Deterministic sampling based on Kullback–Leibler divergence and its applications

Abstract

This paper introduces a new way to extract a set of representative points from a continuous distribution, which focuses on a method where the selection of points is essentially deterministic, with an emphasis on achieving accurate approximation when the size of points is small. These points are generated by minimizing the Kullback–Leibler divergence, which is an information-based measure of the disparity between two probability distributions. We refer to these points as Kullback–Leibler points. Based on the link between the total variation and the Kullback–Leibler divergence, we prove that the empirical distribution of Kullback–Leibler points converges to the target distribution. Additionally, we illustrate that Kullback–Leibler points have advantages in simulations when compared with representative points generated by Monte Carlo or other representative points methods. In addition, to prevent the frequent evaluation of complex functions, a sequential version of Kullback–Leibler points is proposed, which adaptively updates the representative points by learning about the complex or unknown functions sequentially. Two potential applications of Kullback-Leibler points in simulation of complex probability densities and optimization of complex response surfaces are discussed and demonstrated with examples.

Appendix A: Proofs

This appendix material provides proofs of Theorems 1–3. Define $f_n^*$ as the kernel density estimator on $\{\textbf{x}_j^*\}_{j=1}^{n}$, where $\{\textbf{x}_j^*\}_{j=1}^{n} \overset{i.i.d.}{\sim }\ f(\textbf{x})$. The proof of Theorem 1 relies on Lemma 1 below, which indicates that the expectation of $D(f_n^*\Vert f)$ converges to 0 as $n \rightarrow \infty$.

Lemma 1

Suppose probability density function f satisfies conditions (A1)–(A2) and that $\{\textbf{x}_j^*\}_{j=1}^{n} \overset{i.i.d.}{\sim }\ f$. $f_n^*$ is the kernel density estimator on $\{\textbf{x}_j^*\}_{j=1}^{n}$, and the kernel function K and the bandwidth $h_n$ satisfy (K1)–(K5). Then,

$$\begin{aligned} \lim _{n\rightarrow \infty } E[D(f_n^*\Vert f)]=0. \end{aligned}$$

Proof

For simplicity, we take $d=1$ for example, the proof for $d>1$ is similar. Let $\{x_j^*\}_{j=1}^{n} \overset{i.i.d.}{\sim }\ f$, $f_n^*(x)=\frac{1}{nh_{n}}\sum _{i=1}^n K(\frac{x-x_i^*}{h_{n}})$ be the kernel density estimator based on $\{x_j^*\}_{j=1}^{n}$, and kernel function K and the bandwidth $h_n$ satisfy (K1)–(K5).

Since f satisfies (A1)-(A2), then f is continuous over $\mathcal {X}$. And note that f is a density function then we know that there exists constant $f_{\max }$ such $f\le f_{\max } <\infty$ holds. The variance of $f_n^*(x)$ satisfies:

$$\begin{aligned} { \text{ var } }(f_n^{*}(x))= & {} { \text{ var } }\left( \frac{1}{nh_{n}}\sum _{i=1}^n K\left( \frac{x-x_i^*}{h_{n}}\right) \right) \nonumber \\\le & {} \frac{1}{nh_{n}^2}E\left[ K^{2}\left( \frac{x-x_1^*}{h_n}\right) \right] \nonumber \\= & {} \frac{1}{nh_{n}^2}\int _{\mathcal {X}}K^{2}\left( \frac{z-x}{h_n}\right) f(z)dz\nonumber \\= & {} \frac{1}{nh_{n}}\int _{\mathcal {X}}K^{2}(t)f(th_{n}+x)dt\nonumber \\\le & {} \frac{f_{\max }}{nh_{n}}\int _{\mathcal {X}}K^2(t)dt=\frac{C_1}{nh_{n}}, \end{aligned}$$

(A1)

where $C_1=f_{\max } \Vert K\Vert _{2}^{2}$, $\Vert K\Vert _{2}^{2}=\int _{\mathcal {X}}K^{2}(t)dt$.

