4.1. Influence of microscopic properties on the price dynamics of two correlated assets

Our first model to understand the price formation process focuses on two assets. Let $\mu^1, \mu^2 > 0$ , $\alpha \in (1/2,1)$ , $\gamma_1, \gamma_2 \in [0,1]$ , and $H^c_{21}, H^a_{21}, H^c_{12}, H^a_{12} \in [0,1]$ such that the following hold (here $\sqrt{\cdot}$ denotes the principal square root, so that if $x<0$ , then $\sqrt{x}= i \sqrt{-x}$ ):

\begin{align*} 0 &\leq \big(H^c_{12}+H^a_{12}\big)\big(H^c_{21}+H^a_{21}\big) < 1, \\[4pt] 0 &\leq {\mid{{1 - (\gamma_1 + \gamma_2) - \sqrt{(H^c_{12}-H^a_{12})(H^c_{21}-H^a_{21}) + (\gamma_1-\gamma_2)^2}}}\mid} < 1, \\[4pt] 0 &\leq {\mid{{1 - (\gamma_1 + \gamma_2) + \sqrt{(H^c_{12}-H^a_{12})(H^c_{21}-H^a_{21}) + (\gamma_1-\gamma_2)^2}}}\mid} < 1.\end{align*}

In the above, the superscript c (resp. a) stands for continuation (resp. alternation) to describe that after a price move in a given direction, $H^c$ (resp. $H^a$ ) encodes the tendency to trigger other price moves in the same (resp. the opposite) direction. We now have to choose a kernel which satisfies the various assumptions of Section 2 to model the interactions between our two assets. Theorem 1 states that the only relevant parameters for the macroscopic price are $\textbf{K}$ and $\textbf{M}$ . For simplicity we choose the kernel so that $\textbf{M} = \alpha \textbf{I}$ . This leads us to define, for $t \geq 0$ ,

\begin{align*} \phi^T_1(t) &= (1-\gamma_1) \alpha (1 - T^{-\alpha})\mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}, &\phi^{T,c}_{21}(t) = \alpha T^{-\alpha} H^c_{21} \mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}, \\[4pt] \phi^T_2(t) &= \gamma_1 \alpha (1 - T^{-\alpha})\mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}, &\phi^{T,a}_{21}(t) = \alpha T^{-\alpha} H^a_{21} \mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}, \\[4pt] \tilde{\phi}^T_1(t) &= (1- \gamma_2) \alpha (1 - T^{-\alpha})\mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}, &\phi^{T,c}_{12}(t) = \alpha T^{-\alpha} H^c_{12} \mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}, \\[4pt] \tilde{\phi}^T_2(t) &= \gamma_2 \alpha (1 - T^{-\alpha})\mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}, &\phi^{T,a}_{12}(t) = \alpha T^{-\alpha} H^a_{12} \mathbb{1}_{t \geq 1} t^{-(\alpha + 1)}.\end{align*}

For a realistic model, we require the exogenous sources of upward and downward price moves to be equal: $\mu^{1+} = \mu^{1-}$ and $\mu^{2+} = \mu^{2-}$ . Thus, the sequences of baselines and kernels are chosen as

\begin{equation*}\boldsymbol{\mu}^T = T^{\alpha - 1}\begin{pmatrix}\mu^{1} \\[4pt] \mu^{1} \\[4pt] \mu^{2} \\[4pt] \mu^{2}\end{pmatrix}, \quad\boldsymbol{\phi}^T = \begin{pmatrix}\phi_1^T & \quad\phi_2^T & \quad\phi^{T,c}_{12} & \quad \phi^{T,a}_{12} \\[7pt] \phi_2^T & \quad\phi_1^T & \quad\phi^{T,a}_{12} & \quad\phi^{T,c}_{12} \\[7pt] \phi^{T,c}_{21} & \quad\phi^{T,a}_{21} & \quad\tilde{\phi}_{1}^T & \quad\tilde{\phi}_{2}^T \\[7pt] \phi^{T,a}_{21} & \quad\phi^{T,c}_{21} & \quad \tilde{\phi}_{2}^T & \quad \tilde{\phi}_1^T \end{pmatrix}.\end{equation*}

We set

\begin{align*}\boldsymbol{\chi} \,{:}\,{\raise-1.5pt{=}}\, \dfrac{\sqrt{2}}{4 \gamma_1 \gamma_2 - \big(H_{12}^c - H_{12}^a\big)\big(H_{21}^c - H_{21}^a\big)}\begin{pmatrix}2 \gamma_2 & \quad H_{21}^c - H_{21}^a \\[7pt] H_{12}^c - H_{12}^a & \quad 2 \gamma_1\end{pmatrix}, \\[7pt] \boldsymbol{\Gamma} \,{:}\,{\raise-1.5pt{=}}\, \dfrac{1}{1 - \big(H^c_{12}+H^a_{12}\big)\big(H^c_{21}+H^a_{21}\big)} \begin{pmatrix}1 & \quad H^c_{21}+H^a_{21} \\[5pt] H^c_{12}+H^a_{12} & 1\end{pmatrix}.\end{align*}

Applying Theorem 1 yields the following result.

Corollary 1. Consider any limit point $\textbf{P}$ of $\textbf{P}^T$ . Under the above assumptions, it satisfies

(12) \begin{equation} \textbf{P}_t = \boldsymbol{\chi} \int_0^t \begin{pmatrix} \sqrt{V^1_s} dW^1_s \\[7pt] \sqrt{V^2_s} dW^2_s \end{pmatrix},\end{equation}

with

(13) \begin{align} \begin{pmatrix}V^1_t \\[5pt] V^2_t\end{pmatrix} &= \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha - 1} \left( \begin{pmatrix} \mu^1 \\[5pt] \mu^2 \end{pmatrix} - \boldsymbol{\Gamma} \begin{pmatrix}V^1_s \\[5pt] V^2_s\end{pmatrix} \right)ds \nonumber \\[5pt] & + \sqrt{2} \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha - 1} \begin{pmatrix} \sqrt{V^1_s} dZ^1_s \\[8pt] \sqrt{V^2_s} dZ^2_s \end{pmatrix},\end{align}

where $\textbf{W}$ and $\textbf{Z}$ are bi-dimensional independent Brownian motions.

This model helps us understand how microscopic parameters drive the price formation process to generate a macroscopic price and volatility.

