6. Appendix

Jan Kowalski and Jacek Wesołowski

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6.1 Algebra of shift operators

In the first part of Appendix we introduce and analyze an algebraic operator formalism which is crucial for the proof of our main result (given in Subsection 6.2).

For a sequence of vectors $\underset{\_}{\overline{x}}=\left({\underset{\_}{x}}_0\mathrm{,}{\underset{\_}{x}}_1\mathrm{,}{\underset{\_}{x}}_2\mathrm{,}\dots \right),{\underset{\_}{x}}_i\in {ℝ}^N,$  define shifts to the left and to the right by

$$\begin{array}{lll}ℒ\left(\underset{\_}{\overline{x}}\right)\hfill & =\left({\underset{\_}{x}}_1\mathrm{,}{\underset{\_}{x}}_2\mathrm{,}{\underset{\_}{x}}_3\mathrm{,}\dots \right)\hfill & \text{left shift,}\hfill \\ ℛ\left(\underset{\_}{\overline{x}}\right)\hfill & =\left(\underset{\_}{0}\mathrm{,}{\underset{\_}{x}}_0\mathrm{,}{\underset{\_}{x}}_1\mathrm{,}\dots \right)\hfill & \text{right shift}\mathrm{.}\hfill \end{array}$$

Note that $ℒℛ=ℐ$ (identity), but

$$\left(ℐ-ℛℒ\right)\underset{\_}{\overline{x}}=\left({\underset{\_}{x}}_0\mathrm{,}\underset{\_}{0}\mathrm{,}\underset{\_}{0}\mathrm{,}\dots \right)={\underset{\_}{x}}_0\overline{e}\mathrm{,}(6.1)$$

where $\overline{e}=\left(\mathrm{1,0,0,}\dots \right).$

For any $M\times N$  matrix $A$  define

$$A\underset{\_}{\overline{x}}=\left(A{\underset{\_}{x}}_0\mathrm{,}A{\underset{\_}{x}}_1\mathrm{,}A{\underset{\_}{x}}_2\mathrm{,}\dots \right)\mathrm{.}$$

In particular, for a complex (real) number $a,$  taking $A=aI$  we have

$$a\underset{\_}{\overline{x}}=\left(a{\underset{\_}{x}}_0\mathrm{,}a{\underset{\_}{x}}_1\mathrm{,}a{\underset{\_}{x}}_2\mathrm{,}\dots \right)\mathrm{.}$$

Moreover, by the above definitions, for any $i\mathrm{,}j\ge 0$

$${ℛ}^i{ℒ}^{j}A\underset{\_}{\overline{x}}=A{ℛ}^i{ℒ}^j\underset{\_}{\overline{x}}\mathrm{.}$$

For a constant sequence of vectors $\underset{\_}{\overline{x}}=\left(\underset{\_}{x}\mathrm{,}\underset{\_}{x}\mathrm{,}\underset{\_}{x}\mathrm{,}\dots \right)$  we have $ℒ\underset{\_}{\overline{x}}=\underset{\_}{\overline{x}}$  and thus for any $i\mathrm{,}j\ge 0$

$${ℒ}^i{ℛ}^j\underset{\_}{\overline{x}}=\{\begin{array}{ll}\underset{\_}{\overline{x}}\mathrm{,}\hfill & \text{for }i\ge j\mathrm{,}\hfill \\ {ℛ}^{j-i}\underset{\_}{\overline{x}}\mathrm{,}\hfill & \text{for }i<j\mathrm{.}\hfill \end{array}(6.2)$$

If $N=1$  we write $\underset{\_}{\overline{y}}=\overline{y}=\left(y_0\mathrm{,}{y}_1\mathrm{,}{y}_2\dots \right),y_i\in ℝ,$  and $\text{L:=}ℒ,\text{R:=}ℛ.$  Note that, for $\overline{y}={\left(y^n\right)}_{n\ge 0}$  we have

$${\text{L}}^j\overline{y}=y^j\overline{y}(6.3)$$

and thus

$${\text{L}}^j{\text{R}}^i\overline{y}=\{\begin{array}{ll}{y}^{j-i}\overline{y}\mathrm{,}\hfill & \text{for }j\ge i\mathrm{,}\hfill \\ {\text{R}}^{i-j}\overline{y}\mathrm{,}\hfill & \text{for }j<i\mathrm{.}\hfill \end{array}$$

For any $\overline{y}={\left(y_n\right)}_{n\ge 0}$  and any $\underset{\_}{\overline{x}}={\left({\underset{\_}{x}}_n\right)}_{n\ge 0}$  define $\overline{y}\underset{\_}{\overline{x}}={\left(y_n{\underset{\_}{x}}_n\right)}_{n\ge 0}.$  Then for any complex (real) numbers $\alpha ,\beta ,$  any $M\times N$  matrices $A,B,$  any $i\mathrm{,}j\mathrm{,}k\mathrm{,}m\ge 0,$

$$\left(\alpha A{ℛ}^i{ℒ}^j+\beta B{ℒ}^m{ℛ}^k\right)\overline{y}\underset{\_}{\overline{x}}=\left(\alpha {\text{R}}^i{\text{L}}^j\overline{y}\right)\left(A{ℛ}^i{ℒ}^j\underset{\_}{\overline{x}}\right)+\left(\beta {\text{L}}^m{\text{R}}^k\overline{y}\right)\left(B{ℒ}^m{ℛ}^k\underset{\_}{\overline{x}}\right)\mathrm{.}(6.4)$$

