of fixed order. The goal of this paper is to establish the strong consistency of the estimator in case of extreme conditional quantiles. In such a case, the probability of exceeding the quantile tends to zero as the sample size increases, and the extreme conditional quantile is thus located in the distribution tails.

Appendix: Proof of Auxiliary Results

Proof of Lemma 1

Clearly:

$$\displaystyle\begin{array}{rcl} \left \vert \frac{\bar{F}_{n}(y\vert x)} {\bar{F}(y\vert x)} - 1\right \vert \leq \left \vert \frac{\bar{F}_{n}(y\vert x)} {\bar{F}(y\vert x)} - \frac{\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )} {\bar{F}(y\vert x)\mathrm{I\!E}\left (\hat{g}_{n}(x)\right )}\right \vert + \left \vert \frac{\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )} {\bar{F}(y\vert x)\mathrm{I\!E}\left (\hat{g}_{n}(x)\right )} - 1\right \vert.& &{}\end{array}$$

(29)

We have,

$$\displaystyle\begin{array}{rcl} \mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )& =& \frac{1} {n}\sum _{i=1}^{n}\mathrm{I\!E}(K_{ h}(x - X_{i})\mathrm{i}_{Y _{i}>y}) = \mathrm{I\!E}(K_{h}(x - X_{1})\mathrm{i}_{Y _{1}>y}) {}\\ & =& \mathrm{I\!E}(K_{h}(x - X_{1})\mathrm{I\!P}(Y _{1}> y\vert X_{1})) =\int K_{h}(x - z)\mathrm{I\!P}(Y _{1}> y\vert X_{1} = z)g(z)dz {}\\ & =& \int K_{h}(x - z)\bar{F}(y\vert z)g(z)dz, {}\\ \end{array}$$

and $\mathrm{I\!E}\left (\hat{g}_{n}(x)\right ) = \frac{1} {n}\sum _{i=1}^{n}\mathrm{I\!E}(K_{ h}(x - X_{i})) = \mathrm{I\!E}(K_{h}(x - X_{1}))$. Consequently,

$$\displaystyle\begin{array}{rcl} \frac{\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )} {\bar{F}(y\vert x)\mathrm{I\!E}\left (\hat{g}_{n}(x)\right )} - 1& =& \frac{1} {\mathrm{I\!E}(K_{h}(x - X_{1}))}\left (\int K_{h}(x - z)\left [\frac{\bar{F}(y\vert z)} {\bar{F}(y\vert x)} - 1\right ]g(z)dz\right ) {}\\ & =& \frac{1} {\mathrm{I\!E}(K_{h}(x - X_{1}))}\left (\int K_{h}(u)\left [\frac{\bar{F}(y\vert x - u)} {\bar{F}(y\vert x)} - 1\right ]g(x - u)du\right ).{}\\ \end{array}$$

We conclude, since the kernel K is compactly supported, that for some R > 0,

$$\displaystyle{ \left \vert \frac{\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )} {\bar{F}(y\vert x)\mathrm{I\!E}\left (\hat{g}_{n}(x)\right )} - 1\right \vert \leq \sup _{\{x',d(x,x')\leq hR\}}\left \vert \frac{\bar{F}(y\vert x')} {\bar{F}(y\vert x)} - 1\right \vert = A(y,y,x,h). }$$

(30)

Now,

$$\displaystyle\begin{array}{rcl} & & \left \vert \frac{\bar{F}_{n}(y\vert x)} {\bar{F}(y\vert x)} - \frac{\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )} {\bar{F}(y\vert x)\mathrm{I\!E}\left (\hat{g}_{n}(x)\right )}\right \vert {}\\ & &\quad \leq \frac{\left \vert \hat{\psi }_{n}(y,x) -\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )\right \vert } {\bar{F}(y\vert x)\hat{g}_{n}(x)} + \frac{\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )\left \vert \hat{g}_{n}(x) -\mathrm{I\!E}\hat{g}_{n}(x)\right \vert } {\bar{F}(y\vert x)\hat{g}_{n}(x)\mathrm{I\!E}\hat{g}_{n}(x)} {}\\ & & \quad \leq \frac{\left \vert \hat{\psi }_{n}(y,x) -\mathrm{I\!E}\left (\hat{\psi }_{n}(y,x)\right )\right \vert } {\bar{F}(y\vert x)\hat{g}_{n}(x)} + (1 + A(y,y,x,h))\frac{\mathrm{I\!E}\left \vert \hat{g}_{n}(x) -\mathrm{I\!E}\hat{g}_{n}(x)\right \vert } {\hat{g}_{n}(x)}, {}\\ \end{array}$$

by (30). The last bound together with (30) and (29) prove Lemma 1. ■