We begin our remarks with some definitions. We define momentum as the trend (i.e., the imbalance between the number of upward and downward jumps) created by jumps of one asset on itself. For example, momentum is strong when the next price jump after an upward price jump on an asset is more likely to be upward than downward. The opposite effect is referred to as mean-reversion. For example, the parameter $\gamma_1$ controls the intensity of self-induced bid–ask bounce on Asset 1: $\gamma_1$ close to zero corresponds to a strong momentum while $\gamma_1$ close to one corresponds to a strong mean-reversion.

We define cross-asset momentum as the trend created by jumps of one asset on another. For example, cross-asset momentum from Asset 2 to Asset 1 (resp. Asset 1 to Asset 2) appears via $H^c_{21} - H^a_{21}$ (resp. $H^c_{12} - H^a_{12}$ ): when both $H^c_{21} - H^a_{21}$ and $H^c_{12} - H^a_{12}$ are nil, the prices of Asset 1 and Asset 2 are uncorrelated.

We now turn to comments on the volatility process. Because of its role in the single-asset case, we refer to $\textbf{V}$ as the fundamental variance: for example, $V^1$ is the fundamental variance of Asset 1. The equation satisfied by $\textbf{V}$ depends only on the sum of the feedback effects between the two assets through $H^c_{12} + H^a_{12}$ : from a volatility viewpoint, upward and downward jumps have the same impact. Furthermore, we can compute the expected fundamental variance using Mittag-Leffler functions (see Section 5).

Mean-reversion drives down volatility while cross-asset momentum increases it. Indeed, computing $\mathbb{E}[(P^{1}_t)^2]$ , for example, we get

\begin{equation*} \mathbb{E}\big[(P^1_t)^2\big]=2 \dfrac{4\gamma_2^2 \int_0^t \mathbb{E}\big[V^1_s\big] ds + \big(H^{c}_{12} - H^{a}_{12}\big)\big(H^{c}_{21} - H^{a}_{21}\big) \int_0^t \mathbb{E}\big[V^2_s\big] ds }{\big[4 \gamma_1 \gamma_2 - \big(H_{12}^c - H_{12}^a\big)\big(H_{21}^c - H_{21}^a\big)\big]^2}.\end{equation*}

In particular, increasing $\gamma_1$ does not change $\textbf{V}$ but reduces $\mathbb{E}[(P^1_t)^2]$ . This example may be particularly relevant to understanding the contribution of Asset 2 to the volatility of Asset 1 through calibration to market data, since if Asset 2 were removed from the market, we would have

\begin{equation*}\mathbb{E}\big[\big(P^1_t\big)^2\big]=\dfrac{1}{2 \gamma_1}.\end{equation*}

Focusing now on the price formation process, we see that it results from a combination of momentum, mean-reversion, and cross-asset momentum. We illustrate this in two extreme cases: when there is no cross-asset momentum and when cross-asset momentum is strong.

• (i) When there is no cross-asset momentum (i.e. $H^c_{12}=H^a_{12}$ and $ H^c_{21}=H^a_{21}$ ), at the microscopic scale, a price move on Asset 2 has the same impact on the intensity of upward and downward price moves of Asset 1. Thus the difference between the expected number of upward and downward jumps does not change after a price move on Asset 2: the expected microscopic price of Asset 1 is unaffected, and price moves of Asset 2 generate no trend on Asset 1. This results in macroscopic prices being uncorrelated (see Equation (12)).

• (ii) On the other hand, when cross-asset momentum is strong (i.e. $(H^{c}_{12} - H^{a}_{12})(H^{c}_{21} - H^{a}_{21}) \approx 4 \gamma_1 \gamma_2$ , for example if $H^{c}_{12} - H^{a}_{12} = 2\gamma_1\sqrt{1 - \epsilon}$ and $H^{c}_{12} - H^{a}_{12} = 2\gamma_2\sqrt{1 - \epsilon}$ for some small $\epsilon > 0$ ), at the microscopic scale, a price move on Asset 2 significantly increases the probability of a future price move on Asset 1 in the same direction (and vice versa). In this context we have

  \begin{equation*} \boldsymbol{\Delta} + {\textbf{I}} = \dfrac{1}{2\gamma_1 \gamma_2 \epsilon}\begin{pmatrix} \gamma_2 & \quad\gamma_2\sqrt{1 - \epsilon} \\[7pt] \gamma_1\sqrt{1 - \epsilon} & \quad\gamma_1 \end{pmatrix}. \end{equation*}
  Using Equation (12) we can check that
  \begin{equation*}\dfrac{\mathbb{E}\big[P^1_t P^2_t\big]}{\sqrt{\mathbb{E}\big[(P^1_t)^2\big] \mathbb{E}\big[(P^2_t)^2\big]}} \underset{\epsilon \to 0}{\to} 1,\end{equation*}
  and prices evolve in unison.

This example underlines that in our approach (thanks to our no-arbitrage constraint) microscopic features transfer to macroscopic properties in an intuitive way.

4.2. Reproducing realistic correlation matrices of a large number of assets using microscopic properties

It is well known that the correlation matrix of stocks has few large eigenvalues outside of a ‘bulk’ of eigenvalues attributable to noise (see for example [Reference Laloux, Cizeau, Bouchaud and Potters23]). The largest eigenvalue is referred to as the market mode (because the associated eigenvector places a roughly equal weight on each asset) and is much larger than other eigenvalues. Other significant eigenvalues can be related to the presence of sectors: groups of companies with similar characteristics.

How can we provide microstructural foundations for this stylized fact? The large eigenvalue associated to the market mode implies that, in a first approximation, stock prices move together: a price increase on one asset is likely followed by a price increase on all other assets. Translating this into our framework, an upward (resp. downward) jump on an asset increases the probability of an upward (resp. downward) jump on all other assets. We further expect that an upward price move on an asset increases this probability much more on an asset from the same sector than on an unrelated one.

The above remarks lead us to consider a model with the following properties:

• (i) All stocks share some fundamental high-frequency properties because they have similar self-excitement parameters in the kernel.

• (ii) Stocks have a stronger influence on price changes of stocks within the same sector.

• (iii) Within the same sector, all stocks have the same microscopic parameters.