Note also that if $\underset{\_}{\overline{x}}=\left(\underset{\_}{x}\mathrm{,}\underset{\_}{x}\mathrm{,}\dots \right)$  is a constant sequence, then

$${ℛ}^i{ℒ}^j\overline{y}\underset{\_}{\overline{x}}=\left({\text{R}}^i{\text{L}}^j\overline{y}\right)\underset{\_}{\overline{x}}\text{ and }{ℒ}^j{ℛ}^i\overline{y}\underset{\_}{\overline{x}}=\left({\text{L}}^j{\text{R}}^i\overline{y}\right)\underset{\_}{\overline{x}}\mathrm{.}(6.5)$$

Lemma 6.1 Let $v_i,i=1,\dots \mathrm{,}p,$  be functions defined in (3.11), where $a_1\mathrm{,}\dots \mathrm{,}{a}_p$  are arbitrary numbers. Let $\underset{\_}{\overline{x}}=\left(\underset{\_}{x}\mathrm{,}\underset{\_}{x}\mathrm{,}\dots \right)$  and $\overline{y}={\left(y^n\right)}_{n\ge 0}.$  Then for any $i=1,\dots \mathrm{,}p$

$${ℒ}^i\left(ℐ-{\displaystyle \sum _{j=1}^p{a}_j{ℛ}^j}\right)=\left({ℒ}^p-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^{p-j}}\right){ℛ}^{p-i}\mathrm{,}(6.6)$$

$$\left(ℐ-ℛℒ\right)\left({ℒ}^p-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^{p-j}}\right){ℛ}^{p-i}\overline{y}\underset{\_}{\overline{x}}=v_i\left(y\right)\left({\underset{\_}{x}}_0\mathrm{,}\underset{\_}{0}\mathrm{,}\underset{\_}{0}\mathrm{,}\dots \right)(6.7)$$

and

$$\left({ℒ}^p-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^{p-j}}\right)\overline{y}\underset{\_}{\overline{x}}=v_p\left(y\right)\overline{y}\underset{\_}{\overline{x}}\mathrm{.}(6.8)$$

Proof. First, we prove (6.8). By (6.4)

$$\left({ℒ}^p-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^{p-j}}\right)\overline{y}\underset{\_}{\overline{x}}=\left({\text{L}}^p\overline{y}\right){ℒ}^p\underset{\_}{\overline{x}}-{\displaystyle \sum _{j=1}^p{a}_j}\left({\text{L}}^{p-j}\overline{y}\right)\left({ℒ}^{p-j}\underset{\_}{\overline{x}}\right)\mathrm{.}$$

Note that ${\text{L}}^k\overline{y}=y^k\overline{y}$  and ${ℒ}^k\underset{\_}{\overline{x}}=\underset{\_}{\overline{x}}$  for any $k=0,1,\dots .$  Therefore

$$\left({ℒ}^p-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^{p-j}}\right)\overline{y}\underset{\_}{\overline{x}}=\left[\left(y^p-{\displaystyle \sum _{m=1}^p{a}_m{y}^{p-m}}\right)\overline{y}\right]\underset{\_}{\overline{x}}\mathrm{.}$$

Now (6.8) follows by the definition (3.11) for $i=p.$

Again, from (6.2), (6.4) and (6.5) it follows that

$$\left(ℐ-ℛℒ\right)\left({ℒ}^p-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^{p-j}}\right){ℛ}^{p-i}\overline{y}\underset{\_}{\overline{x}}=\left[\left(\text{I}-\text{RL}\right)\left({\text{L}}^p-{\displaystyle \sum _{j=1}^p{a}_j{\text{L}}^{p-j}}\right){\text{R}}^{p-i}\overline{y}\right]\underset{\_}{\overline{x}}\mathrm{.}$$

Since for any $k\in \left\{\mathrm{0,1,}\dots \mathrm{,}p\right\}$

$$\left({\text{L}}^p-{\displaystyle \sum _{j=1}^p{a}_j{\text{L}}^{p-j}}\right){\text{R}}^{p-k}\overline{y}=y^k\overline{y}-{\displaystyle \sum _{j=1}^k{a}_j{y}^{k-j}\overline{y}}-{\displaystyle \sum _{j=k+1}^p{a}_j{\text{R}}^{j-k}\overline{y}}=v_k\left(y\right)\overline{y}-{\displaystyle \sum _{j=k+1}^p{a}_j{\text{R}}^{j-k}\overline{y}}$$

then

$$\left(\text{I}-\text{RL}\right)\left({\text{L}}^p-{\displaystyle \sum _{j=1}^p{a}_j{\text{L}}^{p-j}}\right){\text{R}}^{p-k}\overline{y}=v_k\left(y\right)\overline{e}$$

and thus (6.7) follows.

The identity (6.6) follows by (6.2) since

$${ℒ}^i\left(ℐ-{\displaystyle \sum _{j=1}^p{a}_j{ℛ}^j}\right)={ℒ}^i-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^i{ℛ}^j}={ℒ}^p{ℛ}^{p-i}-{\displaystyle \sum _{j=1}^p{a}_j{ℒ}^{p-j}{ℛ}^{p-i}}\mathrm{.}$$

Lemma 6.2 Let $\mathcal{D}$  be an operator on the space of sequences of vectors from ${ℝ}^N$  defined by

$$\mathcal{D}=ℐ+{\displaystyle \sum _{k=1}^{N-1}}\left(C^k{ℒ}^k+{\left(C^T\right)}^k{ℛ}^k\right)\mathrm{,}(6.9)$$

where $C$  is the covariance matrix defined in Section 2.

