Novel advancements in the Markov chain stock model: analysis and inference

Abstract

In this paper we propose further advancements in the Markov chain stock model. First, we provide a formula for the second order moment of the fundamental price process with transversality conditions that avoid the presence of speculative bubbles. Second, we assume that the process of the dividend growth is governed by a finite state discrete time Markov chain and, under this hypothesis, we are able to compute the moments of the price process. We impose assumptions on the dividend growth process that guarantee finiteness of price and risk and the fulfilment of the transversality conditions. Subsequently, we develop non parametric statistical techniques for the inferential analysis of the model. We propose estimators of price, risk and forecasted prices and for each estimator we demonstrate that they are strongly consistent and that properly centralized and normalized they converge in distribution to normal random variables, then we give also the interval estimators. An application that demonstrate the practical implementation of methods and results to real dividend data concludes the paper.

Appendix: proofs

1.1 Proof of Proposition 2.3

For $k,j,i\in \mathbb N$ with $j>i$ we consider the following expectation:

$$\begin{aligned} \begin{aligned}&\mathbb E_{k}\left[ \prod _{h=1}^{i}\prod _{w=1}^{j}G(k+h)G(k+w)\right] = \mathbb E_{k}\left[ \prod _{h=1}^{i}G^{2}(k+h)\prod _{w=i+1}^{j}G(k+w)\right] \\&\quad =\mathbb E_{k}\left[ \prod _{h=1}^{i}G^{2}(k+h)\prod _{w=i+1}^{j-1}G(k+w)\mathbb E_{k+j-1}[G(k+j)]\right] \\&\quad \le \mathbb E_{k}\left[ \prod _{h=1}^{i}G^{2}(k+h)\prod _{w=i+1}^{j-1}G(k+w)\right] \overline{g}, \end{aligned} \end{aligned}$$

where the last inequality follows from (2.9). If we proceed to compute the expectation by conditioning up to time $i+1$ and we use at each step (2.9) we get

$$\begin{aligned} \begin{aligned}&\mathbb E_{k}\left[ \prod _{h=1}^{i}\prod _{w=1}^{j}G(k+h)G(k+w)\right] \le \mathbb E_{k}\left[ \prod _{h=1}^{i}G^{2}(k+h)\right] (\overline{g})^{j-i}\\&\quad = \mathbb E_{k}\left[ \prod _{h=1}^{i-1}G^{2}(k+h)\mathbb E_{k+i-1}[G^{2}(k+i)]\right] (\overline{g})^{j-i} \\&\quad \le \mathbb E_{k}[\prod _{h=1}^{i-1}G^{2}(k+h)]\overline{g}^{(2)}(\overline{g})^{j-i}, \end{aligned} \end{aligned}$$

(5.1)

where the last inequality follows from (2.11). If we proceed to compute the expectation by conditioning up to time 1 and we use at each step (2.11) we get

$$\begin{aligned} \mathbb E_{k}[\prod _{h=1}^{i}\prod _{w=1}^{j}G(k+h)G(k+w)]\le (\overline{g}^{(2)})^{i}(\overline{g})^{j-i}. \end{aligned}$$

(5.2)

Therefore

$$\begin{aligned} p^{(2)}(k)=\sum _{i=1}^{+\infty }\frac{\mathbb E_{k}[D^{2}(k+i)]}{r^{2i}}+2\sum _{i=1}^{+\infty }\sum _{j>i}\frac{\mathbb E_{k}[D(k+i)D(k+j)]}{r^{i+j}}. \end{aligned}$$

(5.3)

Consequently, we obtain

$$\begin{aligned} \begin{aligned} p^{(2)}(k) =&\sum _{i=1}^{+\infty }\frac{\mathbb E_{k}[\prod _{j=1}^{i}G^{2}(k+j)]}{r^{2i}}d^{2}(k) \\&+2\sum _{i=1}^{+\infty }\sum _{j>i}\frac{\mathbb E_{k}[\prod _{h=1}^{i}G(k+h)\prod _{w=1}^{j}G(k+w)]}{r^{i+j}}d^{2}(k)\\ \le&\sum _{i=1}^{+\infty }\frac{(\overline{g}^{(2)})^{i}}{r^{2i}}d^{2}(k)+2\sum _{i=1}^{+\infty }\sum _{j>i}\frac{(\overline{g}^{(2)})^{i}(\overline{g})^{j-i}}{r^{i+j}}d^{2}(k). \end{aligned} \end{aligned}$$

(5.4)

From (5.4), A1 and A2, using the properties of geometric series, we obtain that $p^{(2)}(k):=\mathbb E_{k}[P^{2}(k)]<+ \infty$.

