*A discrete-time adaptive fading extended Kalman filter with the matrix forgetting factor is given by the following coupled difference equations*
$${\hat{x}}_{n+1}=f({\hat{x}}_{n},{u}_{n})+{K}_{n}({y}_{n}-h({\hat{x}}_{n}))$$(13)

*and Riccati difference equation*:
$${P}_{n+1}={A}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{A}_{n}^{T}+{\mathrm{\Lambda}}_{n}{Q}_{n}{\mathrm{\Lambda}}_{n}^{T}-{K}_{n}({C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}_{n}^{T}+{R}_{n}){K}_{n}^{T},$$(14)

*where K*_{n} is the Kalman gain given by
$${K}_{n}={A}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}_{n}^{T}({C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}_{n}^{T}+{R}_{n}{)}^{-1}.$$(15)

*Moreover, ${\mathrm{\Lambda}}_{n}=\text{\hspace{0.17em}diag\hspace{0.17em}}({\lambda}_{1},{\lambda}_{2},\dots ,{\lambda}_{q})$ is a time varying q × q dimensional diagonal matrix forgetting factor with ${\lambda}_{i}\ge 1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}i=1,2,\dots ,q;$ (see [8, 19] for the computation of Λ*_{n}) . Furthermore, Q_{n} and R_{n} are positive define, symmetric matrices with dimensions q Λ q and m Λ m, respectively, and the covariances matricesfor the currupting noise terms in (1)-(2).

*Consider a nonlinear stochastic system given by (1)-(2) and an extended Kalman filter as stated in Definition 3.1. Let the following assumptions hold*.

There are real numbers $\overline{a},\overline{c},\text{\hspace{0.17em}}\underset{\_}{p},\overline{p}>0\text{\hspace{0.17em}and\hspace{0.17em}}\underset{\_}{\lambda},\overline{\lambda}\ge 1$ such that the following bounds hold for every *n* ∈ ℕ_{0}
$$\parallel {A}_{n}\parallel \le \overline{a},$$(16a)
$$\parallel {C}_{n}\parallel \le \overline{c},$$(16b)
$$\underset{\_}{p}I\le {P}_{n}\le \overline{p}I,$$(16c)
$$\underset{\_}{q}I\le {Q}_{n},$$(16d)
$$\underset{\_}{r}I\le {R}_{n},$$(16e)
$$\underset{\_}{\lambda}I\le {\mathrm{\Lambda}}_{n}\le \overline{\lambda}I,$$(16f)

*where $\underset{\_}{q}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\underset{\_}{r}$ are the smallest eigenvalues of the matrices Q*_{n} and R_{n}, respectively. Moreover, $\underset{\_}{\lambda}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\overline{\lambda}$ are the smallest and the largest diagonal entries of Λ_{n}, respectively

*A*_{n} is nonsingular matrix for every *n* ∈ ℕ_{0}

*There are positive real numbers ${\epsilon}_{\phi},{\epsilon}_{\chi},{\kappa}_{\phi},{\kappa}_{\chi},>0$ such that the nonlinear functions φ, χ in (8) are bounded via*
$$\parallel \phi (x,\hat{x},u)\parallel \le {\kappa}_{\phi}\parallel x-\hat{x}{\parallel}^{2},$$(17)
$$\parallel \chi (x,\hat{x})\parallel \le {\kappa}_{\chi}\parallel x-\hat{x}{\parallel}^{2}.$$(18)

*Then the estimation error ζ*_{n} given by (6) is exponentially bounded in mean square and bounded with probability one, provided that the initial estimation error satisfies
$$\parallel {\zeta}_{0}\parallel \le \in $$(19)

*and the covariance matrices of the noise terms are bounded via*
$${G}_{n}{\mathrm{\Lambda}}_{n}{\mathrm{\Lambda}}_{n}^{T}{G}_{n}^{T}\le \delta I,$$(20)
$${D}_{n}{D}_{n}^{T}\le \delta I$$

for some δ, Ȉ > 0.

To prove Theorem 3.2 we need the following auxiliary results.