The operator $\mathcal{D}$  is invertible and

$${\mathcal{D}}^{-1}=\left(ℐ-C^Tℛ\right)\Delta \left(ℐ-Cℒ\right)\mathrm{.}(6.10)$$

Proof. Note that $I-C{C}^T\text{=diag}\left(1-{\rho }^2\mathrm{,}\dots \mathrm{,1}-{\rho }^2\mathrm{,1}\right).$  Consequently, $\Delta ={\left(I-C{C}^T\right)}^{-1}$  is well defined. Note also that ${\displaystyle {\sum }_{k=0}^{N-1}{C}^k{ℒ}^k}$  is invertible and its inverse is $ℐ-Cℒ\text{.}$  Similarly, ${\displaystyle {\sum }_{k=0}^{N-1}{\left(C^T\right)}^k{ℛ}^k}$  is invertible and its inverse is $ℐ-C^Tℛ\text{.}$

Therefore

$$\begin{array}{lll}{\left[\left(ℐ-C^Tℛ\right)\Delta \left(ℐ-Cℒ\right)\right]}^{-1}\hfill & =\hfill & {\left(ℐ-Cℒ\right)}^{-1}{\Delta }^{-1}{\left(ℐ-C^Tℛ\right)}^{-1}\hfill \\ \hfill & =\hfill & \left({\displaystyle \sum _{k=0}^{N-1}{C}^k{ℒ}^k}\right)\left(I-C{C}^T\right)\left({\displaystyle \sum _{j=0}^{N-1}}{\left(C^T\right)}^j{ℛ}^j\right)\hfill \\ \hfill & =\hfill & {\displaystyle \sum _{k\mathrm{,}j=0}^{N-1}{C}^k}{\left(C^T\right)}^j{ℒ}^k{ℛ}^j-{\displaystyle \sum _{k\mathrm{,}j=1}^{N-1}{C}^k}{\left(C^T\right)}^j{ℒ}^k{ℛ}^j\hfill \\ \hfill & =\hfill & \mathcal{D}+{\displaystyle \sum _{k\mathrm{,}j=1}^{N-1}{C}^k}{\left(C^T\right)}^j{ℒ}^{k-1}\left(ℒℛ-ℐ\right){ℛ}^{j-1}\hfill \\ \hfill & =\hfill & \mathcal{D}\mathrm{.}\hfill \end{array}$$

6.2 Proof of the recurrence

Proof of Theorem 3.1. Note first that since $d_1\mathrm{,}\dots \mathrm{,}{d}_p$  are either real or come in conjugate pairs (see Remark 3.1) it follows from (3.10) that $a_1\mathrm{,}\dots \mathrm{,}{a}_p$  are real numbers.

Recall that ${\underset{\_}{e}}_0=\underset{\_}{1}$  and denote ${\underset{\_}{\overline{e}}}_j=\left({\underset{\_}{e}}_j\mathrm{,}{\underset{\_}{e}}_j\mathrm{,}\dots \right),j\in H^{\prime }=\left\{0\right\}\cup H.$  Recall that the $N\times N$  diagonal matrix $\Delta$  is defined as

$$\Delta ={\left(I-C{C}^T\right)}^{-1}=\frac{1}{1-{\rho }^2}\text{diag}\left(\mathrm{1,}\dots \mathrm{,1,1}-{\rho }^2\right)\mathrm{.}$$

With $d_1\mathrm{,}\dots \mathrm{,}{d}_p$  and $\underset{\_}{c}$  as defined in Theorem 3.1 let (see (6.10))

$$\underset{\_}{\overline{w}}=\left({\underset{\_}{w}}_0\mathrm{,}{\underset{\_}{w}}_1\mathrm{,}\dots \right)={\mathcal{D}}^{-1}{\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}{\overline{d}}_m{\underset{\_}{\overline{e}}}_j}}\mathrm{,}(6.11)$$

where ${\overline{d}}_m=\left(\mathrm{1,}{d}_m\mathrm{,}{d}_m^2\mathrm{,}\dots \right),m=\mathrm{1,}\dots \mathrm{,}p.$  Note that $\Vert {\underset{\_}{w}}_i\Vert$  (the length of the vector ${\underset{\_}{w}}_i)$  is of order ${\left({\max }_{1\le m\le p}\left|d_m\right|\right)}^i,i=0,1,\dots .$  By Remark 3.1 and ASSUMPTION II we have ${\max }_{1\le m\le p}\left|d_m\right|\in \left(\mathrm{0,1}\right).$  Hence (2.1) is a correct definition of a random series (with bounded variance).

Consequently, it suffices to show that:

1. The sequence $\underset{\_}{\overline{w}}$  defined in (6.11) is the sequence of optimal weights. To this end we note that the variance of any linear estimator ${\displaystyle {\sum }_{i=0}^{\infty }{\underset{\_}{u}}_i^T{\underset{\_}{X}}_i},{\underset{\_}{u}}_i\in {ℝ}^N,i=0,1,\dots ,$  has the form

   $$\mathbb{V}\text{ar}{\displaystyle \sum _{i=0}^{\infty }{\underset{\_}{u}}_i^T{\underset{\_}{X}}_i}={\displaystyle \sum _{i=0}^{\infty }{\underset{\_}{u}}_i^T{\underset{\_}{u}}_i}+2{\displaystyle \sum _{i=0}^{\infty }{\displaystyle \sum _{k=1}^{N-1}{\underset{\_}{u}}_i^T{C}^k{\underset{\_}{u}}_{i+k}}}\mathrm{.}(6.12)$$

