${{\mathcal{L}}}_{1}$-Optimal Filtering of Markov Jump Processes. I. Exact Solution and Numerical Implementation Schemes

Abstract

Part I of this research work is devoted to the development of a class of numerical solution algorithms for the filtering problem of Markov jump processes by indirect continuous-time observations corrupted by Wiener noises. The expected ${{\mathcal{L}}}_{1}$ norm of the estimation error is chosen as an optimality criterion. The noise intensity depends on the state being estimated. The numerical solution algorithms involve not the original continuous-time observations, but the ones discretized by time. A feature of the proposed algorithms is that they take into account the probability of several jumps in the estimated state on the time interval of discretization. The main results are the statements on the accuracy of the approximate solution of the filtering problem, depending on the number of jumps taken into account for the estimated state, on the discretization step, and on the numerical integration scheme applied. These statements provide a theoretical basis for the subsequent analysis of particular numerical schemes to implement the solution of the filtering problem.

Appendices

Appendix

Proof of Theorem 1. Using the notations Ξ_r ≜ ξ₁ξ₂…ξ_r and Θ_r ≜ θ₁θ₂…θ_r for the random matrices, together with the estimates ${\widehat{x}}_{r}$ and ${\overline{x}}_{r}(s)$, write the explicit-form expression

$${\widehat{x}}_{r}={\left({\bf{1}}{\Theta }_{r}^{\top }\pi \right)}^{-1}{\Theta }_{r}^{\top }\pi,\qquad {\overline{x}}_{r}(s)={\left({\bf{1}}{\Xi }_{r}^{\top }\pi \right)}^{-1}{\Xi }_{r}^{\top }\pi .$$

From Definition (4.2) it follows that ${\xi }_{q}^{kj}\leqslant {\theta }_{q}^{kj}$. Therefore, the matrix Θ_r − Ξ_r contains nonnegative elements only. For the sake of brevity, the dependence on r in Ξ_r and Θ_r will be omitted. The following chain of inequalities holds:

$$\begin{array}{cccc}{\rm{E}}\left\{{\left\Vert {\widehat{x}}_{r}-{\overline{x}}_{r}(s)\right\Vert }_{1}\right\}={\rm{E}}\left\{{\left\Vert \frac{1}{{\bf{1}}{\Theta }^{\top }\pi }{\Theta }^{\top }\pi -\frac{1}{{\bf{1}}{\Xi }^{\top }\pi }{\Xi }^{\top }\pi \right\Vert }_{1}\right\}\\ ={\rm{E}}\left\{\frac{1}{{\bf{1}}{\Theta }^{\top }\pi {\bf{1}}{\Xi }^{\top }\pi }{\left\Vert {\bf{1}}{\Xi }^{\top }\pi {(\Theta -\Xi )}^{\top }\pi -{\bf{1}}{(\Theta -\Xi )}^{\top }\pi {\Xi }^{\top }\pi \right\Vert }_{1}\right\}\\ \leqslant {\rm{E}}\left\{\frac{1}{{\bf{1}}{\Theta }^{\top }\pi {\bf{1}}{\Xi }^{\top }\pi }\left({\bf{1}}{\Xi }^{\top }\pi \parallel {(\Theta -\Xi )}^{\top }\pi {\parallel }_{1}+{\bf{1}}{(\Theta -\Xi )}^{\top }\pi \parallel {\Xi }^{\top }\pi {\parallel }_{1}\right)\right\}\\ =2{\rm{E}}\left\{\frac{1}{{\bf{1}}{\Theta }^{\top }\pi }{\bf{1}}{(\Theta -\Xi )}^{\top }\pi \right\}.\end{array}$$

(A.1)

Consider the auxiliary estimate

$${\breve{x}}_{r}\triangleq {\rm{E}}\left\{{X}_{{t}_{r}}{{\bf{I}}}_{{A}_{r}^{s}}(\omega )| {{\mathcal{O}}}_{r}\right\}.$$

According to the abstract form of Bayes’s rule,

$${\breve{x}}_{r}=\frac{1}{{\bf{1}}{\Theta }^{\top }\pi }{\Xi }^{\top }\pi $$

and

$${\widehat{x}}_{r}-{\breve{x}}_{r}={\rm{E}}\left\{{X}_{{t}_{r}}{{\bf{I}}}_{{\overline{a}}_{r}^{s}}(\omega )| {{\mathcal{O}}}_{r}\right\}=\frac{1}{{\bf{1}}{\Theta }^{\top }\pi }{(\Theta -\Xi )}^{\top }\pi .$$