The technical details of our setting are presented in Appendix A.4; here we provide only the elements essential to understanding the framework. Let $\mu^1, \dots, \mu^m > 0$ be the baselines of each asset, where we assume $\mu^{i+} = \mu^{i-}$ for all $1 \leq i \leq m$ . Using the same notation as before, take $\gamma \in [0,1]$ , $\alpha \in (1/2,1)$ and $H^c, H^a > 0$ . We consider $R > 0$ different sectors, Sector r having $m_r$ stocks. For a pair of stocks which we dub 1 and 2 in analogy to the previous example, we have the following:

• (i) The self-excitement parameters are equal: $\gamma_1 = \gamma_2 = \gamma$ , where $\gamma$ is the same for all stocks.

• (ii) If Stock 1 and Stock 2 do not belong to the same sector, then $H_{21}^c = H_{12}^c = H^c$ and $H_{21}^a = H_{12}^a = H^a$ , where $H^c,H^a$ are the same for all stocks.

• (iii) If Stock 1 and Stock 2 belong to the same sector r, $H_{21}^c = H_{12}^c = H^c + H^c_r$ , $H_{21}^a = H_{12}^a = H^a + H^a_r$ where $H^c_r,H^a_r$ are the same for all stocks belonging to Sector r.

The asymptotic framework is built as in the previous example, with the details given in the proof of Corollary 2 in Appendix A.4. We write $i_r \,{:}\,{\raise-1.5pt{=}}\, m_0 + m_1 + \cdots + m_{r-1}$ for $1 \leq r \leq R$ (with the convention $m_0=1$ ), so that stocks from Sector r are indexed between $i_{r}$ and $i_{r+1}$ excluded, and define the following vectors:

\begin{align*} \textbf{w} \,{:}\,{\raise-1.5pt{=}}\, \dfrac{1}{\sqrt{m}}(\textbf{e}_1 + \cdots + \textbf{e}_{m}), \\ \textbf{w}_r \,{:}\,{\raise-1.5pt{=}}\, \dfrac{1}{\sqrt{m_r}}\sum_{i_r \leq i < i_{r+1}}\textbf{e}_{i}, \\ \boldsymbol{\theta} \,{:}\,{\raise-1.5pt{=}}\, \sum_{1 \leq i \leq m} \mu^i \textbf{e}_i.\end{align*}

We consider an asymptotic framework where the number of assets will eventually grow to infinity. As will become clear in the proof, the only nontrivial regime appears when

\begin{equation*}H^{c},H^{a},H^{c}_r,H^{a}_r \underset{m \to \infty}{=} \mathcal{O}(m^{-1}).\end{equation*}

Thus we assume that $m H^{c},mH^{a},mH^{c}_r,mH^{a}_r$ converge to $\bar{H}^{c},\bar{H}^{a},\bar{H}^{c}_r,\bar{H}^{a}_r$ as m tends to infinity. We also assume that the proportion of stocks in a given sector relative to the total number of stocks does not vanish: for each $1 \leq r \leq R$ ,

\begin{equation*}\frac{m_r}{m} \underset{m \to \infty}{\to} \eta_r > 0.\end{equation*}

We define the following constants, which will appear in the price and volatility processes: $\lambda^{+} \,{:}\,{\raise-1.5pt{=}}\, \bar{H}^{c} + \bar{H}^{a}$ , $\lambda_r^{+} \,{:}\,{\raise-1.5pt{=}}\, \bar{H}^{c}_r + \bar{H}^{c}_r$ , $\lambda^{-} \,{:}\,{\raise-1.5pt{=}}\, \bar{H}^{c} - \bar{H}^{a}$ , $\lambda^{-}_r \,{:}\,{\raise-1.5pt{=}}\, \bar{H}^{c}_r - \bar{H}^{a}_r$ . Applying Theorem 1 yields the following result.

Corollary 2. Consider any limit point $\textbf{P}$ of $\textbf{P}^T$ . Under the above assumptions, it satisfies

\begin{align*} \textbf{P}_t &= \sqrt{2} \boldsymbol{\Sigma}_{\varepsilon} \int_{0}^{t}{diag}\big(\sqrt{\textbf{V}_s}\big) d\textbf{B}_s,\end{align*}

where $\textbf{B}$ is a Brownian motion;

\begin{equation*}\boldsymbol{\Sigma}_{\varepsilon} \,{:}\,{\raise-1.5pt{=}}\, \Bigg(2 \gamma \textbf{I} - \lambda^{-} \textbf{v} {}^ \top {\bf{v}} - \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r \textbf{v}_{\textbf{r}} {{}^ \top {\bf{v}}_{\textbf{r}}} + \boldsymbol{\varepsilon}\Bigg)^{-1}\end{equation*}

with $\boldsymbol{\epsilon}$ a deterministic $m \times m$ matrix such that

\begin{equation*}{\rho{({\boldsymbol{\epsilon}})}} \underset{m \to \infty}{=} o(m^{-1});\end{equation*}

and $\textbf{V}$ satisfies the stochastic Volterra equation

\begin{align*} \textbf{V}_t &= \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} (\boldsymbol{\theta} - \boldsymbol{\mathcal{V}}_{\epsilon} \textbf{V}_s)ds + \dfrac{\sqrt{2} \alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} {diag}\Big(\sqrt{\textbf{V}_s}\Big)d\textbf{Z}_s,\end{align*}

with $\textbf{Z}$ a Brownian motion independent from $\textbf{W}$ , and

\begin{equation*}\boldsymbol{\mathcal{V}}_{\epsilon} \,{:}\,{\raise-1.5pt{=}}\, \left( \textbf{I} - \lambda^{+} \textbf{v} {}^ \top {\bf{v}} - \sum_{1 \leq r \leq R} \eta_r \lambda^{+}_r \textbf{v}_{\textbf{r}} {{}^ \top {\bf{v}}_{\textbf{r}}} + \boldsymbol{\epsilon} \right)^{-1}\end{equation*}

where $\boldsymbol{\varepsilon}$ is a deterministic $m \times m$ matrix such that

\begin{equation*}{\rho{({\boldsymbol{\varepsilon}})}} \underset{m \to \infty}{=} o(m^{-1}).\end{equation*}

Under the previous corollary, using $\propto$ to denote equality up to a multiplicative constant, the expected fundamental variance can be written as follows using the cumulative Mittag-Leffler function (see Definition 4 in Appendix A.2):

\begin{equation*} \mathbb{E}[\textbf{V}_t ] \propto \textbf{F}^{\alpha, \boldsymbol{\mathcal{V}}_{\epsilon}(t)} \boldsymbol{\theta}.\end{equation*}

