The superposition of Markovian arrival processes: moments and the minimal Laplace transform

Abstract

The superposition of two independent Markovian arrival processes (MAPs) is also a Markovian arrival process of which the Markovian representation is given as the Kronecker sum of the transition rate matrices of the component processes. The moments of stationary intervals of the superposition can be obtained by differentiating the Laplace transform (LT) given in terms of the transition rate matrices. In this paper, we propose a streamlined procedure to determine the minimal LT of the merged process in terms of the minimal LT coefficients of the component processes. Combined with the closed-form transformation between moments and LT coefficients, our result enables us to determine the moments of the superposed process based on the moments of the component processes. The main contribution is that the whole procedure can be implemented without explicit Markovian representations. In order to transform the minimal LT coefficients of the component processes into the minimal LT representation of the merged process, we also introduce another minimal representation. A numerical example is provided to illustrate the procedure.

Appendix

1.1 Proof of Lemma 1

Proof

Let $\tilde{f}(s) \equiv h(s)/g(s)$. Then, by Leibniz’s rule, the k-th order derivative of $\tilde{f}(s) g(s)= h(s)$ is given as

$$\begin{aligned} \sum _{i=0}^{k} \left( {\begin{array}{c}k\\ i\end{array}}\right) \tilde{f}^{(k-i)}(s)g^{(i)}(s)&=h^{(k)}(s) \end{aligned}$$

by which we have

$$\begin{aligned} \sum _{i=0}^{k} \frac{f^{(k-i)}(0) }{(k-i)!} ~ \frac{g^{(i)}(0)}{i!}&= \frac{h^{(k)}(0)}{k!} \end{aligned}$$

and the result follows by Eqs. (2.5) and (3.1).$\square$

1.2 Proof of Corollary 1

Proof

In order to determine $(\varvec{a}, \varvec{b})$, we need $2n-1$ independent equations. For $k = 1, 2,\ldots ,n-1$, Eq. (3.2) can be written as $\textbf{R}_2 \varvec{a} =\varvec{b}$. For $k = n, n+1,\ldots ,2n-1$, Eq. (3.2) can be written as

$$\begin{aligned} \sum _{i=0}^{n-1} (-1)^{k-i} r_{k-i} a_i = (-1)^{k-n+1} r_{k-n} \hspace{0.5in} \end{aligned}$$

or in matrix form

$$\begin{aligned} \left[ \begin{array}{c c c c c c } (-1)^n r_n &{} (-1)^{n-1}r_{n-1} &{} \cdots &{} r_2 &{} -r_1 \\ (-1)^{n+1} r_{n+1} &{} (-1)^n r_n &{} \cdots &{} -r_3 &{} r_2 \\ \vdots &{}\vdots &{} \ddots &{} \vdots &{} \vdots \\ r_{2n-2} &{} -r_{2n-3} &{} \cdots &{} (-1)^n r_n &{} (-1)^{n-1} r_{n-1} \\ -r_{2n-1} &{} r_{2n-2} &{} \cdots &{} (-1)^{n+1} r_{n+1} &{} (-1)^n r_n \\ \end{array} \right] \left[ \begin{array}{c} a_0 \\ a_1 \\ a_2 \\ \vdots \\ a_{n-1} \end{array} \right] = \left[ \begin{array}{c} -r_0 \\ r_1 \\ -r_2 \\ \vdots \\ (-1)^n r_{n-1} \end{array} \right] \end{aligned}$$

where $r_0 =1$. By multiplying every other equation by ($-1$) starting from the first one, we have $\textbf{R}_1 \varvec{a} = \varvec{r}_{0..n-1}$. That is, for $1 \le k \le 2n-1$, Eq. (3.2) can be written

$$\begin{aligned} \left[ \begin{array}{c} \textbf{R}_2 \\ \textbf{R}_1 \end{array} \right] \varvec{a} = \left[ \begin{array}{c} \varvec{b} \\ \varvec{r}_{0..n-1} \end{array} \right] . \end{aligned}$$

