Bayesian Quantile Estimation in Deconvolution

Abstract

Estimating quantiles of a population is a fundamental problem of high practical relevance in nonparametric statistics. This chapter addresses the problem of quantile estimation in deconvolution models with known error distributions taking a Bayesian approach. We develop the analysis for error distributions with characteristic functions decaying polynomially fast, the so-called ordinary smooth error distributions that lead to mildly ill-posed inverse problems. Using Fourier inversion techniques, we derive an inequality relating the sup-norm distance between mixture densities to the Kolmogorov distance between the corresponding mixing cumulative distribution functions. Exploiting this smoothing inequality, we show that a careful choice of the prior law acting as an efficient approximation scheme for the sampling density leads to adaptive posterior contraction rates to the regularity level of the latent mixing density, thus yielding a new adaptive quantile estimation procedure.

Appendix

The following lemma provides sufficient conditions on the true cumulative distribution function $F_{0Y}$ and the prior law $\varPi _n$ so that the posterior measure concentrates on Kolmogorov neighborhoods of $F_{0Y}$. It is a modification of Lemma 1 in [17], pp. 123–125, and of Lemma B.1 in [15], pp. 24–25, with a weaker condition on the prior concentration rate. In fact, Kullback-Leibler type neighborhoods, which involve also the second moment of the log-ratio $\log (f_{0Y}/f_Y)$, can be replaced by Kullback-Leibler neighborhoods.

Lemma 1

Let $F_{0Y}$ be a continuous cumulative distribution function. Let $\varPi _n$ be a prior law on a set $\mathscr {P}_1\subseteq \mathscr {P}_0$ of probability measures with continuous cumulative distribution functions. If, for a constant $C>0$ and a positive sequence $\epsilon _n\rightarrow 0$ such that $n\epsilon _n^2\rightarrow \infty$, we have

$$\begin{aligned} \varPi _n(B_{\textrm{KL}}(P_{0Y};\,\epsilon _n^2))\gtrsim \exp {(-Cn\epsilon _n^2)}, \end{aligned}$$

then, for a sequence $M_n:=\xi (1-\theta )^{-1}(C/2+L_n)^{1/2}$, with $\theta \in (0,\,1)$, $\xi >1$ and $L_n\rightarrow \infty$ such that $L_n^{1/2}\epsilon _n\rightarrow 0$, we have

$$\begin{aligned} \varPi _n(\Vert F_Y-F_{0Y}\Vert _\infty >M_n\epsilon _n\mid Y^{(n)})=o_{\textbf{P}}(1). \end{aligned}$$

(16)

Proof

By Lemma 6.26 of [9], p. 145, with $P_{0Y}^n$-probability at least equal to $(1-L_n^{-1})$, we have

$$\begin{aligned} \int \limits _{\mathscr {P}_1}\prod _{i=1}^n \frac{f_Y}{f_{0Y}}(Y_i)\,\varPi _n(\textrm{d}\mu _Y)\gtrsim \exp {(-(C+2L_n)n\epsilon _n^2)}. \end{aligned}$$

(17)

Following the proofs of Lemma 1 in [17], pp. 123–125, Lemma B.1 in [15], pp. 24–25, and applying the lower bound in (17), the convergence statement in (16) holds true.    $\square$

Remark 1

Lemma 1 shows that, by taking $L_n$ to be a slowly varying sequence, Kullback-Leibler type neighborhoods can be replaced by Kullback-Leibler neighborhoods at the cost of an additional factor in the rate not affecting the power of n, which is thus of the order $L_n^{1/2}\epsilon _n$.

The next lemma assesses the order of the sup-norm of the bias of a cumulative distribution function with density in a Sobolev-type space. It is the sup-norm version of Lemma C.2 in [15], which, instead, considers the $L^1$-norm.