   We need to show that $\underset{\_}{\overline{u}}={\left({\underset{\_}{u}}_i\right)}_{i\ge 0}:=\underset{\_}{\overline{w}}$  with $\underset{\_}{\overline{w}}$  as defined in (6.11) minimize this expression under the constraints (2.2) and (2.3). Since the above variance as a function of $\underset{\_}{\overline{u}}$  is convex then the problem has the unique solution. Using the standard Lagrange method, that is differentiating the Lagrange function (with multipliers ${({\lambda }_{j\mathrm{,}i})}_{j\in H^{\prime }\mathrm{,}i\ge 0})$

   $$V\left(\underset{\_}{\overline{u}}\right)={\displaystyle \sum _{i=0}^{\infty }{\underset{\_}{u}}_i^T{\underset{\_}{u}}_i}+2{\displaystyle \sum _{i=0}^{\infty }{\displaystyle \sum _{k=1}^{N-1}{\underset{\_}{u}}_i^T{C}^k{\underset{\_}{u}}_{i+k}}}-2{\displaystyle \sum _{i=0}^{\infty }{\displaystyle \sum _{j\in H^{\prime }}{\lambda }_{j\mathrm{,}i}{\underset{\_}{u}}_i^t{\underset{\_}{e}}_j}}\mathrm{,}$$

   with respect to ${\left({\underset{\_}{u}}_i\right)}_{i\ge 0}$  and comparing the derivatives to zero, equivalently, we need to show that there exist real numbers (Lagrange multipliers) ${\lambda }_{j\mathrm{,}l},j\in H^{\prime },l=0,1,\dots ,$  such that

   $$\mathcal{D}\underset{\_}{\overline{w}}=\left[ℐ+{\displaystyle \sum _{k=1}^{N-1}}\left(C^k{ℒ}^k+{\left(C^T\right)}^k{ℛ}^k\right)\right]\underset{\_}{\overline{w}}=\underset{\_}{\overline{\Lambda }}\mathrm{,}(6.13)$$

   where $\underset{\_}{\overline{w}}$  is defined in (6.11) and $\underset{\_}{\overline{\Lambda }}=\left({\underset{\_}{\Lambda }}_0\mathrm{,}{\underset{\_}{\Lambda }}_1\mathrm{,}\dots \right)$  with

   $${\underset{\_}{\Lambda }}_l={\displaystyle \sum _{j\in H^{\prime }}{\lambda }_{j\mathrm{,}l}{\underset{\_}{e}}_j}\mathrm{,}l=0,1,\dots$$

2. The constraints (2.2) and (2.3) are satisfied for $\underset{\_}{\overline{w}}$  as defined in (6.11).

3. The basic recurrence (3.9) holds true with $\underset{\_}{\overline{w}}$  defined in (6.11), that is the sequence $\underset{\_}{\overline{r}}$  defined by

   $$\underset{\_}{\overline{r}}:=\left(ℐ-{\displaystyle \sum _{m=1}^p{a}_m{ℛ}^m}\right)\underset{\_}{\overline{w}}(6.14)$$

   has to satisfy

   $${ℒ}^{p+1}\underset{\_}{\overline{r}}=\underset{\_}{\overline{0}}(6.15)$$

   and for any $i=0,1,\dots \mathrm{,}p$

   $$\left(ℐ-ℛℒ\right){ℒ}^i\underset{\_}{\overline{r}}={\displaystyle \sum _{m=1}^p}\left[\left(v_i\left(d_m\right)I-v_{i-1}\left(d_m\right)C^T\right)N\left(d_m\right){\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}{\underset{\_}{e}}_j}\right]\overline{e}\mathrm{,}(6.16)$$

   where $N\left(d\right)=\Delta \left(I-dC\right).$

Ad. 1. We will show that (6.13) holds with

$${\lambda }_{j\mathrm{,}l}={\displaystyle \sum _{m=1}^p{c}_{j\mathrm{,}m}{d}_m^l}\mathrm{,}j\in H^{\prime }\mathrm{,}l=0,1,\dots (6.17)$$

By definition (6.11) of $\underset{\_}{\overline{w}}$  we have

$$\mathcal{D}\underset{\_}{\overline{w}}={\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}{\overline{d}}_m{\underset{\_}{\overline{e}}}_j}}=\left({\displaystyle \sum _{j\in H^{\prime }}{\displaystyle \sum _{m=1}^p{c}_{j\mathrm{,}m}{d}_m^l{\underset{\_}{e}}_j}}\mathrm{,}l=0,1,\dots \right)\mathrm{.}$$

Therefore, by definition of ${\lambda }_{j\mathrm{,}l}’\text{s}$  we obtain

$$\mathcal{D}\underset{\_}{\overline{w}}=\left({\displaystyle \sum _{j\in H^{\prime }}{\lambda }_{j\mathrm{,}l}{\underset{\_}{e}}_j}\right)=\left({\underset{\_}{\Lambda }}_0\mathrm{,}{\underset{\_}{\Lambda }}_1\mathrm{,}\dots \right)=\underset{\_}{\overline{\Lambda }}\mathrm{.}$$