The bias of $f_n^*(x)$ have the following results.

$$\begin{aligned} E[f_n^{*}(x)]-f(x)= & {} \frac{1}{nh_{n}}\sum _{i=1}^{n}E\left[ K\left( \frac{x-x_i^*}{h_{n}}\right) \right] -f(x)\nonumber \\= & {} \frac{1}{h_{n}}\int _{\mathcal {X}}K\left( \frac{z-x}{h_{n}}\right) f(z)dz-f(x)\nonumber \\= & {} \int _{\mathcal {X}}K(t)f(th_{n}+x)dt-f(x)\nonumber \\= & {} \int _{\mathcal {X}}K(t)[f(th_{n}+x)-f(x)]dt. \end{aligned}$$

(A2)

In term of f satisfies $\forall$ $x, y \in \mathcal {X}$, $|f(x)-f(y)|\le L|x- y|$. Hence, we obtain

$$\begin{aligned} |E[f_n^{*}(x)]-f(x)|\le & {} \int _{\mathcal {X}}K(t)|Lth_{n}|dt\nonumber \\= & {} Lh_{n}\int _{\mathcal {X}}|t|K(t) dt\nonumber \\\le & {} Lh_{n}\left\{ \int _{\mathcal {X}}t^2K(t)dt\right\} ^{1/2}=C_2h_{n}, \end{aligned}$$

(A3)

where $C_2=L\left\{ \int _{\mathcal {X}}t^2K(t)dt\right\} ^{1/2}$.

The Kullback–Leibler divergence between f and $f_n^*$ is

$$\begin{aligned} D(f_n^*\Vert f)=\int _{\mathcal {X}} f_n^*(x)\log \frac{f_n^*(x)}{f(x)}dx. \end{aligned}$$

By the inequality $\log x\le (x-1)$, the expectation of $D(f_n^*\Vert f)$ satisfies:

$$\begin{aligned} E[D(f_n^*\Vert f)]= & {} E\left[ \int _{\mathcal {X}} f_n^*(x)\log \frac{f_n^*(x)}{f(x)}dx\right] \nonumber \\\le & {} E\left[ \int _{\mathcal {X}}f_n^*(x)\left( \frac{f_n^*(x)}{f(x)}-1\right) dx\right] \nonumber \\= & {} E\left[ \int _{\mathcal {X}}f_n^*(x)\frac{f_n^*(x)}{f(x)}dx\right] -1\nonumber \\= & {} \int _{\mathcal {X}}E\left[ f_n^*(x)\frac{f_n^*(x)}{f(x)}\right] dx-1\nonumber \\= & {} \int _{\mathcal {X}}\frac{E[f_n^*(x)]^2-f^{2}(x)}{f(x)}dx \nonumber \\\le & {} \int _{\mathcal {X}}\frac{|E[f_n^*(x)]^2-f^{2}(x)|}{f(x)}dx. \end{aligned}$$

(A4)

The penultimate “=” sign holds following from Fubini theorem.

According to (A1) and (A3), we can obtain

$$\begin{aligned} |E[f_n^*(x)]^2-f^{2}(x)|{} & {} =\left|E[f_n^*(x)-f(x)+f(x)]^2-f^{2}(x)\right|\nonumber \\{} & {} =\left|E[f_n^*(x)-f(x)]^2+2f(x)E[f_n^*(x)-f(x)] \right|\nonumber \\{} & {} \le E[f_n^*(x)-f(x)]^2+2f(x)\left|E[f_n^*(x)]-f(x)\right|\nonumber \\{} & {} = E\left\{ f_n^*(x)-E[f_n^*(x)]+E(f_n^*(x))- f(x)\right\} ^2\nonumber \\{} & {} \quad +2f(x)\left|E[f_n^*(x)]-f(x)\right|\nonumber \\{} & {} = { \text{ var } }(f_n^{*}(x)) +\left|E[f_n^{*}(x)]-f(x)\right|^2+2f(x)\left|E[f_n^*(x)]-f(x)\right|\nonumber \\{} & {} \le { \text{ var } }(f_n^{*}(x))+C_{2}^{2}h_{n}^{2}+2f(x)\left|E[f_n^*(x)]-f(x)\right|\nonumber \\{} & {} \le \frac{C_{1}}{nh_n}+C_{2}^{2}h_{n}^{2}+2C_{2}h_{n}f(x) \overset{\Delta }{=}G_n(x). \end{aligned}$$

(A5)

Obviously, $\frac{G_n(x)}{f(x)}$ is monotonically decreasing in n, and $\lim _{n\rightarrow \infty }\frac{G_n(x)}{f(x)}=0$. So, $\forall n$, $\frac{G_n(x)}{f(x)} \le \frac{G_1(x)}{f(x)}$. Due to $\int _{\mathcal {X}}\frac{G_1(x)}{f(x)}dx < \infty$, then, by the Lebesgue’s convergence theorem (Billingsley 2008), we obtain

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _{\mathcal {X}}\frac{G_n(x)}{f(x)}dx=\int _{\mathcal {X}}\lim _{n\rightarrow \infty }\frac{G_n(x)}{f(x)}dx=0. \end{aligned}$$

(A6)

Note that

$$\begin{aligned} \int _{\mathcal {X}}\frac{|E[f_n^*(x)^2]-f^{2}(x)|}{f(x)}dx \le \int _{\mathcal {X}}\frac{G_n(x)}{f(x)}dx \end{aligned}$$

and we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \int _{\mathcal {X}}\frac{|E[f_n^*(x)^2]-f^{2}(x) |}{f(x)}dx=0. \end{aligned}$$

(A7)

Then the conclusion $\lim _{n\rightarrow \infty } E[D(f_n^*\Vert f)]=0$ is established.