Since the expected values in Formula (5.3) do depend only on g(k), the second order moment can be expressed in the compact form $p^{(2)}(k)=\psi _{2}(g(k))d^{2}(k)$. Let us denote the second order price-dividend ratio by

$$\begin{aligned} \psi _{2}(g(k)):=\frac{p^{(2)}(k)}{d^{2}(k)}. \end{aligned}$$

(5.5)

Let $\overline{\psi }_{2}=\max _{i}(\psi _{2}(g_{i}));$ then $0\le \mathbb E_{k}[P^{2}(k)]\le \overline{\psi }_{2}\mathbb E_{k}[D^{2}(k+i)],$ which is equivalent to

$$\begin{aligned} \frac{\mathbb E_{k}[P^{2}(k+i)]}{r^{2i}}\le \overline{\psi }_{2}\frac{\mathbb E_{k}[D^{2}(k+i)]}{r^{2i}}. \end{aligned}$$

(5.6)

Since $p^{(2)}(k)=\mathbb E_{k}[P^{2}(k)]<+ \infty ,$ we have that $\lim _{i\rightarrow +\infty }\frac{\mathbb E_{k}[D^{2}(k+i)]}{r^{2i}}=0$ and hence from (5.6) we get

$$\begin{aligned} \lim _{i\rightarrow +\infty }\frac{\mathbb E_{k}[P^{2}(k+i)]}{r^{2i}}=0. \end{aligned}$$

(5.7)

It remains to prove that $\lim _{N\rightarrow +\infty } \sum _{i=1}^{N}\frac{\mathbb E_{k}[D(k+i)P(k+N)]}{r^{i+N}}=0$. From the Cauchy-Schwarz inequality we have that

$$\begin{aligned} \begin{aligned}&\lim _{N\rightarrow +\infty } \sum _{i=1}^{N}\frac{\mathbb E_{k}[D(k+i)P(k+N)]}{r^{i+N}} \\&\quad \le \lim _{N\rightarrow +\infty } \sum _{i=1}^{N}\Big (\frac{\mathbb E_{k}[D^{2}(k+i)]}{r^{i+N}}\Big )^{\frac{1}{2}}\Big (\frac{\mathbb E_{k}[P^{2}(k+N)]}{r^{2N}}\Big )^{\frac{1}{2}}\\&\quad =\lim _{N\rightarrow +\infty }\Big (\frac{\mathbb E_{k}[P^{2}(k+N)]}{r^{2N}}\Big )^{\frac{1}{2}}\lim _{N\rightarrow +\infty }\sum _{i=1}^{N}\Big (\frac{\mathbb E_{k}[D^{2}(k+i)]}{r^{i+N}}\Big )^{\frac{1}{2}}=0, \end{aligned} \end{aligned}$$

where the last equality holds true because the first factor is zero using (5.7), while the second one is finite as it follows from the finiteness of (5.3).

1.2 Proof of Proposition 2.4

Let $k\in \mathbb {I\!\!N }$ be the current time. At time k the dividend process and the growth dividend process assume two known values denoted by $D(k)=d(k)\in \mathbb {R }$ and $G(k)=g(k)\in E$, respectively. Let us consider first the case when $g(k)=g_{1}$. By combining Equations (5.5), (2.7) and (2.4) we get

$$\begin{aligned} \begin{aligned}&\psi _{2}(g_{1})d^{2}(k)=\mathbb E_{k}\left[ \frac{(G(k+1)d(k)+P(k+1))^{2}}{r^{2}}\right] \\&\quad =\mathbb E_{k}\left[ \frac{G^{2}(k+1)d^{2}(k)}{r^{2}}\right] +\mathbb E_{k}\left[ \frac{P^{2}(k+1)}{r^{2}}\right] +\mathbb E_{k}\left[ \frac{2G(k+1)d(k)P(k+1)}{r^{2}}\right] . \end{aligned}\nonumber \\ \end{aligned}$$