To see that ${\lambda }_{j\mathrm{,}l}$  as defined through (6.17) are real numbers take first conjugates of both sides of $S\underset{\_}{c}=\underset{\_}{e}.$  Note that

$$S^{\ast }=S^{\ast }\left(d_1\mathrm{,}\dots \mathrm{,}{d}_p\right)=S\left(d_1^{\ast }\mathrm{,}\dots \mathrm{,}{d}_p^{\ast }\right)\mathrm{.}$$

Since $d_1\mathrm{,}\dots \mathrm{,}{d}_p$  are either real or come in conjugate pairs (see Rem. 3.1) the equation $S^{\ast }{\underset{\_}{c}}^{\ast }=\underset{\_}{e}$ implies that for any $j\in H^{\prime }$  and any $m=1,\dots \mathrm{,}p$  either $\Im d_m=0$  and then $c_{j\mathrm{,}m}$  is real or $\Im d_m\ne 0$  and then there exists $n\ne m$  (with $d_n^{\ast }=d_m)$  such that $c_{j,n}^{\ast }=c_{j\mathrm{,}m}.$  Therefore the quantities $c_{j\mathrm{,}m}{d}_m^l$  in (6.17) are either real or come in conjugate pairs. Consequently, by (6.17) it follows that ${\lambda }_{j\mathrm{,}l}$  is real.

Ad. 2. Note that applying (6.1) and (6.4) to (6.11) after an easy algebra we get

$${\underset{\_}{w}}_0={\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}N\left(d_m\right){\underset{\_}{e}}_j}}$$

and

$${\underset{\_}{w}}_i={\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}{d}_m^{i-1}}}\left(d_{m}I-C^T\right)N\left(d_m\right){\underset{\_}{e}}_j\mathrm{,}i=1,2,\dots$$

Let us rewrite the constraints (2.2) and (2.3) using the above formulas for ${\underset{\_}{w}}_0$  and ${\underset{\_}{w}}_i,i\ge 1.$  The constraint (2.2) for $i=0$  with ${\underset{\_}{w}}_0$  as defined above takes on the form

$${\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}}}{\underset{\_}{1}}^{T}N\left(d_m\right){\underset{\_}{e}}_j=1(6.18)$$

and for $i\ge 1$

$${\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}{d}_m^{i-1}{\underset{\_}{1}}^T}}\left(d_{m}I-C^T\right)N\left(d_m\right){\underset{\_}{e}}_j=0.(6.19)$$

The constraint (2.3) for $i=0,$  that is for ${\underset{\_}{w}}_0,$  has the form

$${\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}{\underset{\_}{e}}_k^T}}N\left(d_m\right){\underset{\_}{e}}_j=0,k\in H\mathrm{.}(6.20)$$

For $i>0$  it has the form

$${\displaystyle \sum _{m=1}^p{\displaystyle \sum _{j\in H^{\prime }}{c}_{j\mathrm{,}m}{d}_m^{i-1}{\underset{\_}{e}}_k^T}}\left(d_{m}I-C^T\right)N\left(d_m\right){\underset{\_}{e}}_j=0,k\in H\mathrm{.}(6.21)$$

Note that $N\times N$  matrix

$$N\left(d\right)=\frac{1}{1-{\rho }^2}\left[\begin{array}{ccccc}1& -\rho d& \ddots & 0& 0\\ 0& 1& \ddots & \ddots & 0\\ \ddots & \ddots & \ddots & \ddots & \ddots \\ 0& \ddots & \ddots & 1& -\rho d\\ 0& 0& \ddots & 0& 1-{\rho }^2\end{array}\right]$$

and $\left(dI-C^T\right)N\left(d\right)=\frac{d}{1-{\rho }^2}{H}_N\left(d\right)$  - see (3.8). Thus, by elementary computations, we get

$${\underset{\_}{e}}_k^{T}N\left(d\right){\underset{\_}{e}}_j=\frac{1}{1-{\rho }^2}\{\begin{array}{l}\begin{array}{ll}\left(N-1\right)\left(1-d\rho \right)+1-{\rho }^2\mathrm{,}\hfill & k=j=0,\hfill \\ 1-d\rho \mathrm{,}\hfill & k=0,j\in H\text{ or }k\in H\mathrm{,}j=0,\hfill \end{array}\\ \begin{array}{cc}\begin{array}{l}1,\\ -d\rho \mathrm{,}\\ 0,\end{array}& \begin{array}{l}k=j,\\ k=j-\mathrm{1,}\\ \text{otherwise}\mathrm{,}\end{array}\}k,j\in H\end{array}\end{array}(6.22)$$

and

$$\begin{array}{l}{\underset{\_}{e}}_k^T\left(dI-C^T\right)N\left(d\right){\underset{\_}{e}}_j\hfill \\ =\frac{1}{1-{\rho }^2}\{\begin{array}{l}\begin{array}{ll}\left(N-1\right)\left(1-d\rho \right)\left(d-\rho \right)+d\left(1-{\rho }^2\right)\mathrm{,}\hfill & k=j=0,\hfill \\ \left(1-d\rho \right)\left(d-\rho \right)\mathrm{,}\hfill & k=0,j\in H\text{ or }k\in H\mathrm{,}j=0,\hfill \end{array}\\ \begin{array}{cc}\begin{array}{l}-\rho \mathrm{,}\\ d\left(1+{\rho }^2\right)\mathrm{,}\\ -d^2\rho \mathrm{,}\\ 0,\end{array}& \begin{array}{l}k=j+\mathrm{1,}\\ k=j,\\ k=j-1,\\ \text{otherwise}\mathrm{,}\end{array}\}k\mathrm{,}j\in H.\end{array}\end{array}\hfill \end{array}(6.23)$$