Since

\begin{equation*}{\rho{({\boldsymbol{\epsilon}})}} \underset{m \to \infty}{=} o(m^{-1}),\end{equation*}

we neglect it in further comments and use the approximation $\boldsymbol{\mathcal{V}}_{\epsilon} \approx \boldsymbol{\mathcal{V}}_{0}$ . Writing $\xi$ for the largest eigenvalue of $\boldsymbol{\mathcal{V}}_{0}$ and $\textbf{z}$ for the associated eigenvector, and neglecting other eigenvalues (which is reasonable if $\lambda^+ + \sum_{1 \leq r \leq R} \eta_r \lambda^{+}_r \approx 1$ ), from the definition of the Mittag-Leffler function we have

$$\mathbb{E}[{{\bf{V}}_t}] \propto {F^{\alpha ,\xi }}(t){(^ \top }\theta {\bf{z}}){\bf{z}}.$$

Making the further approximation that $\eta_r \lambda^+_r$ is independent of r, we have $\textbf{z} \propto (1, \cdots, 1)$ and

\begin{align*}\mathbb{E}\big[\textbf{P}_t {{}^ \top {{\textbf{P}_t}}}\big] &\propto \boldsymbol{\Sigma_{\varepsilon}} {diag}(\mathbb{E}[\textbf{V}_t ]) {{}^ \top {{\boldsymbol{\Sigma_{\varepsilon}}}}} \\&\propto \boldsymbol{\Sigma_{\varepsilon}} {diag}(\textbf{z}) {{}^ \top {{\boldsymbol{\Sigma_{\varepsilon}}}}} \\& \propto \boldsymbol{\Sigma_{\varepsilon}} {{}^ \top {{\boldsymbol{\Sigma_{\varepsilon}}}}} \propto \boldsymbol{\Sigma_{\varepsilon}}^2.\end{align*}

Therefore the eigenvectors of $\mathbb{E}[\textbf{P}_t {}^ \top {{\textbf{P}_t}}]$ are those of $\boldsymbol{\Sigma_{\varepsilon}}$ . Now, as

\begin{equation*}{\rho{({\boldsymbol{\varepsilon}})}}\underset{m \to \infty}{=} o(m^{-1}),\end{equation*}

we neglect it in further comments and use the approximation $\boldsymbol{\Sigma_{\varepsilon}} \approx \boldsymbol{\Sigma_{0}}$ . When $\lambda^{-} + \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r \approx 2\gamma$ , $\boldsymbol{\Sigma_{0}}$ has one large eigenvalue followed by $R-1$ smaller eigenvalues and much smaller eigenvalues. This is consistent with stylized facts about high-dimensional stock correlation matrices; we have thus built a microscopic model to explain the macroscopic structure of correlation matrices.

The conditions $\lambda^{-} + \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r \approx 1$ and $\lambda^+ + \sum_{1 \leq r \leq R} \eta_r \lambda^{+}_r \approx 1$ correspond to the parameters being close to the point where all directions are critical: when $\lambda^{-} + \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r \approx 2\gamma$ or $\lambda^{-} + \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r \approx 1$ , the spectral radius of ${\int_{0}^{\infty}{{\textbf{C}}}}$ is equal to one and we cannot split the kernel into a critical and a non-critical component.

It would be interesting to study other implications of this model. In particular, we believe that encoding a negative price/volatility correlation into the microscopic parameters could explain the so-called index leverage effect (see [Reference Reigneron, Allez and Bouchaud25] for a definition and empirical analysis of this stylized fact).

5.1. Step 1: C-tightness of $\textbf{(}\textbf{\textit{X}}^{\textbf{\textit{T}}}\textbf{,}\ \textbf{\textit{Y}}^{\textbf{\textit{T}}}\textbf{,}\ \textbf{\textit{Z}}^{\textbf{\textit{T}}}\textbf{)}$

Recall the definition of the rescaled processes

\begin{align*} {\textbf{X}^T_t} \,{:}\,{\raise-1.5pt{=}}\, \dfrac{1}{T^{2\alpha}} \textbf{N}^T_{tT}, \\ {\textbf{Y}^T_t} & \,{:}\,{\raise-1.5pt{=}}\, \dfrac{1}{T^{2\alpha}} \int_0^{tT} \boldsymbol{\lambda}_s ds, \\ {\textbf{Z}^T_t} & \,{:}\,{\raise-1.5pt{=}}\, T^{\alpha} \big(\textbf{X}^T_t - \textbf{Y}^T_t\big) = \dfrac{1}{T^{\alpha}} \textbf{M}^T_{tT}.\end{align*}

As in [Reference El Euch, Fukasawa and Rosenbaum8] and [Reference Jaisson and Rosenbaum21] we show that the limiting processes of $\textbf{X}^T$ and $\textbf{Y}^T$ are the same and that the limiting process of $\textbf{Z}^T$ is the quadratic variation of the limiting process of $\textbf{X}^T$ . We have the following proposition.

Proposition 1. (C-tightness of $(\textbf{X}^T, \textbf{Y}^T, \textbf{Z}^T)$ )

The sequence $(\textbf{X}^T, \textbf{Y}^T, \textbf{Z}^T)$ is C-tight, and if $(\textbf{X},\textbf{Z})$ is a limit point of $(\textbf{X}^T, \textbf{Z}^T)$ , then $\textbf{Z}$ is a continuous martingale with $[\textbf{Z},\textbf{Z}] = {diag}(\textbf{X})$ . Furthermore, we have the convergence in probability

\begin{equation*} \underset{t \in [0,1]}{\sup} {\left\lVert{\textbf{Y}^T_t - \textbf{X}^T_t}\right\rVert_{2}}\overset{\mathbb{P}}{\underset{T \to \infty}{\to}} 0.\end{equation*}