$\square$

1.3 Proof of Lemma 2

Proof

Let $w(s,t) = \sum _{i=1}^{n-1}\sum _{j=1}^{n-1} c_{i,j} s^i t^j$, $h(s) =\sum _{i=1}^{n-1} b_{i} s^i+ a_0$, $g(s)=s^n +\sum _{i=0}^{n-1} a_i s^i$. Then, the joint LT in Eq. (2.12) can be written as

$$\begin{aligned} \tilde{f} (s, t)&= \frac{w(s,t) + a_0 (h(s)+ h(t))- a_0^2}{ g(s)g(t)}. \end{aligned}$$

Let $w_{k, l}(s,t) = \partial ^{k+l} w(s,t)/( \partial s^k \partial t^l)$ for which we have

$$\begin{aligned} \frac{w_{k, l}(0,0)}{k!l!}&= \left\{ \begin{array}{l l} c_{k,l} &{} \text{ for } 0 \le k, l \le n-1 \\ 0 &{} \text{ for } k \ge n \text{ or } l \ge n. \end{array} \right. \end{aligned}$$

By Leibniz’s rule with respect to s and t, it can be shown that

$$\begin{aligned} w_{k, l}(s,t)&= \sum _{i=0}^{k} \sum _{j=0}^{l} \left( {\begin{array}{c}k\\ i\end{array}}\right) \left( {\begin{array}{c}l\\ j\end{array}}\right) \tilde{f}_{i,j}(s,t) g^{(k-i)}(s)g^{(l-j)}(t) \end{aligned}$$

by which

$$\begin{aligned} \frac{w_{k, l}(0,0)}{k!l!}&= \sum _{i=0}^{k} \sum _{j=0}^{l} \frac{\tilde{f}_{i,j}(0,0)}{i! j!} \frac{g^{(k-i)} (0)}{(k-i)!} \frac{g^{(l-j)}(0)}{(l-j)!} \end{aligned}$$

and the result follows by Eqs. (2.6) and (3.1). $\square$

1.4 Proof of Lemma 3

Proof

Note that

$$\begin{aligned} (\textbf{D}_0^\oplus )^k \textbf{D}_1^\oplus&= (\hat{\textbf{D}}_0 \otimes \textbf{I}_n + \textbf{I}_m \otimes \check{\textbf{D}}_0)^k (\hat{\textbf{D}}_1 \otimes \textbf{I}_n + \textbf{I}_m \otimes \check{\textbf{D}}_1) \nonumber \\&=\sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) (\hat{\textbf{D}}_0^{k-i} \otimes \check{\textbf{D}}_0^i)(\hat{\textbf{D}}_1 \otimes \textbf{I}_n + \textbf{I}_m \otimes \check{\textbf{D}}_1) \nonumber \\&=\sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) \left( \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^i + \hat{\textbf{D}}_0^{k-i} \otimes \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \right) . \end{aligned}$$

(A.1)

Let $\varvec{e}^\otimes = \varvec{e}_m \otimes \varvec{e}_n$. The subscript of $\varvec{e}_m$ and $\varvec{e}_n$ shall be suppressed below. Note that $\hat{\textbf{D}}_0 \varvec{e}= -\hat{\textbf{D}}_1 \varvec{e}$ since $\hat{\textbf{Q}} \varvec{e} = (\hat{\textbf{D}}_0+\hat{\textbf{D}}_1) \varvec{e} = \textbf{0}$. By Eq. (2.2) and the definition of $\hat{y}_i$,