(5.8)

Now, let us compute these three expectations:

$$\begin{aligned}&\mathbb E_{k}[G^{2}(k+1)d^{2}(k)]=d^{2}(k)(p_{11}g_{1}^{2}+p_{12}g_{2}^{2}); \end{aligned}$$

(5.9)

$$\begin{aligned}&\mathbb E_{k}[P^{2}(k+1)]=\mathbb E_{k}[\mathbb E_{k+1}[P^{2}(k+1)|G(k+1)]]=\mathbb E_{k}[\psi _{2}(g(k+1))d^{2}(k+1)]\nonumber \\&\quad = \mathbb E_{k}[\psi _{2}(g(k+1))G^{2}(k+1)d^{2}(k)]= d^{2}(k)(p_{11}\psi _{2}(g_{1})g_{1}^{2}+p_{12}\psi _{2}(g_{2})g_{2}^{2});\nonumber \\ \end{aligned}$$

(5.10)

$$\begin{aligned}&\mathbb E_{k}[G(k+1)d(k)P(k+1)]=d(k)\mathbb E_{k}[\mathbb E_{k+1}[G(k+1)P(k+1)|G(k+1)]]\nonumber \\&\quad = d(k)\mathbb E_{k}[G(k+1)\mathbb E_{k+1}[P(k+1)|G(k+1)]]\nonumber \\&\quad = d(k)\mathbb E_{k}[G(k+1)\psi _{1}(g(k+1))d(k+1)]\nonumber \\&\quad = d(k)\mathbb E_{k}[G(k+1)\psi _{1}(g(k+1))G(k+1)d(k)]\nonumber \\&\quad = d^{2}(k)(p_{11}g_{1}\psi _{1}(g_{1})g_{1}+p_{12}g_{2}\psi _{1}(g_{2})g_{2})\nonumber \\&\quad = d^{2}(k)(p_{11}g_{1}^{2}\psi _{1}(g_{1})+p_{12}g_{2}^{2}\psi _{1}(g_{2})). \end{aligned}$$

(5.11)

A substitution of (5.9), (5.10) and (5.11) in (5.8) leads to

$$\begin{aligned} \begin{aligned}&\psi _{2}(g_{1})d^{2}(k)=\frac{1}{r^{2}}\bigg (d^{2}(k)(p_{11}g_{1}^{2}+p_{12}g_{2}^{2})\\&\quad + d^{2}(k)(p_{11}\psi _{2}(g_{1})g_{1}^{2}+p_{12}\psi _{2}(g_{2})g_{2}^{2}) + d^{2}(k)(p_{11}g_{1}^{2}\psi _{1}(g_{1})+p_{12}g_{2}^{2}\psi _{1}(g_{2}))\bigg ). \end{aligned} \end{aligned}$$

Some computations yield

$$\begin{aligned}&\psi _{2}(g_{1})\big (r^{2}-p_{11}g_{1}^{2}\big )-\psi _{2}(g_{2})p_{12}g_{2}^{2}\nonumber \\&\quad =p_{11}g_{1}^{2}\big (1+2\psi _{1}(g_{1})\big )+p_{12}g_{2}^{2}\big (1+2\psi _{1}(g_{2})\big ). \end{aligned}$$

(5.12)

Symmetric arguments produce the second equation of the system (2.13). Concerning the uniqueness of the solution, it is sufficient to note that the matrix of the coefficients of the system has the form $A=\left[ \begin{array}{cc} \ r^{2}-p_{11}g_{1}^{2} &{} -p_{12}g_{2}^{2} \\ \ -p_{21}g_{1}^{2} &{} r^{2}-p_{22}g_{2}^{2} \\ \end{array} \right]$ and then

$$\begin{aligned} det(A)=\left( r^{2}-p_{11}g_{1}^{2}\right) \left( r^{2}-p_{22}g_{2}^{2}\right) -\left( -p_{12}g_{2}^{2}\right) \left( -p_{21}g_{1}^{2}\right) . \end{aligned}$$