Proof. The proof is essentially the same as in [Reference El Euch, Fukasawa and Rosenbaum8], adapted to our structure of Hawkes processes. Given $t \in [0,T]$ , we have

\begin{equation*}\boldsymbol{\lambda}^T_t = \boldsymbol{\mu}^T_t + \int_0^t \boldsymbol{\psi}^T(t-s) \boldsymbol{\mu}^T_s ds+ \int_0^t \boldsymbol{\psi}^T(t-s) d\textbf{M}^T_s,\end{equation*}

and therefore

\begin{align*} \mathbb{E}[\textbf{N}^T_t] &= \mathbb{E}\Big[\int_0^T \boldsymbol{\lambda}_s^T ds\Big] \\[6pt] & = \int_0^T \boldsymbol{\mu}^T_t dt + \int_0^T \int_0^t \boldsymbol{\psi}^T(t-s) \boldsymbol{\mu}^T_s ds dt \leq c T^{2\alpha} {\left\lVert{\boldsymbol{\mu}}\right\rVert_{\infty}},\end{align*}

where we have used the convergence of $T^{1-\alpha} \boldsymbol{\mu}^T_{T \cdot}$ (see Equation (7)) together with the weak convergence of $T^{-\alpha} \boldsymbol{\psi}^T(T \cdot)$ (see Lemma 2). It follows then that

\begin{equation*}\mathbb{E}\big[\textbf{X}^T_1\big] = \mathbb{E}\big[\textbf{Y}^T_1\big] \leq c,\end{equation*}

and since the processes are increasing, $\textbf{X}^T$ and $\textbf{Y}^T$ are tight. As the maximum jump size of $\textbf{X}^T$ and $\textbf{Y}^T$ tends to 0, we have the C-tightness of $(\textbf{X}^T, \textbf{Y}^T)$ . Since $\textbf{N}^T$ is the quadratic variation of $\textbf{M}^T$ , $(M^{T,i})^2 - N^{T,i}$ is an $L^2$ martingale starting at 0, and Doob’s inequality yields

\begin{align*} \sum_{1 \leq i \leq n} \mathbb{E}\bigg[ \underset{t \in [0,1]}{\sup} \big(X^{T,i}_t - Y^{T,i}_t\big)^{2}\bigg] & \leq 4 \sum_{1 \leq i \leq n} \mathbb{E}\Big[\big(X^{T,i}_1 - Y^{T,i}_1\big)^{2}\Big] \\ & \leq 4 T^{-4\alpha} \sum_{1 \leq i \leq n} \mathbb{E}\Big[\big(M^{T,i}_{T}\big)^{2}\Big] \\ & \leq 4 T^{-4\alpha} \sum_{1 \leq i \leq n} \mathbb{E}\Big[N^{T,i}_{T}\Big] \\ & \leq c T^{-2\alpha}.\end{align*}

Using the same approach as in [Reference El Euch, Fukasawa and Rosenbaum8] we conclude that $\textbf{Z}$ is a continuous martingale and $[\textbf{Z},\textbf{Z}]$ is the limit of $[\textbf{Z}^T,\textbf{Z}^T]$ .

5.2. Step 2: Rewriting of limit points of $\textbf{(}\textit{\textbf{X}}^{\textit{\textbf{T}}}, \textit{\textbf{Y}}^{\textit{\textbf{T}}}, \textit{\textbf{Z}}^{\textit{\textbf{T}}}\textbf{)}$

By Proposition 1, for any limit point $(\textbf{X}, \textbf{Y})$ of $(\textbf{X}^T, \textbf{Y}^T)$ , we have $\textbf{X} = \textbf{Y}$ almost surely. We use $\textbf{Y}^T$ to derive an equation for $\textbf{X}$ . As

\begin{equation*}\textbf{Y}^T = \dfrac{1}{T^{2\alpha}}\int_0^{tT} \boldsymbol{\lambda}^T_s ds,\end{equation*}

we first study $\boldsymbol{\lambda}^T_{sT}$ . Using Equation (14), for any $t \in [0,T]$ we have

\begin{align*} \int_0^t \boldsymbol{\lambda}^T_s ds &= \int_0^t \boldsymbol{\mu}^T_s ds + \int_0^t \int_0^u \boldsymbol{\psi}^T(s-u)\boldsymbol{\mu}^T_u du ds + \int_0^t \boldsymbol{\psi}^T(t-s) \textbf{M}^T_s ds \\[5pt] &= \int_0^t \boldsymbol{\mu}^T_s ds + \int_0^t \boldsymbol{\psi}^T(t-s) \int_0^s \boldsymbol{\mu}^T_u du ds + \int_0^t \boldsymbol{\psi}^T(t-s) \textbf{M}^T_s ds.\end{align*}

Thus, for any $t \in [0,1]$ , a change of variables leads to

\begin{align*} \int_0^{tT} \boldsymbol{\lambda}^T_s ds & = \int_0^{tT} \boldsymbol{\mu}^T_s ds + \int_0^{tT} \boldsymbol{\psi}^T(tT-s) \int_0^s \boldsymbol{\mu}^T_u du ds + \int_0^{tT} \boldsymbol{\psi}^T(tT-s) \textbf{M}^T_s ds \\[5pt] & = \int_0^{tT} \boldsymbol{\mu}^T_s ds + T \int_0^{t} \boldsymbol{\psi}^T(tT-sT) \int_0^{sT} \boldsymbol{\mu}^T_u du ds + T \int_0^{t} \boldsymbol{\psi}^T(tT-sT) \textbf{M}^T_{sT} ds \\[5pt] & = T \int_0^t \boldsymbol{\mu}^T_{sT} ds + T \int_0^{t} \boldsymbol{\psi}^T(T(t-s)) \int_0^{sT} \boldsymbol{\mu}^T_u du ds + T\int_0^{t} \boldsymbol{\psi}^T(T(t-s)) \textbf{M}^T_{sT} ds.\end{align*}

Therefore

(17) \begin{align} T^{2\alpha} \textbf{Y}^T_t & = T \int_0^t \boldsymbol{\mu}^T_{sT} ds + T \int_0^{t} \boldsymbol{\psi}^T(T(t-s))\int_0^{sT} \boldsymbol{\mu}^T_u du ds + T\int_0^{t} \boldsymbol{\psi}^T(T(t-s)) \textbf{M}^T_{sT} ds \end{align}

(18) \begin{align} \,{:}\,{\raise-1.5pt{=}}\, T^{2 \alpha}\Big(\textbf{Y}^{T,1}_t + \textbf{Y}^{T,2}_t + \textbf{Y}^{T,3}_t\Big),\end{align}

with obvious notation. Thus, to obtain our limit we use the convergence properties of $\textbf{F}^T$ which we derived earlier. We have the following proposition.