*Under the conditions of Theorem 3.2 there is a real number 0 < α < 1 such that*
$$1-\alpha =\frac{1}{{\underset{\_}{\lambda}}^{2}\left(1+\frac{\underset{\_}{{\lambda}^{2}}\underset{\_}{q}}{{\overline{\lambda}}^{2}\overline{p}{\left(\overline{a}+\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\right)}^{2}}\right)}$$

*and*
$${\mathrm{\Pi}}_{n}=({\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{)}^{-1}$$

*satisfies the inequality*
$$({A}_{n}-{K}_{n}{C}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({A}_{n}-{K}_{n}{C}_{n})\le (1-\alpha ){\mathrm{\Pi}}_{n}$$(22)

*for n ≤ 0 with the Kalman gain K*_{n} given in (15).

The proof mimics Lemma 3.1 in [25]. Substituting (15) in (14) and rearranging the resulting equation yields
$${P}_{n+1}=({A}_{n}-{K}_{n}{C}_{n}){\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}({A}_{n}-{K}_{n}{C}_{n}{)}^{T}+{\mathrm{\Lambda}}_{n}{Q}_{n}{\mathrm{\Lambda}}_{n}^{T}+\text{\hspace{0.17em}}{K}_{n}{C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}({A}_{n}-{K}_{n}{C}_{n}{)}^{T}.$$(23)

Multiplying the factor $({A}_{n}-{K}_{n}{C}_{n}){\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}\text{\hspace{0.17em}by\hspace{0.17em}}{A}_{n}^{-1}$ from left and using Equation (15) yields
$${A}_{n}^{-1}({A}_{n}-{K}_{n}{C}_{n}){\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}={\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}-{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}_{n}^{T}({C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}^{T}+{R}_{n}{)}^{-1}{C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}.$$(24)

Note that the right side of the equation (24) is a symmetric matrix. Thus, applying matrix inversion lemma in [26] we obtain
$${A}_{n}^{-1}({A}_{n}-{K}_{n}{C}_{n}){\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}=({\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}+{C}_{n}{R}_{n}^{-1}{C}_{n}^{T}{)}^{-1}\ge 0.$$(25)

Furthermore,
$${A}_{n}^{-1}{K}_{n}{C}_{n}={\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}_{n}^{T}({C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}_{n}^{T}+{R}_{n}{)}^{-1}{C}_{n}\ge 0.$$(26)

Due to above equations (25) and (26) along with ${\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}=({\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{)}^{T}$ we obtain
$${K}_{n}{C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}({A}_{n}-{K}_{n}{C}_{n}{)}^{T}={A}_{n}\left[{A}_{n}^{-1}{K}_{n}{C}_{n}\right]{\left[{A}_{n}^{-1}({A}_{n}-{K}_{n}{C}_{n}){\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}\right]}^{T}{A}_{n}^{T}\ge 0.$$(27)

From the equations (23) and (26) we have
$${P}_{n+1}\ge ({A}_{n}-{K}_{n}{C}_{n}){\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}({A}_{n}-{K}_{n}{C}_{n}{)}^{T}+{\mathrm{\Lambda}}_{n}{Q}_{n}{\mathrm{\Lambda}}_{n}^{T}.$$(28)

The inequality (25) implies that $({A}_{n}-{K}_{n}{C}_{n}{)}^{-1}$ exists, so we obtain
$${P}_{n+1}\ge ({A}_{n}-{K}_{n}{C}_{n})\left[{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}+({A}_{n}{K}_{n}{C}_{n}{)}^{-1}+{\mathrm{\Lambda}}_{n}{Q}_{n}{\mathrm{\Lambda}}_{n}^{T}{\left(({A}_{n}-{K}_{n}{C}_{n}{)}^{T}\right)}^{-1}\right]({A}_{n}-{K}_{n}{C}_{n}{)}^{T}.$$(29)

From (15) and (16a)-(16f) we have
$$\parallel {K}_{n}\parallel \le \parallel {A}_{n}\parallel \parallel {\mathrm{\Lambda}}_{n}\parallel \parallel {P}_{n}\parallel \parallel {\mathrm{\Lambda}}_{n}^{T}\parallel \parallel {C}_{n}^{T}\parallel \parallel ({C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}^{T}+{R}_{n}{)}^{-1}\parallel \le \overline{a}\overline{\lambda}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}.$$(30)