Using assumption A2, we have that

$$\begin{aligned} \begin{aligned}&det(A){>}(p_{11}g_{1}^{2}{+}p_{12}g_{2}^{2}-p_{11}g_{1}^{2})(p_{21}g_{1}^{2}+p_{22}g_{2}^{2}-p_{22}g_{2}^{2})-(-p_{12}g_{2}^{2})(-p_{21}g_{1}^{2})\\&\quad =p_{12}g_{2}^{2}p_{21}g_{1}^{2}-(-p_{12}g_{2}^{2})(-p_{21}g_{1}^{2})=0. \end{aligned} \end{aligned}$$

The non negativity of the solution is due to the fact that $g_{1},g_{2}\ge 0$.

1.3 Proof of Proposition 2.5

For $n=1$ it is simple to compute the expectation

$$\begin{aligned} E^{(1)}p(d_{k},g_{a})=\sum _{j\in E}p_{aj}p(d_{k}g_{j},g_{j})=\sum _{j\in E}p_{aj}g_{j}p(d_{k},g_{j}). \end{aligned}$$

(5.13)

Consequently, we have the matrix form

$$\begin{aligned} \left( \begin{array}{c} E^{(1)}p(d_{k},g_{1})\\ \vdots \\ E^{(1)}p(d_{k},g_{s}) \end{array}\right)= & {} \mathbf {P} \mathbf {I}_{\mathbf {g}} \left( \begin{array}{c} p(d_{k},g_{1})\\ \vdots \\ p(d_{k},g_{s}) \end{array}\right) , \end{aligned}$$

and, taking into account (cf. (2.8)) that $p(d_{k},g_{a}) = p(d(k), g(k)) = d(k) \psi _1(g(k)) = d_k \psi _1(g_a),$ we obtain (2.22) for $n=1$. For $n=2$ we have

$$\begin{aligned} E^{(2)}p(d_{k},g_{a})= & {} E_{(d(k),g_{a})}[P(D(k+2),g(k+2))]\nonumber \\= & {} E_{(d(k),g_{a})}[E_{(D(k+1),g(k+1))}[P(D(k+2),g(k+2))]\nonumber \\= & {} E_{(d(k),g_{a})}[E^{(1)}P(D(k+1),g(k+1))]\nonumber \\= & {} \sum _{j\in E}p_{aj}E^{(1)}P(g_{j}d_{k},g_{j})=\sum _{j\in E}p_{aj}\sum _{h\in E}p_{jh}p(d_{k}g_{j}g_{h},g_{h})\nonumber \\= & {} \sum _{j\in E}\sum _{h\in E}p_{aj}p_{jh}g_{j}g_{h}p(d_{k},g_{h}). \end{aligned}$$

(5.14)

As we did previously for the case $n=1,$ taking into account that $\left( \mathbf {P} \mathbf {I}_{\mathbf {g}}\right) ^2 = \mathbf {P}^2 \mathbf {I}_{\mathbf {g}^2}$ we obtain the matrix form given in (2.22) for $n=2$. An iteration up to the nth step produces the result given in (2.21) and the corresponding matrix form (2.22).

1.4 Proof of Lemma 3.3

Note that we have

$$\begin{aligned} \left( \mathbf {I}_{r}^{n}-\mathbf {P} \mathbf {I}_{\mathbf {g}}^{n}\right) ^{-1}= & {} \Big (\mathbf {I}_{r}^{n}\cdot \big (\mathbf {I}-\mathbf {P} \mathbf {I}_{\mathbf {g}}^{n}\mathbf {I}_{r}^{-n}\big )\Big )^{-1}\end{aligned}$$

(5.15)

$$\begin{aligned}= & {} \mathbf {I}_{r}^{-n} \sum _{s\ge 0} \mathbf {P}^s \big (\mathbf {I}_{\mathbf {g}}^{n}\big )^{s} \big (\mathbf {I}_{r}^{-n}\big )^{s} \end{aligned}$$