Due to (6.22) and (6.23), the constraints (6.18), (6.19), (6.20) and (6.21) can be rewritten in a matrix form as

$$\left[\begin{array}{cccc}\tilde{G}\left(d_1\right)& \tilde{G}\left(d_2\right)& \cdots & \tilde{G}\left(d_p\right)\\ \overline{G}\left(d_1\right)& \overline{G}\left(d_2\right)& \cdots & \overline{G}\left(d_p\right)\\ d_1\overline{G}\left(d_1\right)& d_2\overline{G}\left(d_2\right)& \cdots & d_p\overline{G}\left(d_p\right)\\ ⋮& ⋮& \ddots & ⋮\\ d_1^i\overline{G}\left(d_1\right)& d_2^i\overline{G}\left(d_2\right)& \cdots & d_p^i\overline{G}\left(d_p\right)\\ ⋮& ⋮& \ddots & ⋮\end{array}\right]\underset{\_}{c}=\overline{e}\mathrm{,}(6.24)$$

where $\tilde{G}\left(d\right)$  is defined through (3.5) and (3.6),

$$\overline{G}\left(d\right)=\frac{d}{1-{\rho }^2}\left[\begin{array}{cc}{H}_{11}\left(d\right)& H_{12}\left(d\right)\\ H_{21}\left(d\right)& H_{22}\left(d\right)\end{array}\right]$$

with

$$\begin{array}{l}{H}_{11}\left(d\right)=\left(N-1\right)\left(1-\rho d\right)\left(1-\rho /d\right)+1-{\rho }^2\mathrm{,}\hfill \\ H_{12}=H_{21}^T=\left(1-\rho d\right)\left(1-\rho /d\right){\underset{\_}{1}}_h^T\mathrm{,}\hfill \\ H_{22}\left(d\right)\text{=diag}\left(H_1\left(d\right)\mathrm{,}\dots \mathrm{,}{H}_s\left(d\right)\right),\hfill \end{array}$$

and matrices $H_i\left(d\right),i=1,\dots \mathrm{,}s,$  are defined in (3.8).

The infinite matrix at the left hand side of (6.24) can be written as

$$\left[\begin{array}{ccccc}I& 0& 0& \cdots & 0\\ 0& I& I& \cdots & I\\ 0& d_{1}I& d_{2}I& \cdots & d_{p}I\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& d_1^{i}I& d_2^{i}I& \cdots & d_p^{i}I\\ ⋮& ⋮& ⋮& \ddots & ⋮\end{array}\right]\left[\begin{array}{cccc}\tilde{G}\left(d_1\right)& \tilde{G}\left(d_2\right)& \cdots & \tilde{G}\left(d_p\right)\\ \overline{G}\left(d_1\right)& 0& \cdots & 0\\ 0& \overline{G}\left(d_2\right)& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & \overline{G}\left(d_p\right)\end{array}\right]\mathrm{,}$$

where $I=I_{h+1}$  and $0=0_{h+1}$  are, respectively, $\left(h+1\right)\times \left(h+1\right)$  unit and zero matrices. Note that the first matrix in the product above is of full rank and can be written as

$$\left[\begin{array}{ccccc}1& 0& 0& \cdots & 0\\ 0& 1& 1& \cdots & 1\\ 0& d_1& d_2& \cdots & d_p\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& d_1^i& d_2^i& \cdots & d_p^i\\ ⋮& ⋮& ⋮& \ddots & ⋮\end{array}\right]\otimes I_{h+1}\mathrm{.}$$

Therefore (6.24) is equivalent to

$$\left[\begin{array}{cccc}\tilde{G}\left(d_1\right)& \tilde{G}\left(d_2\right)& \cdots & \tilde{G}\left(d_p\right)\\ \overline{G}\left(d_1\right)& 0& \cdots & 0\\ 0& \overline{G}\left(d_2\right)& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & \overline{G}\left(d_p\right)\end{array}\right]\underset{\_}{c}={\left(\mathrm{1,0,}\dots \mathrm{,}0\right)}^T\in {ℝ}^{\left(p+1\right)\left(h+1\right)}\mathrm{.}(6.25)$$

Assume that we prove that $\left(h+1\right)\times \left(h+1\right)$  matrices $\overline{G}\left(d_m\right),m=1,\dots \mathrm{,}p,$  are singular. Note that $d\left[H_{21}\left(d\right)\mathrm{,}{H}_{22}\left(d\right)\right]=G\left(d\right)$  due to (3.7). Therefore, the definition (3.4) of $S$  implies that (6.25) is equivalent to $S\underset{\_}{c}=\left(\mathrm{1,0,}\dots \mathrm{,0}\right)\in {ℝ}^{ph+h+1}.$  It is obtained from (6.25) by deleting all rows determined through first rows of matrices $\overline{G}\left(d_m\right),m=1,\dots \mathrm{,}p.$  And the equation $S\underset{\_}{c}=\left(\mathrm{1,0,}\dots \mathrm{,0}\right)$  follows by ASSUMPTION II and the definition of $\underset{\_}{c}.$

Consequently, it suffices to show that $\det \overline{G}\left(d_m\right)=0,m=1,\dots \mathrm{,}p.$  That is, we need to check that

$$0=\det \left[\begin{array}{cc}{H}_{11}\left(d_m\right)& H_{12}\left(d_m\right)\\ H_{21}\left(d_m\right)& H_{22}\left(d_m\right)\end{array}\right]$$

for any $m=1,\dots \mathrm{,}p.$

Note that with $d=d_m$  the right hand side can be written as

$$\det H_{22}\left(d\right)\det \left[H_{11}\left(d\right)-H_{12}\left(d\right)H_{22}^{-1}\left(d\right)H_{21}\left(d\right)\right]$$

and

$$\det H_{22}\left(d\right)={\displaystyle \prod _{i=1}^s}\det H_{m_i}\left(d\right).(6.26)$$