Substituting the inequalities (16a)-(16f) into (29) we obtain
$${P}_{n+1}\ge ({A}_{n}-{K}_{n}{C}_{n})\left[{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}+\frac{\underset{\_}{{\lambda}^{2}}\underset{\_}{q}}{{\left(\overline{a}+\overline{a}\overline{\lambda}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\right)}^{2}}I\right]({A}_{n}-{K}_{n}{C}_{n}{)}^{T}.$$(31)

Multiplying both sides of (31) from left and right with ${\mathrm{\Lambda}}_{n+1}\text{\hspace{0.17em}and\hspace{0.17em}}{\mathrm{\Lambda}}_{n+1}^{T},$ respectively, and using the inequalty (16f) gives$${\mathrm{\Lambda}}_{n+1}{P}_{n+1}{\mathrm{\Lambda}}_{n+1}^{T}\ge \underset{\_}{{\lambda}^{2}}({A}_{n}-{K}_{n}{C}_{n})\left[{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}+\frac{\underset{\_}{{\lambda}^{2}}\underset{\_}{q}}{{\left(\overline{a}+\overline{a}\overline{\lambda}2\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\right)}^{2}}I\right]({A}_{n}-{K}_{n}{C}_{n}{)}^{T}.$$(32)

Taking the inverse of both sides of (32) and multiplying from left and right with $({A}_{n}-{K}_{n}{C}_{n}{)}^{T}\text{\hspace{0.17em}and\hspace{0.17em}}({A}_{n}-{K}_{n}{C}_{n})$ we have,$${\left({A}_{n}-{K}_{n}{C}_{n}\right)}^{T}{\mathrm{\Pi}}_{n+1}({A}_{n}-{K}_{n}{C}_{n})\ge \frac{1}{{\underset{\_}{\lambda}}^{2}}{\left[1+\frac{\underset{\_}{{\lambda}^{2}}\underset{\_}{q}}{{\overline{\lambda}}^{2}\overline{p}{\left(\overline{a}+\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\right)}^{2}}\right]}^{-1}{\mathrm{\Pi}}_{n}.$$(33)

Then the result follows.
$$(1-\alpha )=\frac{1}{{\underset{\_}{\lambda}}^{2}}\frac{1}{\left(1+\frac{\underset{\_}{{\lambda}^{2}}\underset{\_}{q}}{{\overline{\lambda}}^{2}\overline{p}{\left(\overline{a}+\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\right)}^{2}}\right)}.$$(34)

*Let the conditions of Theorem 3.2 be fulfilled, let ${\mathrm{\Pi}}_{n}=({\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{)}^{-1}\text{\hspace{0.17em}and\hspace{0.17em}}{K}_{n},\text{\hspace{0.17em}}{r}_{n}$ given in (15),(8). Then there are positive real numbers ε′, κ*_{nonl} such that

$${r}_{n}^{T}{\mathrm{\Pi}}_{n}\left[2({A}_{n}-{K}_{n}{C}_{n})({x}_{n}-{\hat{x}}_{n})+{r}_{n}\right]\le {\kappa}_{nonl}\parallel {x}_{n}-{\hat{x}}_{n}\parallel {}^{3}$$(35)

*holds for* $\parallel {x}_{n}-{\hat{x}}_{n}\parallel \le {\epsilon}^{\prime}.$

From (15), (16a)-(16f) and ${C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{\prime}{C}_{n}^{\prime}>0,$ we have
$$\parallel {K}_{n}\parallel \le \overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}$$(36)

and using in (8) gives
$$\parallel {r}_{n}\parallel \le \parallel \phi ({x}_{n},{\hat{x}}_{n},{u}_{n})\parallel +\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\parallel \chi ({x}_{n},{\hat{x}}_{n})\parallel .$$(37)

By choosing ${\epsilon}^{\prime}=\text{min}({\epsilon}_{\phi},{\epsilon}_{\chi})$ and using (17), (18) we obtain
$$\parallel {r}_{n}\parallel \le {\kappa}_{\phi}\parallel {x}_{n}-{\hat{x}}_{n}{\parallel}^{2}+\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}{\kappa}_{\chi}\parallel ({x}_{n},{\hat{x}}_{n}){\parallel}^{2}.$$(38)