(5.16)

where we used the fact that matrices like $\mathbf {I}_{r}^{n}$ or $\mathbf {I}_{\mathbf {g}}^{n}$ commute with any matrix. Taking into account Lemma 3.2, we have

$$\begin{aligned}&\frac{\partial \left( \mathbf {I}_{r}^{n}-\mathbf {P} \mathbf {I}_{\mathbf {g}}^{n}\right) ^{-1}}{\partial p_{ij}}\\&\quad = \mathbf {I}_{r}^{-n} \sum _{l\ge 0} \Big (\frac{\partial \mathbf {P}^l }{\partial p_{ij}}\Big ) \big (\mathbf {I}_{\mathbf {g}}^{n}\big )^{l} \big (\mathbf {I}_{r}^{-n}\big )^{l} \\&\quad = \mathbf {I}_{r}^{-n} \sum _{l\ge 0} \sum _{k=1}^{l} \mathbf {P}^{k-1} \Big (\frac{\partial \mathbf {P}}{\partial p_{ij}}\Big ) \mathbf {P}^{l-k} \big (\mathbf {I}_{\mathbf {g}}^{n}\big )^{l} \big (\mathbf {I}_{r}^{-n}\big )^{l}\\&\quad = \mathbf {I}_{r}^{-n} \sum _{k\ge 1}\mathbf {P}^{k-1}\Big (\frac{\partial \mathbf {P}}{\partial p_{ij}}\Big ) \sum _{l\ge k}\mathbf {P}^{l-k} \big (\mathbf {I}_{\mathbf {g}}^{n}\big )^{l} \big (\mathbf {I}_{r}^{-n}\big )^{l}\\&\quad = \mathbf {I}_{r}^{-n} \sum _{k\ge 1}\mathbf {P}^{k-1}\big (\mathbf {I}_{\mathbf {g}}^{n}\big )^{k-1}\big (\mathbf {I}_{r}^{-n}\big )^{k-1}\Big (\frac{\partial \mathbf {P}}{\partial p_{ij}}\Big ) \sum _{l\ge k}\mathbf {P}^{l-k} \big (\mathbf {I}_{\mathbf {g}}^{n}\big )^{l-k} \big (\mathbf {I}_{r}^{-n}\big )^{l-k}\big (\mathbf {I}_{\mathbf {g}}^{n}\big )\big (\mathbf {I}_{r}^{-n}\big )\\&\quad = \mathbf {I}_{\mathbf {g}}^{n} \left( \mathbf {I}_{r}^{n}-\mathbf {P} \mathbf {I}_{\mathbf {g}}^{n}\right) ^{-1}\cdot \frac{\partial \mathbf {P}}{\partial p_{ij}} \cdot \left( \mathbf {I}_{r}^{n}-\mathbf {P} \mathbf {I}_{\mathbf {g}}^{n}\right) ^{-1}. \end{aligned}$$

1.5 Proof of Theorem 3.5

First, note that we have

$$\begin{aligned} \widehat{\varvec{\Psi }}_{1}(m) = \varPhi _{1} (\widehat{p}_{ij}(m), i = 1, \ldots , s, j =1, \ldots , s-1). \end{aligned}$$

(5.17)

Using the strong consistency of $\widehat{p}_{ij}(m),$ as m goes to infinity (cf. Proposition 3.1), and the continuous mapping theorem, we obtain the strong consistency of $\widehat{\varvec{\Psi }}_{1}(m).$

Second, taking into account the expressions (3.3) and (5.17) for $\varvec{\Psi }_{1}$ and $\widehat{\varvec{\Psi }}_{1}(m),$ respectively, together with the asymptotic normality of the vector $(\sqrt{m}\big (\widehat{p}_{ij}(m)-p_{ij}\big ))_{i = 1, \ldots , s, j = 1, \ldots , s-1}$ (cf. Proposition 3.1) with covariance matrix ${\varvec{\varGamma }}$ defined as the restriction of $\widetilde{{\varvec{\varGamma }}}$ given in (3.2) to $s(s-1)\times s(s-1),$ we obtain the asymptotic normality stated in (3.9) using the delta method because $\varvec{\Psi }_{1}$ is a differentiable function of $\mathbf {P}$.