Proposition 2. Let $(\textbf{X}, \textbf{Z})$ be a limit point of $(\textbf{X}^T, \textbf{Z}^T$ ). Then, for any $t \in [0,1]$ , we have

\begin{equation*}\textbf{X}_t = \int_0^{t} \textbf{F}(t-s) \boldsymbol{\mu}_s ds + \int_0^{t} \textbf{F}(t-s)d\textbf{Z}_s.\end{equation*}

Proof. Let $(\textbf{X}, \textbf{Y}, \textbf{Z})$ be a limit point of $(\textbf{X}^T, \textbf{Y}^T, \textbf{Z}^T$ ). First, since

\begin{equation*}T^{1-\alpha} \boldsymbol{\mu}^T_{tT} \underset{T \to \infty}{\to} \boldsymbol{\mu_t}\end{equation*}

(see Equation (7)), $\textbf{Y}^{T,1}_t$ converges to 0 as T tends to infinity. Moving on to $\textbf{Y}^{T,2}$ , by integration by parts, for any $t \in [0,1]$ we obtain

\begin{align*} \textbf{Y}^{T,2}_t &= \int_0^{t} T^{1-\alpha} \boldsymbol{\psi}^T(T(t-s)) T^{-\alpha} \int_0^{sT} \boldsymbol{\mu}^T_u du ds \\[4pt] & = \left[ \textbf{F}^T(t-s) T^{-\alpha} \int_0^{sT} \boldsymbol{\mu}^T_{u} du \right]_{0}^{t} + \int_0^{t} \textbf{F}^T(t-s) T^{1-\alpha} \boldsymbol{\mu}^T_{sT} ds \\[4pt] & = \int_0^{t} \textbf{F}^T(t-s)T^{1-\alpha} \boldsymbol{\mu}^T_{sT} ds.\end{align*}

Using Equation (7) again, together with the uniform convergence of $\textbf{F}^T$ (see Lemma 2), we have the convergence

\begin{equation*}\textbf{Y}^{T,2}_t \underset{T \to \infty}{\to} \int_0^{t} \textbf{F}(t-s) \boldsymbol{\mu}_s ds.\end{equation*}

Finally, $\textbf{Y}^{T,3}_t$ can be written as

\begin{align*} \textbf{Y}^{T,3}_t &= T^{1-2\alpha} \int_0^{t} \boldsymbol{\psi}^T(T(t-s)) \textbf{M}^T_{sT} ds = \int_0^{t} \textbf{F}^T(t-s) d\textbf{Z}^T_s \\[4pt] & = \int_0^{t} \textbf{F}(t-s) d\textbf{Z}_s + \int_0^{t} \textbf{F}(t-s)(d\textbf{Z}^T_s - d\textbf{Z}_s) + \int_0^{t} (\textbf{F}^T(t-s) - \textbf{F}(t-s)) d\textbf{Z}^T_s.\end{align*}

The Skorokhod representation theorem applied to $(\textbf{Z}^T, \textbf{Z})$ yields the existence of copies in law $$(\mathop {{{\bf{Z}}^{^T}}}\limits^ \sim ,{\rm{ }}\mathop {\bf{Z}}\limits^ \sim )$$ , with $$\mathop {{{\bf{Z}}^T}}\limits^ \sim $$ converging almost surely to $$\mathop {\bf{Z}}\limits^ \sim $$ . We proceed with $$(\mathop {{{\bf{Z}}^{^T}}}\limits^ \sim ,{\rm{ }}\mathop {\bf{Z}}\limits^ \sim )$$ and keep the previous notation. The stochastic Fubini theorem [Reference Veraar27] gives, almost surely,

\begin{equation*}\int_0^{t} \textbf{F}(t-s)(d\textbf{Z}^T_s - d\textbf{Z}_s) = \int_0^{t} \textbf{f}(s) (\textbf{Z}^T_{t-s} - \textbf{Z}_{t-s}) ds.\end{equation*}

From the dominated convergence theorem we obtain the almost sure convergence

\begin{equation*}\int_0^{t} \textbf{f}(s) (\textbf{Z}^T_{t-s} - \textbf{Z}_{t-s}) ds \underset{T \to \infty}{\to} 0.\end{equation*}

Furthermore, since $[\textbf{Z}^T,\textbf{Z}^T] = {diag}(\textbf{X}^T)$ we have

\begin{align*}\sum_{1 \leq i \leq n} \mathbb{E}\left[ \left( \int_0^{t} (\textbf{F}^T(t-s) - \textbf{F}(t-s)) d\textbf{Z}^T_s \right)_{i}^2 \right] & \leq \sum_{1 \leq i,j \leq n} \int_0^{t} \big(F_{ij}^T(t-s)\\ & \ \ - F_{ij}(t-s)\big)^2 T^{1-\alpha} \mathbb{E}\big[\lambda^{T,j}_{sT}\big] ds.\end{align*}

Using Equation (14) together with Lemma 1, we can bound $\mathbb{E}[\lambda^{T,j}_{sT}]$ independently of T, and

\begin{align*}\sum_{1 \leq i \leq n} \mathbb{E}\left[ \left( \int_0^{t} (\textbf{F}^T(t-s) - \textbf{F}(t-s)) d\textbf{Z}^T_s \right)_{i}^2 \right] \leq c \sum_{1 \leq i,j \leq n} \int_0^{t} \Big(F_{ij}^T(t-s) - F_{ij}(t-s)\Big)^2 ds.\end{align*}

The right-hand side converges to 0 by the dominated convergence theorem together with the uniform convergence of $\textbf{F}^T$ to $\textbf{F}$ (see Lemma 2). From Proposition 1 we know that $\textbf{Y}=\textbf{X}$ almost surely. Putting everything together, almost surely,

\begin{equation*}\textbf{X}_t = \int_0^{t} \textbf{F}(t-s) \boldsymbol{\mu}_s ds + \int_0^{t} \textbf{F}(t-s)d\textbf{Z}_s.\end{equation*}

This is valid for any limit point $(\textbf{X},\textbf{Z})$ of $(\textbf{X}^T,\textbf{Z}^T)$ , which concludes the proof.

The previous proposition gives suitable martingale properties of limit points of $\textbf{Z}^T$ to apply the martingale representation theorem, which is the topic of the following proposition.

