Regularization of Positive Signal Nonparametric Filtering in Multiplicative Observation Model

Abstract

A solution to the problem of useful random signal extraction from a mixture with a noise in the multiplicative observation model is proposed. Unlike conventional filtering tasks, in the problem under consideration it is supposed that the distribution (and the model) of the useful signal is unknown. Therefore, in this case one cannot apply such well-known techniques like Kalman filter or posterior Stratonovich-Kushner evolution equation. The new paper is a continuation and development of the author’s article, reported at the First ISNPS Conference (Halkidiki’2012), where the filtering problem of positive signal with the unknown distribution had been solved using the generalized filtering equation and nonparametric kernel techniques. In the present study, new findings are added concerning the construction of stable procedures for filtering, the search for optimal smoothing parameter in the multidimensional case and some of the convergence results of the proposed techniques. The main feature of the problem is the positive distribution support. In this case, the classical methods of nonparametric estimation with symmetric kernels are not applicable because of large estimator bias at the support boundary. To overcome this drawback, we use asymmetric gamma kernel functions. To have stable estimators, we propose a regularization procedure with a data-driven optimal regularization parameter. Similar filtering algorithms can be used, for instance, in the problems of volatility estimation in statistical models of financial and actuarial mathematics.

Appendix

Proof of Lemma 2.1.3. We give here only outline of the proof. Provided the condition (1), from formula (13) it follows that for a sequence of statistically dependent random variables the variance of their sum is expressed through the covariance

$$\begin{aligned} C= & {} 2N^{-1} \sum \limits _{i=1}^{N-1} \left( 1-\frac{1}{N} \right) Cov\left( \tilde{K}(\mathbf x,\mathbf X_1), \tilde{K}(\mathbf x,\mathbf X_{1+i}) \right) . \end{aligned}$$

(41)

This covariance is estimated from above by using the Davydov’s inequality [14]

$$\begin{aligned} \quad \quad |Cov\left( \tilde{K}(\mathbf x,\mathbf X_1),\tilde{K}(\mathbf x,\mathbf X_{1+i})\right) |\le 2\pi \alpha (i)^{1/r}\parallel \tilde{K}(\mathbf x,\mathbf X_1) \parallel _q \parallel \tilde{K}(\mathbf x,\mathbf X_{1+i}) \parallel _p, \end{aligned}$$

where $p^{-1}+q^{-1}+r^{-1}=1$, $\alpha (i)$ is a strong mixing coefficient of the process $(X_n)$, and $\parallel \cdot \parallel _q$ is a norm in the space $L_q$. This norm can be estimated by the following expression

$$\begin{aligned} \parallel \tilde{K}(\mathbf x,\mathbf X_1) \parallel _q= & {} \left( \int \left( \prod \limits _{j=1}^dK_{{\rho _1(x_j),b_j}}\left( t_{j}\right) \right) ^q f(t_1^d)dt_1\ldots d t_{d}\right) ^{1/q} \nonumber \\= & {} \left( \mathsf E\left( \prod \limits _{j=1}^dK^{q-1}_{{\rho _1(x_j),b_j}}\left( \xi _{j}\right) f(\xi _1^d)\right) \right) ^{1/q}, \end{aligned}$$

(42)

where a kernel $\prod _{j=1}^dK_{{\rho _1(x_j),b_j}}\left( \xi _{j}\right)$ is used as density function and random variables $\xi _{j}$ is distributed like $Gamma(\rho _1(x_j),b_j)$-distribution with expectation $\mu _j=x_j$ and variance $\sigma _{\xi }^2 = x_jb_j$. Expectation in parentheses of (42) is calculated by expanding the function $f(\xi _1^d)$ in a Taylor series at the point $\mathbf {\mu } = \mu _1^d$. After some algebra we get

$$\begin{aligned} |C|=|C(\hat{f}(\mathbf {x})|\leqslant \frac{D(\upsilon ,\mathbf x)}{N}b^{-d\frac{1+\upsilon }{2}} \int _{1}^{\infty }\alpha (\tau )^{\upsilon } d\tau ,\quad 0<\upsilon<1, \quad D(\upsilon ,\mathbf x)<\infty . \end{aligned}$$

Provided the condition (2), we use the technique of the proof from [15]. To do this, we divide (41) into two terms $|C|=(2/N)\sum _{i=1}^{N-1}(\cdot )=(2/N)\big (\sum _{i=1}^{c(N)}(\cdot )+\sum _{i=c(N)+1}^{N-1}(\cdot )\big )=I_1+I_2$ and estimate each at $N\rightarrow \infty .$ As a result we get

$$\begin{aligned} I_1= O\left( \frac{c(N)b^{d/2}}{nb^{d/2}}\right) ,\quad I_2= O\left( \frac{1}{nb^{d/2}c(N)b^{d\upsilon /2}}\right) . \end{aligned}$$

It remains to find c(N) satisfying conditions $c(N)b^{d/2}\rightarrow 0$ and $c(N)b^{d\upsilon /2}\rightarrow \infty .$ Let $c(N)=b^{-\varepsilon }$. Then, these conditions are met simultaneously if $\frac{d}{2}\!>\!\varepsilon \!> \!\frac{d}{2}\upsilon >0, \quad 0< \upsilon < 1,$ and $|C|=o\left( Nb^{d/2}\right) ^{-1}$. $\square$

Proof of Lemma 2.2.3. Proof of this Lemma is carried out, in principle, in the same way as Lemma 2.1.3, but it occupies a lot of space because of a complex estimator expression containing a special functions (see [10]). $\square$