(A.2)

From (A.1) and (A.2) it follows that, for r = 1 and for all π ∈ Π,

$$\begin{array}{cc}{\rm{E}}\left\{{\left\Vert {\widehat{x}}_{1}-{\overline{x}}_{1}(s)\right\Vert }_{1}\right\}\leqslant 2{\rm{E}}\left\{{\left\Vert {\rm{E}}\left\{{X}_{{t}_{1}}{{\bf{I}}}_{{\overline{a}}_{1}^{s}}(\omega )| {{\mathcal{O}}}_{1}\right\}\right\Vert }_{1}\right\}\\ =2{\rm{E}}\left\{\mathop{\sum }\limits_{n=1}^{N}{\rm{E}}\left\{{X}_{{t}_{1}}^{n}{{\bf{I}}}_{{\overline{a}}_{1}^{s}}(\omega )| {{\mathcal{O}}}_{1}\right\}\right\}=2{\rm{E}}\left\{{\rm{E}}\left\{{{\bf{I}}}_{{\overline{a}}_{1}^{s}}(\omega )| {{\mathcal{O}}}_{1}\right\}\right\}=2{\rm{P}}\left\{{\overline{a}}_{1}^{s}\right\}.\end{array}$$

(A.3)

The process ${N}_{t}^{X}$ representing the total number of state jumps for X_t is a counting process, and its quadratic characteristic has the form

$${\langle {N}^{X},{N}^{X}\rangle }_{t}=-\mathop{\int}\limits_{0}^{t}\mathop{\sum }\limits_{n=1}^{N}{\lambda }_{nn}{X}_{s}^{n}ds.$$

Therefore, the requisite probability can be estimated from above as

$${\rm{P}}\left\{{\overline{a}}_{1}^{s}\right\}\leqslant {e}^{-\overline{\lambda }h}\mathop{\sum }\limits_{k=s+1}^{\infty }\frac{{(\overline{\lambda }h)}^{k}}{k!}={C}_{1}\frac{{(\overline{\lambda }h)}^{s+1}}{(s+1)!}.$$

(A.4)

In combination with (A.3), this inequality implies $\sigma (s)\leqslant 2{C}_{1}\frac{{(\overline{\lambda }h)}^{s+1}}{(s+1)!},$ i.e., the local accuracy estimate (4.7) is true.

The Markov property of the pair $({X}_{t},{N}_{t}^{X})$ and (A.4) can be used for obtaining an upper bound for the probability ${\rm{P}}\left\{{\overline{a}}_{r}^{s}\right\}$ as well:

$${\rm{P}}\left\{{\overline{a}}_{r}^{s}\right\}\leqslant 1-{\left(1-{C}_{1}\frac{{(\overline{\lambda }h)}^{s+1}}{(s+1)!}\right)}^{r}.$$

This result finally yields the global accuracy estimate (4.8). The proof of Theorem 1 is complete.

Proof of Theorem 2. Well,

$${\widetilde{x}}_{1}={({\bf{1}}{\psi }_{1}^{\top }\pi )}^{-1}{\psi }_{1}^{\top }\pi,\quad {\overline{x}}_{1}={({\bf{1}}{\xi }_{1}^{\top }\pi )}^{-1}{\xi }_{1}^{\top }\pi \quad {\rm{and}}\quad {\Delta }_{1}={\widetilde{x}}_{1}-{\overline{x}}_{1}.$$

Using the properties of matrix operations, write

$$[{\gamma }^{\top }\pi {\bf{1}}-{\bf{1}}{\gamma }^{\top }\pi I]{\gamma }^{\top }\pi =0.$$