To prove Theorems 1 and 2, we also need the following lemma. $\square$

Lemma 2

Let f and g be two density functions supported on $\mathcal {X}\subseteq R^d$, then

• (a)
  $$\begin{aligned} V(f, g)\overset{\textrm{def}}{=}\sup _{A\in \mathcal {B}}\left|\int _{A} (f(\textbf{x})-g(\textbf{x}))d\textbf{x}\right|=\frac{1}{2}\int _{\mathcal {X}}|f(\textbf{x})-g(\textbf{x})|d\textbf{x}, \end{aligned}$$

  where $\mathcal {B}$ is the Borel $\sigma$-algebra of $\mathcal {X}$.

• (b)
  $$\begin{aligned} 2V^{2}(f, g) \le D(g \Vert f). \end{aligned}$$

This Lemma can be obtained by Scheffé’s theorem (refer to Tsybakov 2009 p.84) and Pinsker’s inequality [refer to Tsybakov (2009, p. 88)], respectively.

Proof of Theorem 1

Define the sequence of random variables $\{\textbf{x}_j^*\}_{j=1}^{\infty } \overset{i.i.d.}{\sim }\ f$, and let $f_n^*$ denote the kernel density estimator on $\{\textbf{x}_j^*\}_{j=1}^{n}$. According the Lemma 1, $\lim _{n\rightarrow \infty } E[D(f_n^*\Vert f)]=0$.

Consider now the kernel density estimator $f_{n}^{KL}$ on the KL points $\{\varvec{\xi }_{i}\}_{i=1}^n$. By the definition of KL points,

$$\begin{aligned} D(f_{n}^{KL}\Vert f) \le E[D(f_n^*\Vert f)], \end{aligned}$$

so $\lim _{n\rightarrow \infty }D(f_{n}^{KL}\Vert f)=0$.

Using Pinsker’s inequality in Lemma 2 (b),

$$\begin{aligned} \frac{1}{2}\left( \int _{\mathcal {X}} |f_{n}^{KL}(\textbf{x})-f(\textbf{x})|d\textbf{x}\right) ^2\le D(f_{n}^{KL}\Vert f), \end{aligned}$$

then conclusion $\lim _{n\rightarrow \infty } \int _{\mathcal {X}} |f_{n}^{KL} (\textbf{x})-f(\textbf{x})|d\textbf{x}=0$ follows.

Proof of Theorem 2

Due to $f_{n}^{KL}$ is the kernel density estimator on KL points $\{\varvec{\xi }_{i}\}_{i=1}^n$ and satisfies (K1)–(K5), then

$$\begin{aligned} \sup _{A \in \mathcal {B}} \left|\int _{A} [f_{n}^{KL}(\textbf{x})-f(x)]d\textbf{x}\right|=\frac{1}{2} \int _{\mathcal {X}} |f_{n}^{KL}(\textbf{x})-f(\textbf{x})|d\textbf{x} \end{aligned}$$

(A8)

based on Lemma 2 (a). Combing Theorem 1 and (A8), we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{A \in \mathcal {B}} \left|\int _{A} [f_{n}^{KL}(\textbf{x})-f(\textbf{x})]d\textbf{x}\right|=0. \end{aligned}$$

(A9)

Set $A=(-\infty , \textbf{x}]\in \mathcal {B}$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup _{\textbf{x}}|F_{n}(\textbf{x})-F(\textbf{x})|=0. \end{aligned}$$

where $F_n$ is the cumulative distribution function of density function $f_{n}^{KL}$, and F is the cumulative distribution function of f. $\square$

Two lemmas will be needed for the proof of Theorem 3.

Lemma 3

Suppose kernel K satisfys (K1)–(K5). Then,

• (a) $\forall$ $\epsilon >0$, there exist $M>0$, such that $\int _{[-M, M]^d}K(\textbf{t})d\textbf{t}\ge 1-\epsilon$.