Since $\parallel {x}_{n}-{\hat{x}}_{n}{\parallel}^{2}\le {\epsilon}_{n}^{\prime},$ we have
$$\parallel {r}_{n}\parallel \le {\kappa}_{\phi}\parallel {x}_{n}-{\hat{x}}_{n}{\parallel}^{2}.$$(39)

Define
$${\kappa}^{\prime}={\kappa}_{\phi}+(\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}){\kappa}_{\chi}.$$(40)

Then, for $\parallel {x}_{n}-{\hat{x}}_{n}{\parallel}^{2}\le {\epsilon}^{\prime},$ from (38) by taking ${\mathrm{\Pi}}_{n}=({\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{)}^{-1}$ and using (16a)-( 16f) we obtain
$${r}_{n}^{T}{\mathrm{\Pi}}_{n}\left[2({A}_{n}-{K}_{n}{C}_{n})({x}_{n}-{\hat{x}}_{n})+{r}_{n}\right]\le {\kappa}^{\prime}\parallel {x}_{n}-{\hat{x}}_{n}{\parallel}^{2}\frac{1}{{\underset{\_}{\lambda}}^{2}\underset{\_}{p}}\left[2\left(\overline{a}+\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\right)\right]\times \parallel {x}_{n}-{\hat{x}}_{n}\parallel +{\kappa}^{\prime}{\epsilon}^{\prime}\parallel {x}_{n}-{\hat{x}}_{n}\parallel .$$(41)

Rearranging (41) gives
$${r}_{n}^{T}{\mathrm{\Pi}}_{n}\left[2({A}_{n}-{K}_{n}{C}_{n})({x}_{n}-{\hat{x}}_{n})+{r}_{n}\right]\le {\kappa}^{\prime}\frac{1}{{\underset{\_}{\lambda}}^{2}\underset{\_}{p}}\left[2\left(\overline{a}+\overline{a}{\overline{\lambda}}^{2}\overline{p}{\overline{c}}^{2}\frac{1}{\underset{\_}{r}}\right)+{\kappa}^{\prime}{\epsilon}^{\prime}\right]\parallel {x}_{n}-{\hat{x}}_{n}{\parallel}^{3}={\kappa}_{nonl}\parallel {x}_{n}-{\hat{x}}_{n}{\parallel}^{3}$$(42)

where
$${\kappa}_{nonl}={\kappa}^{\prime}\frac{1}{{\underset{\_}{\lambda}}^{2}\underset{\_}{p}}\left[2\left(\overline{a}+\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}\right)+{\kappa}^{\prime}{\epsilon}^{\prime}\right].$$(43)

*Let the conditions of Theorem 3.2 be fulfilled, let ${\mathrm{\Pi}}_{n}=({\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{)}^{-1}$ given in (15), (9). Then there is a positive real number κ*_{noise} independent of δ such that
$$E\left\{{s}_{n}^{T}{\mathrm{\Pi}}_{n+{1}^{S}n}\right\}\le {\kappa}_{noise}\delta $$(44)

*holds*.

Using the equation (9) and after matrix distribution we obtain
$${s}_{n}^{T}{\mathrm{\Pi}}_{n+{1}^{S}n}=\left\{({G}_{n}{w}_{n}-{K}_{n}{D}_{n}{v}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({G}_{n}{w}_{n}-{K}_{n}{D}_{n}{v}_{n})\right\}$$(45)

$$=\left\{({G}_{n}{w}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({G}_{n}{w}_{n})-({G}_{n}{w}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({K}_{n}{D}_{n}{v}_{n})-({K}_{n}{D}_{n}{v}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({G}_{n}{w}_{n})+({K}_{n}{D}_{n}{v}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({K}_{n}{D}_{n}{v}_{n})\right\}.$$(46)

Recall that the vectors *w*_{n} and *v*_{n} are uncorrelated, the terms containing both vanish so we have
$${s}_{n}^{T}{\mathrm{\Pi}}_{n+{1}^{S}n}=\{({G}_{n}{w}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({G}_{n}{w}_{n})+({K}_{n}{D}_{n}{v}_{n}{)}^{T}{\mathrm{\Pi}}_{n+1}({K}_{n}{D}_{n}{v}_{n})\}.$$(47)