The last point we have to deal with is to give an expression of the partial derivative matrix $\varvec{\Phi }_{1}'.$ Note that we have

$$\begin{aligned} \varvec{\Phi }_{1}' =\begin{pmatrix} \frac{\partial \varPhi _1^1}{\partial p_{11}} &{} \cdots &{} \frac{\partial \varPhi _1^1}{\partial p_{1 (s-1)}} &{} \cdots &{} \frac{\partial \varPhi _1^1}{\partial p_{s1}} &{} \cdots &{} \frac{\partial \varPhi _1^1}{\partial p_{s (s-1)}}\\ \vdots &{} &{} \vdots &{} &{} \vdots &{} &{} \vdots \\ \frac{\partial \varPhi _1^s}{\partial p_{11}} &{} \cdots &{} \frac{\partial \varPhi _1^s}{\partial p_{1 (s-1)}} &{} \cdots &{} \frac{\partial \varPhi _1^s}{\partial p_{s1}} &{} \cdots &{} \frac{\partial \varPhi _1^s}{\partial p_{s (s-1)}} \end{pmatrix} \in {\mathcal {M}}_{s \times s(s-1)}. \end{aligned}$$

(5.18)

Note that in the computation of the derivative of $\varPhi _{1}$ with respect to its argument $(p_{ij}, i = 1, \ldots , s, j =1, \ldots , s-1),$ this argument is ordered in the lexicographic order: $(p_{11}, \ldots , p_{1(s-1)}, p_{21}, \ldots , p_{2(s-1)}, \ldots , p_{s 1}, \ldots , p_{s(s-1)}) \in \mathbb R^{s(s-1)}.$

An arbitrary column of this matrix $\varvec{\Phi }_{1}'$ corresponding to $(i, j), i = 1, \ldots , s, j =1, \ldots , s-1,$ is given by

$$\begin{aligned} \frac{\partial \varPhi _1}{\partial p_{ij}}= & {} \frac{\partial }{\partial p_{ij}} \left( \left( \mathbf {I}_{r}-\mathbf {P} \mathbf {I}_{\mathbf {g}}\right) ^{-1} \mathbf {P} \mathbf {g} \right) \nonumber \\= & {} \frac{\partial \left( \mathbf {I}_{r}-\mathbf {P} \mathbf {I}_{\mathbf {g}}\right) ^{-1}}{\partial p_{ij}} \mathbf {P} \mathbf {g} + \left( \mathbf {I}_{r}-\mathbf {P} \mathbf {I}_{\mathbf {g}}\right) ^{-1} \frac{\partial \mathbf {P} }{\partial p_{ij}} \mathbf {g}. \end{aligned}$$

(5.19)

Using the expression of $\frac{\partial \mathbf {P} }{\partial p_{ij}}$ given in (3.5), together with the computation of $\frac{\partial \left( \mathbf {I}_{r}-\mathbf {P} \mathbf {I}_{\mathbf {g}}\right) ^{-1}}{\partial p_{ij}}$ given in Lemma 3.3 (for $n=1$), allows us to completely compute $\frac{\partial \varPhi _1}{\partial p_{ij}}$ and thus $\varvec{\Phi }_{1}'.$

1.6 Proof of Theorem 3.7

First, note that we have

$$\begin{aligned} \widehat{\varvec{\Psi }}_{2}(m) = \varPhi _{2} (\widehat{p}_{ij}(m), i = 1, \ldots , s, j =1, \ldots , s-1). \end{aligned}$$

(5.20)

Using the strong consistency of estimator $\widehat{\varvec{\Psi }}_{1}(m)$ (cf. Theorem 3.5), the strong consistency of $\widehat{p}_{ij}(m),$ as m goes to infinity (cf. Proposition 3.1), and the continuous mapping theorem, we obtain the strong consistency of $\widehat{\varvec{\Psi }}_{2}(m).$