$$\begin{aligned} \hat{\varvec{p}}\hat{\textbf{D}}_0^i \varvec{e}&= -\hat{\varvec{p}}\hat{\textbf{D}}_0^{i-1} \hat{\textbf{D}}_1 \varvec{e} = - \hat{y}_{i-1},\\ \hat{\varvec{\pi }}\hat{\textbf{D}}_0^i \varvec{e}&= \hat{\varvec{\pi }}\hat{\textbf{D}}_0 \hat{\textbf{D}}_0^{i-1} \varvec{e} = -\hat{\lambda }_A \hat{\varvec{p}} \hat{\textbf{D}}_0^{i-1} \varvec{e} = \hat{\lambda }_A \hat{y}_{i-2}, \\ \hat{\varvec{\pi }}\hat{\textbf{D}}_0^i \hat{\textbf{D}}_1 \varvec{e}&= -\hat{\varvec{\pi }}\hat{\textbf{D}}_0^{i+1} \varvec{e} = - \hat{\lambda }_A \hat{y}_{i-1}. \end{aligned}$$

Likewise, $\check{\varvec{p}}\check{\textbf{D}}_0^i \varvec{e} = - \check{y}_{i-1}$, $\check{\varvec{\pi }}\check{\textbf{D}}_0^i \varvec{e} = \check{\lambda }_A \check{y}_{i-2}$, and $\check{\varvec{\pi }}\check{\textbf{D}}_0^i \check{\textbf{D}}_1 \varvec{e} = - \check{\lambda }_A \check{y}_{i-1}$. By these identities and the definition of $\hat{y}_k$ and $\check{y}_k$,

$$\begin{aligned} (\hat{\varvec{p}}\otimes \check{\varvec{\pi }}) \left( \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^i + \hat{\textbf{D}}_0^{k-i} \otimes \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \right) \varvec{e}^\otimes&=\hat{\varvec{p}}\hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \varvec{e} \check{\varvec{\pi }} \check{\textbf{D}}_0^i \varvec{e} + \hat{\varvec{p}} \hat{\textbf{D}}_0^{k-i}\varvec{e} \check{\varvec{\pi }} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \varvec{e} \nonumber \\&=\hat{y}_{k-i} \check{\lambda }_A \check{y}_{i-2}+ \hat{y}_{k-i-1}\check{\lambda }_A \check{y}_{i-1}, \end{aligned}$$

(A.2)

$$\begin{aligned} (\hat{\varvec{\pi }}\otimes \check{\varvec{p}}) \left( \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^i + \hat{\textbf{D}}_0^{k-i} \otimes \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \right) \varvec{e}^\otimes&=\hat{\varvec{\pi }}\hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \varvec{e} \check{\varvec{p}} \check{\textbf{D}}_0^i \varvec{e} + \hat{\varvec{\pi }}\hat{\textbf{D}}_0^{k-i} \varvec{e} \check{\varvec{p}} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \varvec{e} \nonumber \\&= \hat{\lambda }_A \hat{y}_{k-i-1} \check{y}_{i-1} + \hat{\lambda }_A \hat{y}_{k-i-2} \check{y}_i. \end{aligned}$$

(A.3)

Let $\left( {\begin{array}{c}n\\ i\end{array}}\right) = 0$ for $i<0$ or $i> n$. Then, by Pascal’s identity,

$$\begin{aligned} \sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) \left( \hat{y}_{k-i} \check{y}_{i-2} + \hat{y}_{k-i-1} \check{y}_{i-1} \right)&=\sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) \hat{y}_{k-i} \check{y}_{i-2} + \sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) \hat{y}_{k-i-1} \check{y}_{i-1} \nonumber \\&=\sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) \hat{y}_{k-i} \check{y}_{i-2} + \sum _{i=1}^{k+1} \left( {\begin{array}{c}k\\ i-1\end{array}}\right) \hat{y}_{k-i} \check{y}_{i-2} \nonumber \\&=\sum _{i=0}^{k+1} \left( {\begin{array}{c}k\\ i\end{array}}\right) \hat{y}_{k-i} \check{y}_{i-2} + \sum _{i=0}^{k+1} \left( {\begin{array}{c}k\\ i-1\end{array}}\right) \hat{y}_{k-i} \check{y}_{i-2} \nonumber \\&=\sum _{i=0}^{k+1} \left( {\begin{array}{c}k+1\\ i\end{array}}\right) \hat{y}_{k-i} \check{y}_{i-2}. \end{aligned}$$