• (b) For the above $M>0$, $\exists$ $\textbf{y}_0$, such that $\varphi (\textbf{x}, \textbf{y}_0)\ge 1/3$, where

  $$\begin{aligned} \varphi (\textbf{x}, \textbf{y})=\int _{\prod _{i=1}^d[x_i-y_i, x_i+y_i]}K(\textbf{t})d\textbf{t}, \end{aligned}$$

  $\textbf{x}=(x_1, \ldots , x_d)\in [-M, M]^d$ and $\textbf{y}=(y_1, \ldots , y_d)$.

Proof

(a) follows from is a density function.

(b) Note that, $\lim _{\textbf{y}\rightarrow +\infty } \varphi (\textbf{x}, \textbf{y})\ge 1/2$ for any $\textbf{x}=(x_1, \ldots , x_d)\in [-M, M]^d$, then conclusion follows. $\square$

Lemma 4

Let $\{\varvec{\xi }_{i}\}_{i=1}^n$ be the KL points of f. Then

$$\begin{aligned} \lim _{n\rightarrow \infty } \sup \limits _{\textbf{x}\in \mathcal {X}}\frac{N_{\textbf{x}}}{n}=0, \end{aligned}$$

where

$$\begin{aligned} N_{\textbf{x}}=\sum _{i=1}^{n}I\left( \frac{\textbf{x}-\varvec{\xi }_{i}}{h_n}\in [-M, M]^d\right) , \end{aligned}$$

$h_n$ is the bandwidth used to generate KL points and M is defined in Lemma 3.

Proof

For simplicity, we take $d = 1$ for example. Due to $\forall$ $\delta >0$, $\lim _{n\rightarrow \infty }\frac{\delta }{h_n}=+\infty$, so $\exists$ $n_0$, such that when $n> n_0$, $\frac{\delta }{h_{n}}\ge y_0$ holds, where $y_0$ satisfies $I(|x|\le M)\varphi (x, y_0)\ge 1/3$ defined in Lemma 3.

We use reduction to absurdity to prove this result. If Lemma 4 doesn’t hold, then $\exists$ $x^{*}$, for $\forall$ $N>n_0$, $\exists$ $n_{k}>N$, such that $\frac{N_{x^*}}{n_k}\ge c_0$, which means

$$\begin{aligned} \frac{\sum _{i=1}^{n_k}I(|\frac{x^*-\xi _i}{h_{n_k}}|\le M)}{n_k}\ge c_0, \end{aligned}$$

where $c_0$ is a positive constant.

Then, we have

$$\begin{aligned} F_{n_k}(x^*+\delta )-F_{n_k}(x^*-\delta )= & {} \frac{1}{n_k}\sum _{i=1}^{n_k}\int _{x^*-\delta }^{x^*+\delta }\frac{1}{h_{n_k}}K\left( \frac{t-\xi _i}{h_{n_k}}\right) dt\nonumber \\= & {} \frac{1}{n_k}\sum _{i=1}^{n_k}\int _{\frac{x^*-\xi _i}{h_{n_k}}-\frac{\delta }{h_{n_k}}}^{\frac{x^*-\xi _i}{h_{n_k}}+\frac{\delta }{h_{n_k}}}K(z)dz\nonumber \\\ge & {} \frac{1}{n_k}\sum _{i=1}^{n_k}I\left( \left|\frac{x^*-\xi _i}{h_{n_k}}\right|\le M\right) \int _{\frac{x^*-\xi _i}{h_{n_k}}-y_0}^{\frac{x^*-\xi _i}{h_{n_k}}+y_0}K(z)dz\nonumber \\\ge & {} \frac{\sum _{i=1}^{n_k}I(|\frac{x^{*}-\xi _i}{h_{n_k}}|\le M)}{3n_k}\nonumber \\\ge & {} \frac{c_0}{3}, \end{aligned}$$

where $F_{n_k}$ is the cumulative distribution function correspond the kernel density estimator $f_{n_k}$, which used to generated KL points. The penultimate “$\ge$" holds by Lemma 3. This contradicts to $F_{n_k}$ is a continuous distribution. We conclude the proof. $\square$

Proof of Theorem 3

Let $F_n^{^{KL}}$ denote the standard empirical distribution of KL ponits $\{\varvec{\xi }_{i}\}_{i=1}^n$, $F_{n}$ denote the cumulative distribution function of $f_{n}^{KL}$. We first prove that

$$\begin{aligned} \lim _{n \rightarrow \infty }\sup _\textbf{x}|F_n(\textbf{x})-F_n^{^{KL}}(\textbf{x})|=0. \end{aligned}$$