By the group equations (16) and the inequality ${C}_{n}{\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{C}_{n}^{T}>0$ we have
$$\parallel {K}_{n}\parallel <\overline{a}{\overline{\lambda}}^{2}\overline{p}\overline{c}\frac{1}{\underset{\_}{r}}.$$(48)

This inequality yields
$${s}_{n}^{T}{\mathrm{\Pi}}_{n+1}{s}_{n}\le \frac{1}{{\underset{\_}{\lambda}}^{2}\underset{\_}{p}}{w}_{n}^{T}{G}_{n}^{T}{G}_{n}{w}_{n}+\frac{{\overline{a}}^{2}{\overline{p}}^{2}{\overline{c}}^{2}{\overline{\lambda}}^{2}}{\underset{\_}{p}{\underset{\_}{r}}^{2}}{v}_{n}^{T}{D}_{n}^{T}{D}_{n}{v}_{n}.$$(49)

Taking the trace of the above inequality we get
$${s}_{n}^{T}{\mathrm{\Pi}}_{n+1}{s}_{n}\le \frac{1}{{\underset{\_}{\lambda}}^{2}\underset{\_}{p}}\text{\hspace{0.17em}tr\hspace{0.17em}}\left({w}_{n}^{T}{G}_{n}^{T}{G}_{n}{w}_{n}\right)+\frac{{\overline{a}}^{2}{\overline{p}}^{2}{\overline{c}}^{2}{\overline{\lambda}}^{2}}{\underset{\_}{p}{\underset{\_}{r}}^{2}}\text{\hspace{0.17em}tr\hspace{0.17em}}\left({v}_{n}^{T}{D}_{n}^{T}{D}_{n}{v}_{n}\right).$$(50)

Since $\text{tr\hspace{0.17em}}(\mathrm{\Gamma}\mathrm{\Delta})=\text{\hspace{0.17em}tr\hspace{0.17em}}(\mathrm{\Delta}\mathrm{\Gamma}),$ using (50) we obtain
$${s}_{n}^{T}{\mathrm{\Pi}}_{n+1}{s}_{n}\le \frac{1}{{\underset{\_}{\lambda}}^{2}\underset{\_}{p}}\text{\hspace{0.17em}tr\hspace{0.17em}}({G}_{n}{w}_{n}{w}_{n}^{T}{G}_{n}^{T})+\frac{{\overline{a}}^{2}{\overline{p}}^{2}{\overline{c}}^{2}{\overline{\lambda}}^{2}}{\underset{\_}{p}{\underset{\_}{r}}^{2}}\text{\hspace{0.17em}tr\hspace{0.17em}}({D}_{n}{v}_{n}{v}_{n}^{T}{D}_{n}^{T}),$$(51)

where *D*_{n} and *G*_{n} are deterministic matrices. Remember that *w*_{n} and *v*_{n} are vector valued white noise process, thus,
$$E\{{v}_{n}{v}_{n}^{T}\}=I$$(52)

and
$$E\{{w}_{n}{w}_{n}^{T}\}=I$$(53)

hold. Thus we have
$$E\left\{{s}_{n}^{T}{\mathrm{\Pi}}_{n+{1}^{S}n}\right\}\le \frac{1}{{\underset{\_}{\lambda}}^{2}\text{tr}({G}_{n}{\mathrm{\Lambda}}_{n}{\mathrm{\Lambda}}_{n}^{T}{G}_{n}^{T})\underset{\_}{p}}+\frac{{\overline{a}}^{2}{\overline{p}}^{2}{\overline{c}}^{2}{\overline{\lambda}}^{2}}{\underset{\_}{p}{\underset{\_}{r}}^{2}}\text{\hspace{0.17em}tr\hspace{0.17em}}\left({D}_{n}{v}_{n}{v}_{n}^{T}{D}_{n}^{T}\right).$$(54)