Both estimates have stability, and consequently $\parallel {\widetilde{x}}_{1}{\parallel }_{1}=\parallel {\overline{x}}_{1}{\parallel }_{1}=1$. The following chain of inequalities holds:

$$\begin{array}{cccc}\parallel {\Delta }_{1}{\parallel }_{1}=\frac{1}{{\bf{1}}{\psi }_{1}^{\top }\pi {\bf{1}}{\xi }_{1}^{\top }\pi }{\left\Vert {\bf{1}}{\xi }_{1}^{\top }\pi {\psi }_{1}^{\top }\pi -{\bf{1}}{\psi }_{1}^{\top }\pi {\xi }_{1}^{\top }\pi \right\Vert }_{1}=\frac{1}{{\bf{1}}{\psi }_{1}^{\top }\pi {\bf{1}}{\xi }_{1}^{\top }\pi }{\left\Vert {\bf{1}}{\xi }_{1}^{\top }\pi {\gamma }_{1}^{\top }\pi -{\bf{1}}{\gamma }_{1}^{\top }\pi {\xi }_{1}^{\top }\pi \right\Vert }_{1}\\ =\frac{1}{{\bf{1}}{\psi }_{1}^{\top }\pi {\bf{1}}{\xi }_{1}^{\top }\pi }{\left\Vert \left[{\gamma }_{1}^{\top }\pi {\bf{1}}-{\bf{1}}{\gamma }_{1}^{\top }\pi I\right]{\xi }_{1}^{\top }\pi \right\Vert }_{1}=\frac{1}{{\bf{1}}{\psi }_{1}^{\top }\pi {\bf{1}}{\xi }_{1}^{\top }\pi }{\left\Vert \left[{\gamma }_{1}^{\top }\pi {\bf{1}}-{\bf{1}}{\gamma }_{1}^{\top }\pi I\right]\left[{\xi }_{1}^{\top }\pi +{\gamma }_{1}^{\top }\pi \right]\right\Vert }_{1}\\ =\frac{1}{{\bf{1}}{\xi }_{1}^{\top }\pi }{\left\Vert \left[{\gamma }_{1}^{\top }\pi {\bf{1}}-{\bf{1}}{\gamma }_{1}^{\top }\pi I\right]{\widetilde{X}}_{1}\right\Vert }_{1}\leqslant \frac{1}{{\bf{1}}{\xi }_{1}^{\top }\pi }{\left\Vert \left[{\gamma }_{1}^{\top }\pi {\bf{1}}-{\bf{1}}{\gamma }_{1}^{\top }\pi I\right]\right\Vert }_{1}{\left\Vert {\widetilde{X}}_{1}\right\Vert }_{1}\\ \leqslant 2\frac{{\bf{1}}{\overline{\gamma }}_{1}^{\top }\pi }{{\bf{1}}{\xi }_{1}^{\top }\pi }=\mathop{\sum }\limits_{i=1}^{N}{\pi }_{i}\frac{\mathop{\sum }\limits_{j=1}^{N}{\overline{\gamma }}_{1}^{ij}}{\mathop{\sum }\limits_{k,\ell =1}^{N}{\xi }_{1}^{k\ell }{\pi }_{k}}.\end{array}$$

In combination with (5.7) and (A.7), this inequality leads to

$${\rm{E}}\left\{{{\bf{I}}}_{{a}_{1}^{s}}(\omega )\parallel {\Delta }_{1}{\parallel }_{1}\right\}\leqslant 2\mathop{\sum }\limits_{i=1}^{N}{\pi }_{i}\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\mathop{\sum }\limits_{i=1}^{N}{\overline{\gamma }}^{ij}(y)dy\leqslant 2\delta .$$

The first inequality in (5.8) follows from the fact that the inequality above is valid for any π ∈ Π.

Define random matrices representing functions of ξ_r and ψ_r:

$$\begin{array}{ccc}{\Xi }_{q,r}\triangleq \left\{\begin{array}{ll}{\xi }_{q}{\xi }_{q+1}\ldots {\xi }_{r}&\,\text{if}\,\,q\leqslant r\\ I&\,\text{otherwise}\,,\end{array}\right.\\ {\Psi }_{q,r}\triangleq \left\{\begin{array}{ll}{\psi }_{q}{\xi }_{q+1}\ldots {\psi }_{r}&\,\text{if}\,\,q\leqslant r\\ I&\,\text{otherwise}\,,\end{array}\right.\\ {\Gamma }_{q,r}\triangleq {\Psi }_{q,r}-{\Xi }_{q,r}.\end{array}$$

For proving Theorem 2, an auxiliary result will be needed.