Proposition 3. Let $(\textbf{X},\textbf{Z})$ be a limit point of $(\textbf{X}^T, \textbf{Z}^T)$ . There exists, up to an extension of the original probability space, an n-dimensional Brownian motion $\textbf{B}$ and a nonnegative process $\textbf{V}$ such that, for any $t \in [0,1]$ , one has

\begin{align*} \textbf{X}_t &= \int_0^{t} \textbf{V}_s ds, \\[4pt] \textbf{Z}_t & = \int_0^{t} {diag}\Big(\sqrt{\textbf{V}_s}\Big) d\textbf{B}_s, \\[4pt] \textbf{V}_t &= \int_0^t \textbf{f}(t-s) \boldsymbol{\mu}_sds + \int_0^{t} \textbf{f}(t-s) {diag}\Big(\sqrt{\textbf{V}_s}\Big) d\textbf{B}_s.\end{align*}

Proof. This proof relies on the martingale representation theorem applied to $\textbf{Z}$ . Consider a limit point $(\textbf{X},\textbf{Z})$ of $(\textbf{X}^T, \textbf{Z}^T)$ . Following the proof of Theorem 3.2 in [Reference Jaisson and Rosenbaum21] and using Proposition 2, $\textbf{X}$ can be written as the integral of a process $\textbf{V}$ , i.e.

\begin{equation*}\textbf{X}_t = \int_0^{t} \textbf{V}_s ds,\end{equation*}

with $\textbf{V}$ satisfying the equation

\begin{equation*}\textbf{V}_t = \int_0^t \textbf{f}(t-s) \boldsymbol{\mu}_s ds + \int_0^{t} \textbf{f}(t-s) d\textbf{Z}_s.\end{equation*}

Therefore, as $[\textbf{Z}, \textbf{Z}]_t = {diag}(\textbf{X}_t) = {diag}(\int_0^{t} \textbf{V}_s ds) $ and $\textbf{Z}$ is a continuous martingale, by the martingale representation theorem (see for example Theorem 3.9 from [Reference Revuz and Yor26]), there exists (up to an enlargement of the probability space) a multivariate Brownian motion $\textbf{B}$ and a predictable square-integrable process $\textbf{H}$ such that

\begin{equation*}\textbf{Z}_t = \int_0^t \textbf{H}_s d\textbf{B}_s.\end{equation*}

Furthermore, note that $\textbf{V}$ is a nonnegative process (as $\textbf{X}$ is a nondecreasing process), and we have

\begin{equation*}\textbf{Z}_t = \int_0^t {diag}\big(\sqrt{\textbf{V}_s}\big) {diag}\big(\sqrt{\textbf{V}_s}\big)^{-1} \textbf{H}_s d\textbf{B}_s.\end{equation*}

A simple computation shows that, since $${[{\bf{Z}},{\bf{Z}}]_t} = {\int_0^t {{{\bf{H}}_s}} ^ \top }{{\bf{H}}_s}ds = {{\bf{X}}_t} = \int_0^t {{{\bf{V}}_s}} ds$$ , the process $${\mathop {\rm{B}}\limits^ \sim }_{\bf{t}} \,{:}\,{\raise-1.5pt{=}}\, \int_0^t {diag}\Big(\sqrt{\textbf{V}_s}\Big)^{-1} \textbf{H}_s d\textbf{B}_s$$ is a Brownian motion. Finally,

$${{\bf{V}}_t} = \int_0^t {\bf{f}} (t - s){\mu _s}ds + \int_0^t {\bf{f}} (t - s)diag(\sqrt {{{\bf{V}}_s}} )d{\mathop {\bf{B}}\limits^ \sim } _s.$$

A straightforward application of Lemma 4.4 and Lemma 4.5 in [Reference Jaisson and Rosenbaum21] yields the following lemma.

Lemma 3. Consider a (weak) nonnegative solution $\textbf{V}$ of the stochastic Volterra equation

\begin{equation*}\textbf{V}_t = \int_0^t \textbf{f}(t-s) \boldsymbol{\mu}_s ds + \int_0^{t} \textbf{f}(t-s) {diag}\big(\sqrt{\textbf{V}_s}\big) d\textbf{B}_s,\end{equation*}

where $\textbf{B}$ is a Brownian motion. Then every component of $\textbf{V}$ has pathwise Hölder regularity $\alpha - 1/2 - \epsilon$ for any $\epsilon > 0$ .

5.3. Step 3: Proof of Equation (11)

Properties of the Mittag-Leffler function (as in [Reference El Euch, Fukasawa and Rosenbaum8]) enable us to rewrite the previous stochastic differential equation using power-law kernels, which is the subject of the next proposition. Let

\begin{align*}\boldsymbol{\Theta}^1 \,{:}\,{\raise-1.5pt{=}}\, \bigg(\textbf{O}_{\textbf{11}} + \textbf{O}_{\textbf{12}}\bigg(\textbf{I} - {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds\bigg)^{-1} {\int_{0}^{\infty}{{\textbf{B}}}}(s)ds\bigg) \textbf{K}^{-1},\\[5pt] \boldsymbol{\Theta}^2 \,{:}\,{\raise-1.5pt{=}}\, \bigg(\textbf{O}_{\textbf{21}} + \textbf{O}_{\textbf{22}} \bigg(\textbf{I} - {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds\bigg)^{-1} {\int_{0}^{\infty}{{\textbf{B}}}}(s)ds\bigg)\textbf{K}^{-1},\\[5pt] \boldsymbol{\Lambda} \,{:}\,{\raise-1.5pt{=}}\, \dfrac{\alpha}{\Gamma(1 - \alpha)} \textbf{K} \textbf{M}^{-1}. \end{align*}