(A.4)

Likewise,

$$\begin{aligned} \sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) \left( \hat{y}_{k-i-1} \check{y}_{i-1} + \hat{y}_{k-i-2} \check{y}_i \right)&=\sum _{i=0}^{k+1} \left( {\begin{array}{c}k+1\\ i\end{array}}\right) \hat{y}_{k-i-1} \check{y}_{i-1}, \end{aligned}$$

(A.5)

By Eqs. (5.1), (A.1), (A.2), (A.3), (A.4), and (A.5),

$$\begin{aligned} y^\oplus _k = \varvec{p}^\oplus (\textbf{D}_0^\oplus )^k \textbf{D}_1^\oplus \varvec{e}&=\frac{1}{\lambda _A^\oplus } \left( \hat{\lambda }_A (\hat{\varvec{p}}\otimes \check{\varvec{\pi }}) + \check{\lambda }_A (\hat{\varvec{\pi }}\otimes \check{\varvec{p}}) \right) (\textbf{D}_0^\oplus )^k \textbf{D}_1^\oplus \varvec{e} \\&=\frac{\hat{\lambda }_A \check{\lambda }_A}{\lambda _A^\oplus } \sum _{i=0}^k \left( {\begin{array}{c}k\\ i\end{array}}\right) \left( \hat{y}_{k-i} \check{y}_{i-2} + 2\hat{y}_{k-i-1} \check{y}_{i-1} + \hat{y}_{k-i-2} \check{y}_i \right) \\&=\frac{\hat{\lambda }_A \check{\lambda }_A}{\lambda _A^\oplus } \sum _{i=0}^{k+1} \left( {\begin{array}{c}k+1\\ i\end{array}}\right) \left( \hat{y}_{k-i} \check{y}_{i-2} + \hat{y}_{k-i-1} \check{y}_{i-1}\right) \\&=\frac{\hat{\lambda }_A \check{\lambda }_A}{\lambda _A^\oplus } \sum _{i=0}^{k+2} \left( {\begin{array}{c}k+2\\ i\end{array}}\right) \hat{y}_{k-i} \check{y}_{i-2} \end{aligned}$$

where the last equality is due to the same argument used in (A.4). $\square$

1.5 Proof of Lemma 4

Proof

By Eq. (A.1),

$$\begin{aligned} (\textbf{D}_0^\oplus )^k \textbf{D}_1^\oplus (\textbf{D}_0^\oplus )^l \textbf{D}_1^\oplus&= \sum _{i=0}^k \sum _{j=0}^l \left( {\begin{array}{c}k\\ i\end{array}}\right) \left( {\begin{array}{c}l\\ j\end{array}}\right) \left( \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^{i+j} +\hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \otimes \check{\textbf{D}}_0^{i+j} \check{\textbf{D}}_1 \right. \nonumber \\&~~~~~~~~~~~~~~~ \left. + \hat{\textbf{D}}_0^{k-i+l-j} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j + \hat{\textbf{D}}_0^{k-i+l-j} \otimes \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \check{\textbf{D}}_1 \right) . \end{aligned}$$