Second, taking into account the expressions (3.12) and (5.20) for $\varvec{\Psi }_{2}$ and $\widehat{\varvec{\Psi }}_{2}(m),$ respectively, together with the asymptotic normality of the vector $(\sqrt{m}\big (\widehat{p}_{ij}(m)-p_{ij}\big ))_{i = 1, \ldots , s, j = 1, \ldots , s-1}$ (cf. Proposition 3.1) with covariance matrix ${\varvec{\varGamma }}$ defined as the restriction of $\widetilde{{\varvec{\varGamma }}}$ given in (3.2) to $s(s-1)\times s(s-1),$ we obtain the asymptotic normality stated in (3.14) using the delta method because $\varvec{\Psi }_{2}$ is a differentiable function of $\mathbf {P}$ being $\varvec{\Psi }_{1}$ a differentiable function of $\mathbf {P}$.

The last point we have to deal with is to give an expression of the partial derivative matrix $\varvec{\Phi }_{2}'.$ Note that we have

$$\begin{aligned} \varvec{\Phi }_{2}' =\begin{pmatrix} \frac{\partial \varPhi _2^1}{\partial p_{11}} &{} \cdots &{} \frac{\partial \varPhi _2^1}{\partial p_{1 (s-1)}} &{} \cdots &{} \frac{\partial \varPhi _2^1}{\partial p_{s1}} &{} \cdots &{} \frac{\partial \varPhi _2^1}{\partial p_{s (s-1)}}\\ \vdots &{} &{} \vdots &{} &{} \vdots &{} &{} \vdots \\ \frac{\partial \varPhi _2^s}{\partial p_{11}} &{} \cdots &{} \frac{\partial \varPhi _2^s}{\partial p_{1 (s-1)}} &{} \cdots &{} \frac{\partial \varPhi _2^s}{\partial p_{s1}} &{} \cdots &{} \frac{\partial \varPhi _2^s}{\partial p_{s (s-1)}} \end{pmatrix} \in {\mathcal {M}}_{s \times s(s-1)}. \end{aligned}$$

(5.21)

An arbitrary column of this matrix corresponding to $(i, j), i = 1, \ldots , s, j =1, \ldots , s-1,$ is given by

$$\begin{aligned} \frac{\partial \varPhi _2}{\partial p_{ij}}= & {} \frac{\partial }{\partial p_{ij}} \left( \left( \mathbf {I}_{r}^{2}-\mathbf {P} \cdot \mathbf {I}_{\mathbf {g}}^{2}\right) ^{-1} \cdot \mathbf {P}\cdot \left( \mathbf {g} \diamond \mathbf {g}+2\varvec{\Psi }_{1}\diamond \mathbf {g} \diamond \mathbf {g}\right) \right) \nonumber \\= & {} \frac{\partial \left( \mathbf {I}_{r}^{2}-\mathbf {P} \cdot \mathbf {I}_{\mathbf {g}}^{2}\right) ^{-1}}{\partial p_{ij}} \cdot \mathbf {P}\cdot \left( \mathbf {g} \diamond \mathbf {g}+2\varvec{\Psi }_{1}\diamond \mathbf {g} \diamond \mathbf {g}\right) \nonumber \\&+ \left( \mathbf {I}_{r}^{2}-\mathbf {P} \cdot \mathbf {I}_{\mathbf {g}}^{2}\right) ^{-1} \cdot \frac{\partial \mathbf {P}}{\partial p_{ij}}\cdot \left( \mathbf {g} \diamond \mathbf {g}+2\varvec{\Psi }_{1}\diamond \mathbf {g} \diamond \mathbf {g}\right) \nonumber \\&+ \left( \mathbf {I}_{r}^{2}-\mathbf {P} \cdot \mathbf {I}_{\mathbf {g}}^{2}\right) ^{-1} \cdot \mathbf {P}\cdot \left( \mathbf {g} \diamond \mathbf {g}+2 \frac{\partial \varvec{\Psi }_{1}}{\partial p_{ij}}\diamond \mathbf {g} \diamond \mathbf {g}\right) . \end{aligned}$$

(5.22)