Lemma 3

If ${\phi }_{r}\triangleq {\phi }_{r}({{\mathcal{Y}}}_{1},\ldots,{{\mathcal{Y}}}_{r})$ is a nonnegative ${{\mathcal{O}}}_{r}$-measurable random variable and ${\Phi }_{r}\triangleq \frac{{\phi }_{r}}{{\bf{1}}{\Xi }_{1,r}^{\top }\pi }$, then

$${\rm{E}}\left\{{{\bf{I}}}_{{A}_{r}^{s}}(\omega ){\Phi }_{r}\right\}=\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\ldots \mathop{\int}\limits_{{{\mathbb{R}}}^{M}}{\phi }_{r}({y}_{1},\ldots,{y}_{r})d{y}_{r}\ldots d{y}_{1}.$$

(A.5)

Proof of Lemma 3. Consider a nonnegative integrable function ${\phi }_{1}={\phi }_{1}(y):{{\mathbb{R}}}^{M}\to {{\mathbb{R}}}_{+}$ and an ${{\mathcal{O}}}_{1}$-measurable random variable of the form

$${\Phi }_{1}\triangleq \frac{{\phi }_{1}({{\mathcal{Y}}}_{1})}{{\bf{1}}{\xi }_{1}^{\top }({{\mathcal{Y}}}_{1})\pi }=\frac{{\phi }_{1}({{\mathcal{Y}}}_{1})}{\mathop{\sum }\limits_{i,j=1}^{N}\mathop{\sum }\limits_{m=0}^{s}\mathop{\int}\limits_{{\mathcal{D}}}{\mathcal{N}}\left({{\mathcal{Y}}}_{1},fu,\mathop{\sum }\limits_{p=1}^{N}{u}^{p}{g}_{p}\right){\rho }^{i,j,m}(du){\pi }_{i}}.$$

(A.6)

Find ${\rm{E}}\left\{{{\bf{I}}}_{{a}_{1}^{s}}(\omega ){\Phi }_{1}\right\}$:

$$\begin{array}{cc}{\rm{E}}\left\{{{\bf{I}}}_{{a}_{1}^{s}}(\omega ){\Phi }_{1}\right\}=\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\mathop{\int}\limits_{{\mathcal{D}}}\frac{{\phi }_{1}(y)\mathop{\sum }\limits_{k,\ell =1}^{N}\mathop{\sum }\limits_{n=0}^{s}{\mathcal{N}}\left(y,fv,\mathop{\sum }\limits_{q=1}^{N}{v}^{q}{g}_{q}\right){\rho }^{k,\ell,n}(dv){\pi }_{k}}{\mathop{\sum }\limits_{i,j=1}^{N}\mathop{\sum }\limits_{m=0}^{s}\mathop{\int}\limits_{{\mathcal{D}}}{\mathcal{N}}\left(y,fu,\mathop{\sum }\limits_{p=1}^{N}{u}^{p}{g}_{p}\right){\rho }^{i,j,m}(du){\pi }_{i}}dy\\ =\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}{\phi }_{1}(y)\frac{\mathop{\sum }\limits_{k,\ell =1}^{N}\mathop{\sum }\limits_{n=0}^{s}\mathop{\int}\limits_{{\mathcal{D}}}{\mathcal{N}}\left(y,fv,\mathop{\sum }\limits_{q=1}^{N}{v}^{q}{g}_{q}\right){\rho }^{k,\ell,n}(dv){\pi }_{k}}{\mathop{\sum }\limits_{i,j=1}^{N}\mathop{\sum }\limits_{m=0}^{s}\mathop{\int}\limits_{{\mathcal{D}}}{\mathcal{N}}\left(y,fu,\mathop{\sum }\limits_{p=1}^{N}{u}^{p}{g}_{p}\right){\rho }^{i,j,m}(du){\pi }_{i}}dy=\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}{\phi }_{1}(y)dy.\end{array}$$

(A.7)

Consider a nonnegative integrable function ${\phi }_{2}={\phi }_{1}({y}_{1},{y}_{2}):\,{{\mathbb{R}}}^{2M}\to {{\mathbb{R}}}_{+}$ and an ${{\mathcal{O}}}_{2}$-measurable random variable of the form