Since $H_m\left(d\right)$  can be decomposed as

$$H_m\left(d\right)=D_m^{-1}{R}_m{D}_m\mathrm{,}(6.27)$$

where $D_m\text{=diag}\left(\mathrm{1,}d\mathrm{,}{d}^2\mathrm{,}\dots \mathrm{,}{d}^{m-1}\right)$  and $R_m$  is defined in (3.2) we see that

$$\det H_m\left(d\right)=1+{\rho }^2+\dots +{\rho }^{2m}\ne 0.$$

Now, from (6.26) it follows that $\det H_{22}\ne 0.$

On the other hand

$$\det \left[H_{11}\left(d\right)-H_{12}\left(d\right)H_{22}^{-1}\left(d\right)H_{21}\left(d\right)\right]=\left(N-1\right)\alpha \left(\rho \mathrm{,}d\right)+1-{\rho }^2-{\alpha }^2\left(\rho \mathrm{,}d\right){\displaystyle \sum _{j=1}^s{\underset{\_}{1}}^T}{H}_{m_j}^{-1}\underset{\_}{1}\mathrm{,}(6.28)$$

where $\alpha \left(\rho \mathrm{,}d\right)=1+{\rho }^2-\left(d+d^{-1}\right)\rho .$

The decomposition (6.27) of $H_m$  gives

$${\underset{\_}{1}}^T{H}_m^{-1}\underset{\_}{1}=\text{tr}\left({\underset{\_}{1}}^T{D}_m^{-1}{R}_m^{-1}{D}_m\underset{\_}{1}\right)\text{=tr}\left(D_m{\underset{\_}{11}}^T{D}_m^{-1}{R}_m^{-1}\right).$$

Moreover, since $\text{tr}\left(A\right)\text{=tr}\left(A^T\right)$

$${\underset{\_}{1}}^T{H}_m^{-1}\underset{\_}{1}=\text{tr}\left({\left(D_m{\underset{\_}{11}}^T{D}_m^{-1}{R}_m^{-1}\right)}^T\right)\text{=tr}\left(R_m^{-1}{D}_m^{-1}{\underset{\_}{11}}^T{D}_m\right)\text{=tr}\left(D_m^{-1}{\underset{\_}{11}}^T{D}_m{R}_m^{-1}\right)\mathrm{.}$$

Combining the last two expressions for ${\underset{\_}{1}}^T{H}_m^{-1}\underset{\_}{1}$  we get

$${\underset{\_}{1}}^T{H}_m^{-1}\underset{\_}{1}=\text{tr}\left(\frac{1}{2}\left(D_m{\underset{\_}{11}}^T{D}_m^{-1}+D_m^{-1}{\underset{\_}{11}}^T{D}_m\right)R_m^{-1}\right)\mathrm{.}$$

Note that

$${\left(D_m{\underset{\_}{11}}^T{D}_m^{-1}+D_m^{-1}{\underset{\_}{11}}^T{D}_m\right)}_{ij}=d^{\left|i-j\right|}+d^{-\left|i-j\right|}\mathrm{,}$$

and that

$$\frac{1}{2}\left(d^k+d^{-k}\right)=T_k\left(\frac{1}{2}\left(d+d^{-1}\right)\right)\mathrm{,}k=0,1,\dots \mathrm{,}$$

where $\left(T_k\right)$  is the $k^{\text{th}}$  Chebyshev polynomials of the first type.

Thus

$${\underset{\_}{1}}^T{H}_m^{-1}\underset{\_}{1}=\text{tr}{T}_m\left(x\right)R_m^{-1}\mathrm{,}$$

where $x=x\left(d\right)=\frac{1}{2}\left(d+d^{-1}\right)$  and the matrix $T_m$  is defined in (3.1). Plugging this expression to (6.28) we find out that

$$\det \left(H_{11}\left(d\right)-H_{12}\left(d\right)H_{22}^{-1}\left(d\right)H_{21}\left(d\right)\right)=Q_p\left(x\left(d\right)\right)\mathrm{,}$$

where $Q_p$  is the polynomial defined in (3.3). By ASSUMPTION I $Q_p\left(x\left(d_m\right)\right)=0,$  thus the above equality gives $\det \overline{G}\left(d_m\right)=0,m=1,\dots \mathrm{,}p.$  Finally, we conclude that the constraints (2.2) and (2.3) are satisfied and thus the proof of point 2 is completed.