Using the computation of $\frac{\partial \left( \mathbf {I}_{r}^{2}-\mathbf {P} \mathbf {I}_{\mathbf {g}}^{2}\right) ^{-1}}{\partial p_{ij}}$ given in Lemma 3.3 (for $n=2$), the expression of $\frac{\partial \mathbf {P} }{\partial p_{ij}}$ given in (3.5), together with the computation of $\frac{\partial \varvec{\Psi }_{1}}{\partial p_{ij}}$ obtained in (5.19), we completely compute the value of $\frac{\partial \varPhi _2}{\partial p_{ij}}$ and thus of $\varvec{\Phi }_{2}'.$

1.7 Proof of Theorem 3.8

First, note that we have

$$\begin{aligned}&\left( \widehat{E^{(n)}}(m)p(d_{k},g_{1}), \ldots , \widehat{E^{(n)}}(m)p(d_{k},g_{s})\right) ^\top \nonumber \\&\quad = \varTheta (\widehat{p}_{ij}(m), i = 1, \ldots , s, j =1, \ldots , s-1). \end{aligned}$$

(5.23)

Using the strong consistency of $\widehat{p}_{ij}(m),$ as m goes to infinity (cf. Proposition 3.1), and the continuous mapping theorem, we obtain the strong consistency of estimator (3.16).

Second, taking into account the expressions (3.17) and (5.23) for the expected forecast fundamental prices and the corresponding estimator, together with the asymptotic normality of the vector $(\sqrt{m}\left( \widehat{p}_{ij}(m)-p_{ij}\right) )_{i = 1, \ldots , s, j = 1, \ldots , s-1}$ (cf. Proposition 3.1) with covariance matrix ${\varvec{\varGamma }}$ defined as the restriction of $\widetilde{{\varvec{\varGamma }}}$ given in (3.2) to $s(s-1)\times s(s-1),$ we obtain the asymptotic normality stated in (3.19) using the delta method.

The last point we have to deal with is to give an expression of the partial derivative matrix $\varvec{\Theta }'.$ Note that we have

$$\begin{aligned} \varvec{\Theta }' =\begin{pmatrix} \frac{\partial \varTheta ^1}{\partial p_{11}} &{} \cdots &{} \frac{\partial \varTheta ^1}{\partial p_{1 (s-1)}} &{} \cdots &{} \frac{\partial \varTheta ^1}{\partial p_{s1}} &{} \cdots &{} \frac{\partial \varTheta ^1}{\partial p_{s (s-1)}}\\ \vdots &{} &{} \vdots &{} &{} \vdots &{} &{} \vdots \\ \frac{\partial \varTheta ^s}{\partial p_{11}} &{} \cdots &{} \frac{\partial \varTheta ^s}{\partial p_{1 (s-1)}} &{} \cdots &{} \frac{\partial \varTheta ^s}{\partial p_{s1}} &{} \cdots &{} \frac{\partial \varTheta ^s}{\partial p_{s (s-1)}} \end{pmatrix} \in {\mathcal {M}}_{s \times s(s-1)}. \end{aligned}$$

(5.24)

An arbitrary column of this matrix corresponding to $(i, j), i = 1, \ldots , s, j =1, \ldots , s-1,$ is given by

$$\begin{aligned} \frac{\partial \varTheta }{\partial p_{ij}}= & {} \frac{\partial }{\partial p_{ij}} \left( d_k \mathbf {P}^n \mathbf {I}_{\mathbf {g}^n} \varvec{\Psi }_{1}\right) \nonumber \\= & {} d_k \frac{\partial \mathbf {P}^n}{\partial p_{ij}} \mathbf {I}_{\mathbf {g}^n} \varvec{\Psi }_{1} + d_k \mathbf {P}^n \mathbf {I}_{\mathbf {g}^n} \frac{\partial \varvec{\Psi }_{1}}{\partial p_{ij}}. \end{aligned}$$

(5.25)

Using the computation of $\frac{\partial \mathbf {P}^n}{\partial p_{ij}}$ given in Lemma 3.2, of $\frac{\partial \mathbf {P}}{\partial p_{ij}}$ given in (3.5) and of $\frac{\partial \varvec{\Psi }_{1}}{\partial p_{ij}}$ obtained in (5.19), we are able to compute the value of $\frac{\partial \varTheta }{\partial p_{ij}}$ and thus of $\varTheta '.$