By Eq. (2.2) and the definition of $\hat{z}_{ij}$,

$$\begin{aligned} \hat{\varvec{p}} \hat{\textbf{D}}_0^i \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^j\varvec{e}&=-\hat{\varvec{p}} \hat{\textbf{D}}_0^i \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{j-1}\hat{\textbf{D}}_1 \varvec{e} =- z_{i,j-1},\\ \hat{\varvec{\pi }}\hat{\textbf{D}}_0^i \varvec{e}&= \hat{\varvec{\pi }}\hat{\textbf{D}}_0 \hat{\textbf{D}}_0^{i-1} \varvec{e} = -\lambda _A \varvec{p}\hat{\textbf{D}}_0^{i-1} \varvec{e} = \lambda _A y_{i-2}, \\ \hat{\varvec{\pi }}\hat{\textbf{D}}_0^i \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^j\varvec{e}&= \lambda _A \varvec{p} \hat{\textbf{D}}_0^{i-1}\hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{j-1} \hat{\textbf{D}}_1 \varvec{e} = \lambda _A z_{i-1,j-1},\\ \hat{\varvec{\pi }}\hat{\textbf{D}}_0^i \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^j \hat{\textbf{D}}_1 \varvec{e}&= -\lambda _A \varvec{p} \hat{\textbf{D}}_0^{i-1}\hat{\textbf{D}}_1 \hat{\textbf{D}}_0^j \hat{\textbf{D}}_1 \varvec{e} = -\lambda _A z_{i-1,j}. \end{aligned}$$

Likewise, $\check{\varvec{p}} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j\varvec{e} =-z_{i,j-1}$, $\check{\varvec{\pi }}\check{\textbf{D}}_0^i \varvec{e} = \lambda _A y_{i-2}$, $\check{\varvec{\pi }}\check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j\varvec{e} = \lambda _A z_{i-1,j-1}$, and $\check{\varvec{\pi }}\check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \check{\textbf{D}}_1 \varvec{e} = -\lambda _A z_{i-1,j}$. By these identities and the definition of $\hat{y}_k$, $\check{y}_k$, $\hat{z}_{ij}$, and $\check{z}_{ij}$, we have

$$\begin{aligned} (\hat{\varvec{p}}\otimes \check{\varvec{\pi }}) ( \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^{i+j} )\varvec{e}^\otimes&= \hat{\varvec{p}} \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \hat{\textbf{D}}_1 \varvec{e} \check{\varvec{\pi }} \check{\textbf{D}}_0^{i+j} \varvec{e} = \hat{z}_{k-i,l-j} \check{\lambda }_A \check{y}_{i+j-2},\\ (\hat{\varvec{p}}\otimes \check{\varvec{\pi }})(\hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \otimes \check{\textbf{D}}_0^{i+j} \check{\textbf{D}}_1)\varvec{e}^\otimes&=\hat{\varvec{p}} \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \varvec{e} \check{\varvec{\pi }} \check{\textbf{D}}_0^{i+j} \check{\textbf{D}}_1 \varvec{e} =\hat{z}_{k-i,l-j-1} \check{\lambda }_A \check{y}_{i+j-1},\\ (\hat{\varvec{p}}\otimes \check{\varvec{\pi }})(\hat{\textbf{D}}_0^{k-i+l-j} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j )\varvec{e}^\otimes&= \hat{\varvec{p}}\hat{\textbf{D}}_0^{k-i+l-j} \hat{\textbf{D}}_1 \varvec{e} \check{\varvec{\pi }} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \varvec{e} = \hat{y}_{k-i+l-j} \check{\lambda }_A \check{z}_{i-1,j-1},\\ (\hat{\varvec{p}}\otimes \check{\varvec{\pi }})(\hat{\textbf{D}}_0^{k-i+l-j} \otimes \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \check{\textbf{D}}_1 ) \varvec{e}^\otimes&= \hat{\varvec{p}} \hat{\textbf{D}}_0^{k-i+l-j} \varvec{e} \check{\varvec{\pi }} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \check{\textbf{D}}_1 \varvec{e} = \hat{y}_{k-i+l-j-1} \check{\lambda }_A \check{z}_{i-1,j},\\ (\hat{\varvec{\pi }}\otimes \check{\varvec{p}}) ( \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \hat{\textbf{D}}_1 \otimes \check{\textbf{D}}_0^{i+j} )\varvec{e}^\otimes&= \hat{\varvec{\pi }} \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \hat{\textbf{D}}_1 \varvec{e} \check{\varvec{p}} \check{\textbf{D}}_0^{i+j} \varvec{e}= \hat{\lambda }_A \hat{z}_{k-i-1,l-j} \check{y}_{i+j-1},\\ (\hat{\varvec{\pi }}\otimes \check{\varvec{p}}) (\hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \varvec{e} \check{\textbf{D}}_0^{i+j} \check{\textbf{D}}_1)\varvec{e}^\otimes&= \hat{\varvec{\pi }} \hat{\textbf{D}}_0^{k-i} \hat{\textbf{D}}_1 \hat{\textbf{D}}_0^{l-j} \varvec{e} \check{\varvec{p}} \check{\textbf{D}}_0^{i+j} \check{\textbf{D}}_1 \varvec{e} = \hat{\lambda }_A \hat{z}_{k-i-1,l-j-1} \check{y}_{i+j} ,\\ (\hat{\varvec{\pi }}\otimes \check{\varvec{p}}) (\hat{\textbf{D}}_0^{k-i+l-j} \hat{\textbf{D}}_1 \varvec{e} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j )\varvec{e}^\otimes&= \hat{\varvec{\pi }} \hat{\textbf{D}}_0^{k-i+l-j} \hat{\textbf{D}}_1 \varvec{e} \check{\varvec{p}} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \varvec{e}= \hat{\lambda }_A \hat{y}_{k-i+l-j-1} \check{z}_{i,j-1}, \\ (\hat{\varvec{\pi }}\otimes \check{\varvec{p}}) (\hat{\textbf{D}}_0^{k-i+l-j} \varvec{e} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \check{\textbf{D}}_1 )\varvec{e}^\otimes&= \hat{\varvec{\pi }}\hat{\textbf{D}}_0^{k-i+l-j} \varvec{e} \check{\varvec{p}} \check{\textbf{D}}_0^i \check{\textbf{D}}_1 \check{\textbf{D}}_0^j \check{\textbf{D}}_1 \varvec{e} =\hat{\lambda }_A \hat{y}_{k-i+l-j-2} \check{z}_{i,j}. \end{aligned}$$