Lemma 2

Let $F_{0X}$ be the cumulative distribution function of a probability measure $\mu _{0X}\in \mathscr {P}_0$ with density $f_{0X}$. Suppose that there exists $\alpha >0$ such that $\int _{\mathbb R}|t|^\alpha |\hat{f}_{0X}(t)|\,\textrm{d}t<\infty$. Let $K\in L^1(\mathbb R)$ be symmetric, with $\hat{K}\in L^1(\mathbb R)$ such that $\hat{K}\equiv 1$ on $[-1,\,1]$. Then, for every $b>0$,

$$\begin{aligned} \Vert F_{0X}*K_b-F_{0X}\Vert _\infty =O(b^{\alpha +1}). \end{aligned}$$

Proof

Let $b_{F_{0X}}:=(F_{0X}*K_b-F_{0X})$ be the bias of $F_{0X}$. By the same arguments used for the function $G_{2,b}$ in [6], pp. 251–252, we have

$$\begin{aligned}\begin{aligned} \Vert b_{F_{0X}}\Vert _\infty :=\sup _{x\in \mathbb R}|b_{F_{0X}}(x)|&=\sup _{x\in \mathbb R} \left| \frac{1}{2\pi }\int \limits _{|t|>1/b}\exp {(-\imath t x)} \frac{[1-\hat{K}(bt)]}{(-\imath t)}\hat{f}_{0X}(t)\,\textrm{d}t\right| \\&\le \frac{1}{2\pi }\int \limits _{|t|>1/b} \frac{|1-\hat{K}(bt)|}{|(-\imath t)|}|\hat{f}_{0X}(t)|\,\textrm{d}t, \end{aligned}\end{aligned}$$

where the mapping $t\mapsto [1-\hat{K}(bt)][\hat{f}_{0X}(t)\textbf{1}_{[-1,\,1]^c}(bt)/t]$ is in $L^1(\mathbb R)$ by the assumption that $(|\cdot |^\alpha \hat{f}_{0X})\in L^1(\mathbb R)$. Note that

$$\begin{aligned}\begin{aligned} \Vert b_{F_{0X}}\Vert _\infty&\le \frac{1}{2\pi }\int \limits _{|t|>1/b} \frac{|1-\hat{K}(bt)|}{|(-\imath t)|^{\alpha +1}}\underbrace{|(-\imath t)^{\alpha }\hat{f}_{0X}(t)|}_ {=|\widehat{D^{\alpha }f_{0X}}(t)|}\,\textrm{d}t\\&< \frac{ b^{\alpha +1}}{2\pi }\int \limits _{|t|>1/b} [1+|\hat{K}(bt)|]|t|^\alpha |\hat{f}_{0X}(t)|\,\textrm{d}t\lesssim b^{\alpha +1}\int \limits _{|t|>1/b} |t|^\alpha |\hat{f}_{0X}(t)|\,\textrm{d}t \lesssim b^{\alpha +1} \end{aligned} \end{aligned}$$

because $\Vert \hat{K}\Vert _\infty \le \Vert K\Vert _1<\infty$. The assertion follows.    $\square$

The following lemma establishes the order of the sup-norm of the bias of the cumulative distribution function of a Gaussian mixture, when the mixing distribution is any probability measure on the real line and the scale parameter is chosen as a multiple of the kernel bandwidth, up to a logarithmic factor. It is analogous to Lemma G.1 in [15], p. 46, which, instead, considers the $L^1$-norm. Both results rely upon the fact that a Gaussian density has exponentially decaying tails.

Lemma 3

Let $F_{X}$ be the cumulative distribution function of $\mu _X=\mu _H*\phi _\sigma$, with $\mu _{H}\in \mathscr {P}$ and $\sigma >0$. Let $K\in L^1(\mathbb R)$ be symmetric, with $\hat{K}\in L^1(\mathbb R)$ such that $\hat{K}\equiv 1$ on $[-1,\,1]$. Given $\alpha >0$ and a sufficiently small $b>0$, for $\sigma =O(2b|\log b^{\alpha +1}|^{1/2})$, we have

$$\begin{aligned} \Vert F_{X}*K_b-F_{X}\Vert _\infty =O(b^{\alpha +1}). \end{aligned}$$