For simplicity, we take $d=1$, $\forall$ $x \in \mathcal {X}$,

$$\begin{aligned} |F_{n}(x)-F_n^{^{KL}}(x)|= & {} \frac{1}{n}\sum _{i=1}^{n}\left|\int _{-\infty }^{x}\frac{1}{h_{n}}K\left( \frac{t-\xi _{i}}{h_n}\right) dt-I(\xi _i \le x)\right|\\= & {} \frac{1}{n}\sum _{i=1}^{n}\left|\int _{-\infty }^{ \frac{x-\xi _{i}}{h_n}}K(z)dz-I(\xi _i \le x) \right|. \end{aligned}$$

If $\xi _{i} \le x$, then $\frac{x-\xi _{i}}{h_n} \ge 0$, and

$$\begin{aligned} \left|\int _{-\infty }^{ \frac{x-\xi _{i}}{h_n}}K(z)dz-I(\xi _i \le x)\right|= & {} \left|\int _{-\infty }^{ \frac{x-\xi _{i}}{h_n}}K(z)dz-1\right|\\= & {} \int _{\frac{x-\xi _{i}}{h_n}}^{ +\infty }K(z)dz\\\le & {} I\left( 0\le \frac{x-\xi _{i}}{h_n}\le M\right) \int _{ \frac{x-\xi _{i}}{h_n}}^ {M}K(z)dz+\frac{ \epsilon }{2}, \end{aligned}$$

where M and $\epsilon$ are defined as in Lemma 3, i.e. $\forall$ $\epsilon >0$, there exist $M>0$, such that $\int _{-M}^{M}K(t)dt\ge 1-\epsilon$, and $\int _{-\infty }^{-M}K(t)dt=\int _{M}^{+\infty }K(t)dt<\frac{\epsilon }{2}$.

If $\xi _{i} > x$, then $\frac{x-\xi _{i}}{h_n} <0$, and

$$\begin{aligned} \left|\int _{-\infty }^{ \frac{x-\xi _{i}}{h_n}}K(z)dz-I(\xi _i \le x)\right|= & {} \left|\int _{-\infty }^{ \frac{x-\xi _{i}}{h_n}}K(z)dz-0\right|\\\le & {} I\left( -M\le \frac{x-\xi _{i}}{h_n}< 0\right) \int _{-M}^{ \frac{x-\xi _{i}}{h_n}}K(z)dz+\frac{ \epsilon }{2}. \end{aligned}$$

In summary, for $\forall$ $x\in \mathcal {X}$ the absolute error between $F_{n}(x)$ and $F_n^{^{KL}}(x)$ is

$$\begin{aligned} 0\le |F_{n}(x)-F_n^{^{KL}}(x)|\le & {} \frac{1}{n}\sum _{i=1}^{n}I\left( \left|\frac{x-\xi _{i}}{h_n}\right|\le M\right) \int _{ |\frac{x-\xi _{i}}{h_n}|}^{ M}K(z)dz+\frac{ \epsilon }{2}\nonumber \\\le & {} \frac{1}{2n}\sum _{i=1}^{n}I\left( \left|\frac{x-\xi _{i}}{h_n}\right|\le M\right) +\frac{ \epsilon }{2}. \end{aligned}$$

(A10)

In term of Lemma 4,

$$\begin{aligned} \lim _{n \rightarrow \infty } \sup _x \frac{1}{n}\sum _{i=1}^{n}I\left( \left|\frac{x-\xi _{i}}{h_n}\right|\le M\right) =0. \end{aligned}$$

Hence, $\forall$ $\epsilon > 0$, there exist N, such that when $n\ge N$, for $\forall$ $x\in \mathcal {X}$, we have

$$\begin{aligned} 0\le |F_{n}(x)-F_n^{^{KL}}(x)|\le \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon . \end{aligned}$$

Consequently, we obtain $\lim _{n \rightarrow \infty }\sup _{\textbf{x}} |F_{n}(x)-F_n^{^{KL}}(x)|=0$. The proof for $d > 1$ is similar, hence

$$\begin{aligned} \lim _{n \rightarrow \infty } \sup _{\textbf{x}} |F_{n}(\textbf{x})-F_n^{^{KL}}(\textbf{x})|=0. \end{aligned}$$

(A11)

Due to

$$\begin{aligned} \sup _{\textbf{x}}|F_n^{^{KL}}(\textbf{x})-F(\textbf{x})|\le \sup _{\textbf{x}}|F_n^{^{KL}}(\textbf{x})-F_{n}(\textbf{x})|+\sup _{\textbf{x}}|F_{n}(\textbf{x})-F(\textbf{x})|, \end{aligned}$$

and combing with Theorem 2, we have

$$\begin{aligned} \lim _{n \rightarrow \infty } \sup _{\textbf{x}} |F_n^{^{KL}}(\textbf{x})-F(\textbf{x})|=0. \end{aligned}$$

$\square$