Then, by the same argument as used in (A.4) with Pascal’s identity,

$$\begin{aligned} y^\oplus _{k,l}&=\varvec{p}^\oplus (\textbf{D}_0^\oplus )^k \textbf{D}_1^\oplus (\textbf{D}_0^\oplus )^l \textbf{D}_1^\oplus \varvec{e}\\&=\frac{\hat{\lambda }_A \check{\lambda }_A}{\lambda _A^\oplus } \sum _{i=0}^k \sum _{j=0}^l \left( {\begin{array}{c}k\\ i\end{array}}\right) \left( {\begin{array}{c}l\\ j\end{array}}\right) \left( \hat{z}_{k-i,l-j} \check{y}_{i+j-2}+ \hat{z}_{k-i,l-j-1} \check{y}_{i+j-1} \right. \\&\left. + \hat{y}_{k-i+l-j} \check{z}_{i-1,j-1}+ \hat{y}_{k-i+l-j-1} \check{z}_{i-1,j} \right. \\&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \left. + \hat{z}_{k-i-1,l-j} \check{y}_{i+j-1}+ \hat{z}_{k-i-1,l-j-1} \check{y}_{i+j} \right. \\&\left. + \hat{y}_{k-i+l-j-1} \check{z}_{i,j-1}+ \hat{y}_{k-i+l-j-2} \check{z}_{i,j} \right) \\&=\frac{\hat{\lambda }_A \check{\lambda }_A}{\lambda _A^\oplus } \sum _{i=0}^{k+1} \sum _{j=0}^{l+1} \left( {\begin{array}{c}k+1\\ i\end{array}}\right) \left( {\begin{array}{c}l+1\\ j\end{array}}\right) \left( \hat{z}_{k-i,l-j} \check{y}_{i+j-2} + \hat{y}_{k-i+l-j} \check{z}_{i-1,j-1}\right) . \end{aligned}$$

$\square$