Ad. 3. First, we will show that for $\underset{\_}{\overline{r}}$  defined by (6.14) the identity (6.15) holds. To this end observe that by (6.6) for $i=p,$  (6.10) and (6.13)

$${ℒ}^{p+1}\left(ℐ-{\displaystyle \sum _{m=1}^p{a}_m{ℛ}^m}\right)\underset{\_}{\overline{w}}=ℒ\left({ℒ}^p-{\displaystyle \sum _{m=1}^p{a}_m{ℒ}^{p-m}}\right){\mathcal{D}}^{-1}\underset{\_}{\overline{\Lambda }}=ℒ{\mathcal{D}}^{-1}\left({ℒ}^p-{\displaystyle \sum _{m=1}^p{a}_m{ℒ}^{p-m}}\right)\underset{\_}{\overline{\Lambda }}\mathrm{.}$$

Note also that for any $j=1,\dots \mathrm{,}p$  by (6.8)

$$\left({ℒ}^p-{\displaystyle \sum _{m=1}^p{a}_m{ℒ}^{p-m}}\right){\overline{d}}_j=v_p\left(d_j\right){\overline{d}}_j\mathrm{.}$$

By the definition (3.10) of $a_m,m=1,\dots \mathrm{,}p$  it follows that $v_p\left(d_j\right)=0.$  Due to the definition of $\Lambda$  through (6.17) we conclude that ${ℒ}^{p+1}\underset{\_}{\overline{r}}=\underset{\_}{\overline{0}}.$

In order to check (6.16) first we note that due to (6.10) it follows from (6.3) and (6.5) that for $\overline{y}={\left(y^n\right)}_{n\ge 0}$  and $\underset{\_}{\overline{x}}=\left(\underset{\_}{x}\mathrm{,}\underset{\_}{x}\mathrm{,}\dots \right)$

$${\mathcal{D}}^{-1}\overline{y}\underset{\_}{\overline{x}}=\left(ℐ-C^Tℛ\right)N\left(y\right)\overline{y}\underset{\_}{\overline{x}}\mathrm{.}$$

Therefore for any $i\ge 0$  any $d_j$  and ${\underset{\_}{e}}_{j_k}$  by (6.6)

$$\begin{array}{lll}{ℒ}^i\left(ℐ-{\displaystyle \sum _{m=1}^p{a}_m{ℛ}^m}\right){\mathcal{D}}^{-1}{\overline{d}}_j{\underset{\_}{\overline{e}}}_{j_k}\hfill & =\hfill & \left({ℒ}^p-{\displaystyle \sum _{m=1}^p{a}_m{ℒ}^{p-m}}\right){ℛ}^{p-i}{\overline{d}}_{j}N\left(d_j\right){\underset{\_}{\overline{e}}}_{j_k}\hfill \\ \hfill & -\hfill & \left({ℒ}^p-{\displaystyle \sum _{m=1}^p{a}_m{ℒ}^{p-m}}\right){ℛ}^{p-\left(i-1\right)}{\overline{d}}_j{C}^{T}N\left(d_j\right){\underset{\_}{\overline{e}}}_{j_k}\mathrm{.}\hfill \end{array}$$

Finally, we use (6.7) with $\overline{y}={\overline{d}}_j,\underset{\_}{\overline{x}}=N\left(d_j\right){\underset{\_}{\overline{e}}}_{j_k}$  to the first part and with $\overline{y}={\overline{d}}_j,$   $\underset{\_}{\overline{x}}=C^{T}N\left(d_j\right){\underset{\_}{\overline{e}}}_{j_k}$  to the second part of the expression at the right hand side of the equation above arriving at

$$\left(ℐ-ℛℒ\right){ℒ}^i\left(ℐ-{\displaystyle \sum _{m=1}^p{a}_m{ℛ}^m}\right){\mathcal{D}}^{-1}{\overline{d}}_j{\underset{\_}{\overline{e}}}_{j_k}=\left(v_i\left(d_j\right)I-v_{i-1}\left(d_j\right)C^T\right)N\left(d_j\right)\left({\underset{\_}{e}}_{j_k},\underset{\_}{0}\mathrm{,}\underset{\_}{0}\mathrm{,}\dots \right)\mathrm{.}$$

Thus (6.16) holds true.

Finally we will prove the formula (3.12) for the variance of the BLUE ${\widehat{\mu }}_t.$  To this end we observe first that

$$ℂ\text{ov}\left({\widehat{\mu }}_t\mathrm{,}{\underset{\_}{X}}_{t-i}\right)={\underset{\_}{w}}_i+{\displaystyle \sum _{k=1}^{N-1}{C}^k{\underset{\_}{w}}_{i+k}}+{\displaystyle \sum _{k=1}^{i\wedge \left(N-1\right)}}{\left(C^T\right)}^k{\underset{\_}{w}}_{i-k}$$

for any $i=0,1,\dots .$  On the other hand, due to (6.13), we see that the right hand side of the above equality is equal to ${\underset{\_}{\Lambda }}_i.$  That is, for any $i=0,1,\dots$

$$ℂ\text{ov}\left({\widehat{\mu }}_t\mathrm{,}{\underset{\_}{X}}_{t-i}\right)={\displaystyle \sum _{j\in H^{\prime }}{\lambda }_{j\mathrm{,}i}{\underset{\_}{e}}_j}\mathrm{.}$$

Now, we write

$$\mathbb{V}\text{ar}{\widehat{\mu }}_t={\displaystyle \sum _{i=0}^{\infty }{\underset{\_}{w}}_i^T}ℂ\text{ov}\left({\widehat{\mu }}_t\mathrm{,}{\underset{\_}{X}}_{t-i}\right)={\displaystyle \sum _{i=0}^{\infty }{\displaystyle \sum _{j\in H^{\prime }}{\lambda }_{j\mathrm{,}i}{\underset{\_}{w}}_i^T{\underset{\_}{e}}_j}}\mathrm{.}$$

Due to the constraints (2.2) and (2.3) it follows from the above formula that $\mathbb{V}\text{ar}{\widehat{\mu }}_t={\lambda }_{\mathrm{0,0}}.$  Thus, (3.12) follows from (6.17).

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