$$\begin{array}{cc}{\Phi }_{2}\triangleq \frac{{\phi }_{1}({{\mathcal{Y}}}_{1},{{\mathcal{Y}}}_{2})}{{\bf{1}}{\Xi }_{1,2}^{\top }({{\mathcal{Y}}}_{1},{{\mathcal{Y}}}_{2})\pi }\\ =\frac{{\phi }_{2}({{\mathcal{Y}}}_{1},{{\mathcal{Y}}}_{2})}{\mathop{\sum }\limits_{i,{i}_{2},j=1}^{N}\mathop{\sum }\limits_{{m}_{1},{m}_{2}=0}^{s}\mathop{\int}\limits_{{\mathcal{D}}}\mathop{\int}\limits_{{\mathcal{D}}}{\mathcal{N}}\left({{\mathcal{Y}}}_{1},f{u}_{1},\mathop{\sum }\limits_{{p}_{1}=1}^{N}{u}^{{p}_{1}}{g}_{{p}_{1}}\right){\mathcal{N}}\left({{\mathcal{Y}}}_{2},f{u}_{2},\mathop{\sum }\limits_{{p}_{2}=1}^{N}{u}^{{p}_{2}}{g}_{{p}_{2}}\right){\rho }^{i,{i}_{2},{m}_{1}}(d{u}_{1}){\rho }^{{i}_{2},j,{m}_{2}}(d{u}_{2}){\pi }_{i}}.\end{array}$$

Find ${\rm{E}}\left\{{{\bf{I}}}_{{A}_{2}^{s}}(\omega ){\Phi }_{2}\right\}$:

$$\begin{array}{ccc}{\rm{E}}\left\{{{\bf{I}}}_{{A}_{2}^{s}}(\omega ){\Phi }_{2}\right\}=\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}{\phi }_{2}({y}_{1},{y}_{2})\\ \times \frac{\mathop{\sum }\limits_{k,{k}_{2},\ell =1}^{N}\mathop{\sum }\limits_{{n}_{1},{n}_{2}=0}^{s}\mathop{\int}\limits_{{\mathcal{D}}}\mathop{\int}\limits_{{\mathcal{D}}}{\mathcal{N}}\left({y}_{1},f{v}_{1},\mathop{\sum }\limits_{{q}_{1}=1}^{N}{v}^{{q}_{1}}{g}_{{q}_{1}}\right){\mathcal{N}}\left({y}_{2},f{v}_{2},\mathop{\sum }\limits_{{q}_{2}=1}^{N}{v}^{{q}_{2}}{g}_{{q}_{2}}\right){\rho }^{k,{k}_{2},{n}_{1}}(d{v}_{1}){\rho }^{{k}_{2},\ell,{n}_{2}}(d{v}_{2}){\pi }_{k}}{\mathop{\sum }\limits_{i,{i}_{2},j=1}^{N}\mathop{\sum }\limits_{{m}_{1},{m}_{2}=0}^{s}\mathop{\int}\limits_{{\mathcal{D}}}\mathop{\int}\limits_{{\mathcal{D}}}{\mathcal{N}}\left({y}_{1},f{u}_{1},\mathop{\sum }\limits_{{p}_{1}=1}^{N}{u}^{{p}_{1}}{g}_{{p}_{1}}\right){\mathcal{N}}\left({y}_{2},f{u}_{2},\mathop{\sum }\limits_{{p}_{2}=1}^{N}{u}^{{p}_{2}}{g}_{{p}_{2}}\right){\rho }^{i,{i}_{2},{m}_{1}}(d{u}_{1}){\rho }^{{i}_{2},j,{m}_{2}}(d{u}_{2}){\pi }_{i}}\\ \times d{y}_{2}d{y}_{1}=\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}{\phi }_{2}({y}_{1},{y}_{2})d{y}_{2}d{y}_{1}.\end{array}$$

The general case ${\rm{E}}\left\{{{\bf{I}}}_{{A}_{r}^{s}}(\omega ){\Phi }_{r}\right\}$ is considered by analogy. The proof of Lemma 3 is complete.

Now estimate from above the norm of the error ${\Delta }_{r}={\widetilde{x}}_{r}-{\overline{x}}_{r}$. From the definition of the matrices Ξ, Ψ, and Γ it follows that

$${\Gamma }_{1,r}\triangleq {\Psi }_{1,r}-{\Xi }_{1,r}=\mathop{\sum }\limits_{t=1}^{r}{\Psi }_{1,t-1}{\gamma }_{t}{\Psi }_{t+1,r}.$$

(A.8)