From the equations (20) and (21) we have
$$\text{tr\hspace{0.17em}}({G}_{n}{\mathrm{\Lambda}}_{n}{\mathrm{\Lambda}}_{n}^{T}{G}_{n}^{T})\le \delta \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{tr}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(I)=q\delta $$(55)

and
$$\text{tr\hspace{0.17em}}({D}_{n}{D}_{n}^{T})\le \delta \text{\hspace{0.17em}tr\hspace{0.17em}}(I)=m\delta ,$$(56)

where *q* and *m* are the number of rows *G*_{n} and *D*_{n}, respectively. Defining
$${\kappa}_{noise}=\frac{q}{{\underset{\_}{\lambda}}^{2}\underset{\_}{p}}+\frac{{\overline{a}}^{2}{\overline{c}}^{2}{\overline{p}}^{2}{\overline{\lambda}}^{2}m}{\underset{\_}{p}{\underset{\_}{r}}^{2}}$$(57)

yields
$$E\left\{{s}_{n}^{T}{\mathrm{\Pi}}_{n+{1}^{S}n}\right\}\le {\kappa}_{noise}\delta ,$$(58)

This completes the proof.

We are now ready to prove the main result stated in Theorem 3.2 of the paper.

There exists a function depending on error estimate
$${V}_{n}({\zeta}_{n})={\zeta}_{n}^{T}{\mathrm{\Pi}}_{n}{\zeta}_{n}$$(59)

with ${\mathrm{\Pi}}_{n}=({\mathrm{\Lambda}}_{n}{P}_{n}{\mathrm{\Lambda}}_{n}^{T}{)}^{-1}$ since *P*_{n} is positive definite. From the inequalities (16c)-(16f) we have
$$\frac{1}{\overline{p}{\overline{\lambda}}^{2}}\parallel {\zeta}_{n}{\parallel}^{2}\le {V}_{n}({\zeta}_{n})\le \frac{1}{\underset{\_}{p}\underset{\_}{{\lambda}^{2}}}\parallel {\zeta}_{n}{\parallel}^{2},$$(60)

which is similar to (10) with $\underset{\_}{v}=\frac{1}{\overline{p}{\overline{\lambda}}^{2}}\text{\hspace{0.17em}and\hspace{0.17em}}\overline{v}=\frac{1}{\underset{\_}{p}{\underset{\_}{\lambda}}^{2}}.$ We need an upper bound on $E\{{V}_{n+1}({\zeta}_{n+1})|{\zeta}_{n}\}$ as stated in (11) to meet the requirements of Lemma 2.1. From (7) we obtain
$${V}_{n}({\zeta}_{n+1})=[({A}_{n}-{K}_{n}{C}_{n}){\zeta}_{n}+{r}_{n}+{s}_{n}{]}^{T}{\mathrm{\Pi}}_{n+1}[({A}_{n}-{K}_{n}{C}_{n}){\zeta}_{n}+{r}_{n}+{s}_{n}].$$(61)

Using Lemma 3.3 we obtain
$${V}_{n}({\zeta}_{n+1})\le (1-\alpha ){V}_{n}({\zeta}_{n})+{r}_{n}^{T}{\mathrm{\Pi}}_{n+1}(2({A}_{n}-{K}_{n}{C}_{n}){\zeta}_{n}+{r}_{n})+2{s}_{n}^{T}{\mathrm{\Pi}}_{n+1}(({A}_{n}-{K}_{n}{C}_{n}){\zeta}_{n}+{r}_{n})+{s}_{n}^{T}{\mathrm{\Pi}}_{n+{1}^{S}n}.$$(62)