Proof

Let $b_{F_X}:=(F_X*K_b-F_X)$ be the bias of $F_X$. Defined for every $b,\,\sigma >0$ the function

$$\begin{aligned}\widehat{f_{b,\sigma }}(t):=\frac{1-\hat{K}(bt)}{t}\hat{\phi }(\sigma t/\sqrt{2})\textbf{1}_{[-1,\,1]^c}(bt), \quad t\in \mathbb R,\end{aligned}$$

since $t\mapsto [\hat{\mu }_H(t)\hat{\phi }(\sigma t/\sqrt{2})]\widehat{f_{b,\sigma }}(t)$ is in $L^1(\mathbb R)$, arguing as for $G_{2,b}$ in [6], pp. 251–252, we have that

$$\begin{aligned}\begin{aligned} \Vert b_{F_X}\Vert _\infty :=\sup _{x\in \mathbb R}|b_{F_X}(x)|&=\sup _{x\in \mathbb R}\left| \frac{1}{2\pi }\int \limits _{|t|>1/b} \exp {(-\imath t x)}\frac{[1-\hat{K}(bt)]}{(-\imath t)} \hat{\mu }_H(t)\hat{\phi }(\sigma t)\,\textrm{d}t\right| \\&= \sup _{x\in \mathbb R}\left| \frac{1}{2\pi }\int \limits _{|t|>1/b} \exp {(-\imath t x)}\hat{\mu }_H(t)\hat{\phi }(\sigma t/\sqrt{2})\widehat{f_{b,\sigma }}(t)\,\textrm{d}t\right| \\&=\Vert \mu _H*\phi _{\sigma /\sqrt{2}}*f_{b,\sigma }\Vert _\infty , \end{aligned} \end{aligned}$$

where $f_{b,\sigma }(\cdot ):=(2\pi )^{-1}\int _{\mathbb R}\exp {(-\imath t \cdot )}\widehat{f_{b,\sigma }}(t)\,\textrm{d}t$ because $\widehat{f_{b,\sigma }}\in L^1(\mathbb R)$. Since $\Vert \mu _H*\phi _{\sigma /\sqrt{2}}\Vert _1=1$ and $\Vert f_{b,\sigma }\Vert _\infty \le \Vert \widehat{f_{b,\sigma }}\Vert _1<\infty$ for all $\mu _H\in \mathscr {P}$ and $\sigma >0$, by Young’s convolution inequality,

$$\begin{aligned} \Vert b_{F_X}\Vert _\infty =\Vert \mu _H*\phi _{\sigma /\sqrt{2}}*f_{b,\sigma }\Vert _\infty \le \Vert \mu _H*\phi _{\sigma /\sqrt{2}}\Vert _1\times \Vert f_{b,\sigma }\Vert _\infty =\Vert f_{b,\sigma }\Vert _\infty , \end{aligned}$$

where

$$\begin{aligned} \begin{aligned}\Vert f_{b,\sigma }\Vert _\infty \le \Vert \widehat{f_{b,\sigma }}\Vert _1&\le \int \limits _{|t|>1/b} \frac{1+|\hat{K}(bt)|}{|t|}\hat{\phi }(\sigma t/\sqrt{2})\,\textrm{d}t\\&\lesssim b \int \limits _{|t|>1/b} \hat{\phi }(\sigma t/\sqrt{2})\,\textrm{d}t \lesssim (b/\sigma )^2\hat{\phi }(\sigma /(\sqrt{2}b))\lesssim b^{\alpha +1} \end{aligned} \end{aligned}$$

because $\Vert \hat{K}\Vert _\infty \le \Vert K\Vert _1<\infty$, the upper tail of a Gaussian distribution is bounded above by

$$\begin{aligned}\int \limits _{1/b}^\infty \hat{\phi }(\sigma t)\,\textrm{d}t\lesssim \frac{b}{\sigma ^2}\hat{\phi }(\sigma /b)\end{aligned}$$

and $(\sigma /b)^2=O(\log (1/b^{\alpha +1}))$ by assumption. The assertion follows.    $\square$