Performing transformations similar to those for Δ₁, write

$$\parallel {\Delta }_{r}{\parallel }_{1}\leqslant \frac{1}{{\bf{1}}{\Xi }_{1,r}^{\top }\pi }{\left\Vert \left[{\Gamma }_{1,r}^{\top }\pi {\bf{1}}-{\bf{1}}{\Gamma }_{1,r}^{\top }\pi I\right]\right\Vert }_{1}\leqslant 2\mathop{\sum }\limits_{t=1}^{r}\frac{1}{{\bf{1}}{\Xi }_{1,r}^{\top }\pi }{\bf{1}}{\Psi }_{t+1,r}^{\top }{\overline{\gamma }}_{t}^{\top }{\Psi }_{1,t-1}^{\top }\pi .$$

(A.9)

For estimating the contribution of each term in (A.9), take advantage of (A.5). For the sake of simplicity, consider the case r = 3, a function $\phi ({y}_{1},{y}_{2},{y}_{3}):{{\mathbb{R}}}^{3M}\to {{\mathbb{R}}}_{+}$ of the form

$$\phi ({y}_{1},{y}_{2},{y}_{3})={\bf{1}}{\psi }^{\top }({y}_{3}){\overline{\gamma }}^{\top }({y}_{2}){\psi }^{\top }({y}_{1})\pi,$$

and an O₃-measurable random variable of the form

$$\Phi \triangleq \frac{\phi ({{\mathcal{Y}}}_{1},{{\mathcal{Y}}}_{2},{{\mathcal{Y}}}_{3})}{{\bf{1}}{\Xi }_{1,3}^{\top }({{\mathcal{Y}}}_{1},{{\mathcal{Y}}}_{2},{{\mathcal{Y}}}_{3})\pi }.$$

Estimate the following expression from above:

$$\begin{array}{ccc}{\rm{E}}\left\{{{\bf{I}}}_{{A}_{3}^{s}}(\omega )\Phi \right\}=\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}\mathop{\sum }\limits_{i,j,k,m=1}^{N}{\pi }_{i}{\psi }^{ij}({y}_{1}){\overline{\gamma }}^{jk}({y}_{2}){\psi }^{km}({y}_{3})d{y}_{3}d{y}_{2}d{y}_{1}\\ =\mathop{\sum }\limits_{i,j,k=1}^{N}{\pi }_{i}\mathop{\sum }\limits_{\ell =1}^{L}{\varrho }_{\ell }^{ij}\mathop{\int}\limits_{{{\mathbb{R}}}^{M}}{\overline{\gamma }}^{jk}({y}_{2})d{y}_{2}\mathop{\sum }\limits_{m=1}^{N}\mathop{\sum }\limits_{n=1}^{L}{\varrho }_{n}^{km}\\ ={\mathfrak{W}}\mathop{\sum }\limits_{i,j=1}^{N}{\pi }_{i}\mathop{\sum }\limits_{\ell =1}^{L}{\varrho }_{\ell }^{ij}\mathop{\sum }\limits_{k=1}^{N}\ \mathop{\int}\limits_{{{\mathbb{R}}}^{M}}{\overline{\gamma }}^{jk}({y}_{2})d{y}_{2}\leqslant {\mathfrak{W}}\delta \mathop{\sum }\limits_{i=1}^{N}{\pi }_{i}\mathop{\sum }\limits_{j=1}^{N}\mathop{\sum }\limits_{\ell =1}^{L}{\varrho }_{\ell }^{ij}\leqslant {{\mathfrak{W}}}^{2}\delta .\end{array}$$

By analogy, it can be demonstrated that for an arbitrary r ⩾ 2,

$${\rm{E}}\left\{{{\bf{I}}}_{{A}_{r}^{s}}(\omega )\frac{{\bf{1}}{\Psi }_{t+1,r}^{\top }{\overline{\gamma }}_{t}^{\top }{\Psi }_{1,t-1}^{\top }\pi }{{\bf{1}}{\Xi }_{1,r}^{\top }\pi }\right\}\leqslant {{\mathfrak{W}}}^{r-1}\delta .$$

This finally yields

$${\rm{E}}\left\{{{\bf{I}}}_{{A}_{r}^{s}}(\omega )\parallel {\Delta }_{r}{\parallel }_{1}\right\}\leqslant 2r{{\mathfrak{W}}}^{r-1}\delta,$$

the second inequality of (5.8) is immediate from the fact that the inequality above holds for any π ∈ Π.

The proof of Theorem 2 is complete.

Funding

This work was supported in part by the Russian Foundation for Basic Research, project no. 19-07-00187 A.