Taking the conditional expectation $E\{{V}_{n+1}({\zeta}_{n+1})|{\zeta}_{n}\}$ and considering the white noise property it can be seen that the term $E\{{s}_{n}^{T}{\mathrm{\Pi}}_{n+1}(({A}_{n}-{K}_{n}{C}_{n}){\zeta}_{n}+{r}_{n})|{\zeta}_{n}\}$ vanishes since neither ${\mathrm{\Pi}}_{n+1}\text{\hspace{0.17em}nor\hspace{0.17em}}{A}_{n},\text{\hspace{0.17em}}{K}_{n},\text{\hspace{0.17em}}{C}_{n},\text{\hspace{0.17em}}{r}_{n},\text{\hspace{0.17em}}{s}_{n},\text{\hspace{0.17em}}{\zeta}_{n}\text{\hspace{0.17em}depend on\hspace{0.17em}}{v}_{n}\text{\hspace{0.17em}or\hspace{0.17em}}{w}_{n}.$ The remaining terms are estimated by Lemma 3.4 and Lemma 3.5 as
$$E\{{V}_{n+1}({\zeta}_{n+1})|{\zeta}_{n}\}-{V}_{n}({\zeta}_{n})\le -\alpha {V}_{n}({\zeta}_{n})+{\kappa}_{nonl}\parallel {\zeta}_{n}\parallel {}^{3}+{\kappa}_{noise}\delta $$

for $\parallel {\zeta}_{n}\parallel \le {\epsilon}^{\prime}.$ We define$$\epsilon =\text{min}\left({\epsilon}^{\prime},\frac{\alpha}{2\overline{p}{\overline{\lambda}}^{2}{\kappa}_{noise}}\right)$$(64)

Then from (59), (60) under condition $\parallel {\zeta}_{n}\parallel \le \epsilon $ we obtain
$${\kappa}_{nonl}\parallel {\zeta}_{n}\parallel \parallel {\zeta}_{n}{\parallel}^{2}\le \frac{\alpha}{2\overline{p}{\overline{\lambda}}^{2}}\parallel {\zeta}_{n}{\parallel}^{2}\le \frac{\alpha}{2}{V}_{n}({\zeta}_{n}).$$(65)

Substituting into (63) yields
$$E\{{V}_{n+1}({\zeta}_{n+1})|{\zeta}_{n}\}-{V}_{n}({\zeta}_{n})\le -\alpha {V}_{n}({\zeta}_{n})+\text{\hspace{0.17em}}\underset{\u2a7d\frac{\alpha}{2}{V}_{n}({\zeta}_{n})}{\underset{\u23df}{{\kappa}_{nonl}\parallel {\zeta}_{n}{\parallel}^{3}}}+{\kappa}_{noise}\delta \le -\frac{\alpha}{2}{V}_{n}({\zeta}_{n})+{\kappa}_{noise}\delta $$(66)

for $\parallel {\zeta}_{n}\parallel \le \epsilon .$ Therefore we are able to apply Lemma 2.1 with $\parallel {\zeta}_{0}\parallel \le \epsilon ,\text{\hspace{0.17em}}\underset{\_}{v}=\frac{1}{\overline{p}\overline{\lambda}2},\overline{v}=\frac{1}{\underset{\_}{p}{\underset{\_}{\lambda}}^{2}}\text{\hspace{0.17em}and\hspace{0.17em}}\mu ={\kappa}_{noise}\delta .$ However, with some $\stackrel{~}{\epsilon}\le \epsilon \text{\hspace{0.17em}for\hspace{0.17em}}\stackrel{~}{\epsilon}\le \parallel {\zeta}_{n}\parallel \le \epsilon $ we have to guarantee the inequality
$$E\{{V}_{n+1}({\zeta}_{n+1})|{\zeta}_{n}\}-{V}_{n}({\zeta}_{n})\le -\frac{\alpha}{2}{V}_{n}({\zeta}_{n})+{\kappa}_{noise}\delta \le 0.$$(67)

Choosing with the aid of (64)
$$\delta =\frac{\alpha {\stackrel{~}{\epsilon}}^{2}}{2\overline{p}{\overline{\lambda}}^{2}{\kappa}_{noise}}$$(68)

with some
$\stackrel{~}{\epsilon}\le \epsilon \text{\hspace{0.17em}we have for\hspace{0.17em}}\parallel {\zeta}_{n}\parallel \underset{\_}{>}\stackrel{~}{\epsilon}$
$${\kappa}_{noise}\le \frac{\alpha}{2\overline{p}{\overline{\lambda}}^{2}}\parallel {\zeta}_{n}{\parallel}^{2}\le \frac{\alpha}{2}{V}_{n}({\zeta}_{n}),$$(69)

which says that (67) holds. In result we conclude that the estimation error remains bounded if the initial error and noise terms are bounded as stated in (19)-(21).

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