Proposition 4. Given an m-dimensional Brownian motion $\textbf{B}$ , a nonnegative process $\textbf{V}$ is a solution of the stochastic differential equation

\begin{equation*}\textbf{V}_t = \int_0^t \textbf{f}(t-s) \boldsymbol{\mu}_s ds + \int_0^{t} \textbf{f}(t-s) {diag}\big(\sqrt{\textbf{V}_s}\big) d\textbf{B}_s\end{equation*}

if and only if there exists a process $$\mathop {\bf{V}}\limits^ \sim $$ of Hölder regularity $\alpha - 1/2 - \epsilon$ for any $\epsilon > 0$ such that $\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim ~}_t = (V^1, \cdots, V^{n_c})$ and $\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }_t = (V^{n_c+1}, \cdots, V^{2m})$ are nonnegative processes and $$\mathop {\bf{V}}\limits^ \sim $$ is solution of the following stochastic Volterra equation:

\begin{align*}{\mathop {\bf{V}}\limits^ \sim }_t =& \dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1}\Big(\textbf{O}_{\textbf{11}}^{({-}1)}\boldsymbol{\mu^1}_s + \textbf{O}_{\textbf{12}}^{({-}1)}\boldsymbol{\mu^2}_s - {\mathop {\bf{V}}\limits^ \sim }_s\Big) ds \\[4pt] &+\dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1} \textbf{O}_{\textbf{11}}^{({-}1)}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{1}}_s \\[4pt] &+\dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1} \textbf{O}_{\textbf{12}}^{({-}1)}{diag}\Big(\sqrt{\boldsymbol{\Theta}^2}{\mathop {\bf{V}}\limits^ \sim }_s\Big) d\textbf{W}^{\textbf{2}}_s,\end{align*}

where $\textbf{W}^{\textbf{1}} \,{:}\,{\raise-1.5pt{=}}\, (B^1, \cdots, B^{n_c})$ and $\textbf{W}^{\textbf{2}} \,{:}\,{\raise-1.5pt{=}}\, (B^{n_c+1}, \cdots, B^{2m})$ .

Proof. We begin by showing the first implication. Starting from Proposition 3 we have

\begin{equation*}\textbf{V}_t = \int_0^t \textbf{f}(t-s) \boldsymbol{\mu}_s ds + \int_0^{t} \textbf{f}(t-s) {diag}\Big(\sqrt{\textbf{V}_s}\Big) d\textbf{B}_s.\end{equation*}

Developing from the definition of $\textbf{f}$ in Equation (16), for any $t \in [0,1]$ , $\textbf{f}$ can be written

\begin{align*}\textbf{f}(t) &= \begin{pmatrix} \big(\textbf{O}_{\textbf{11}} + \textbf{O}_{\textbf{12}}\big(\textbf{I} - {\int_{0}^{\infty}{\textbf{C}}}(s)ds\big)^{-1} {\int_{0}^{\infty}{\textbf{B}}}(s)ds\big) \textbf{K}^{-1} \textbf{f}^{\alpha,\boldsymbol{\Lambda}}(t) & \quad \textbf{0} \\[8pt] \big(\textbf{O}_{\textbf{21}} + \textbf{O}_{\textbf{22}} (\textbf{I} - {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds)^{-1} {\int_{0}^{\infty}{{\textbf{B}}}}(s)ds\big)\textbf{K}^{-1} \textbf{f}^{\alpha,\boldsymbol{\Lambda}}(t) & \quad \textbf{0} \end{pmatrix} \begin{pmatrix} \textbf{O}_{\textbf{11}}^{({-}1)} & \quad \textbf{O}_{\textbf{12}}^{({-}1)} \\[8pt] \textbf{O}_{\textbf{21}}^{({-}1)} & \quad \textbf{O}_{\textbf{22}}^{({-}1)} \end{pmatrix}.\end{align*}

Defining $\textbf{V}^{\textbf{1}} \,{:}\,{\raise-1.5pt{=}}\, (V^1, \cdots, V^{n_c})$ and $\textbf{V}^{\textbf{2}} \,{:}\,{\raise-1.5pt{=}}\, (V^{n_c+1}, \cdots, V^{2m})$ , we have

\begin{align*} \textbf{V}^{\textbf{1}}_t &=\boldsymbol{\Theta^1} \int_0^t \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu}^{\boldsymbol{1}}_s ds + \boldsymbol{\Theta^1} \int_0^t \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu}^{\boldsymbol{2}}_s ds \\[5pt] & + \boldsymbol{\Theta^1} \int_0^{t} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{1}}_s}\Big) d\textbf{W}^{\textbf{1}}_s+ \boldsymbol{\Theta^1} \int_0^{t} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{2}}_s}\Big) d\textbf{W}^{\textbf{2}}_s.\end{align*}

If $\boldsymbol{\Theta^1}$ were nonsingular, we could express $\textbf{V}^{\textbf{1}}$ with power-law kernels using the same approach as in [Reference El Euch, Fukasawa and Rosenbaum8]. In general we define

\begin{align*} {\mathop {\bf{V}}\limits^ \sim }_t \,{:}\,{\raise-1.5pt{=}}\, \int_0^t \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) (\textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu^1}_s + \textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu^2}_s)ds \\[5pt] &+ \int_0^{t} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{1}}_s}\Big) d\textbf{W}^{\textbf{1}}_s + \int_0^{t} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{2}}_s}\Big) d\textbf{W}^{\textbf{2}}_s.\end{align*}

From the same arguments as in Lemma 3, Hölder regularity of $\textbf{V}$ carries over to ${\mathop {\bf{V}}\limits^ \sim }$ , and the components of ${\mathop {\bf{V}}\limits^ \sim }$ are of Hölder regularity $\alpha - 1/2 - \epsilon$ for any $\epsilon > 0$ ; hence Lemma 3 shows $\mathcal{{K}} \,{:}\,{\raise-1.5pt{=}}\, I^{1 - \alpha} {\mathop {\bf{V}}\limits^ \sim }$ is well-defined, where $I^{1-\alpha}$ is the fractional integration operator of order $1-\alpha$ (see Definition 1 in Appendix A.2). Note that for any t in [0, 1], using Lemma 4 of Appendix A.2, we have

\begin{align*} \mathcal{{K}}_t =& \int_0^t \boldsymbol{\Lambda} (\textbf{I} - \textbf{F}^{\alpha, \boldsymbol{\Lambda}}(t-s)) (\textbf{O}^{({-}1)}_{\textbf{11}} \boldsymbol{\mu^1}_s + \textbf{O}^{({-}1)}_{\textbf{12}} \boldsymbol{\mu^2}_s) ds \\[4pt] &+ \int_0^t \boldsymbol{\Lambda} (\textbf{I} - \textbf{F}^{\alpha, \boldsymbol{\Lambda}}(t-s)) \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{1}}_s}\Big) d\textbf{W}^{\textbf{1}}_s \\[4pt] &+ \int_0^t \boldsymbol{\Lambda} (\textbf{I} - \textbf{F}^{\alpha, \boldsymbol{\Lambda}}(t-s)) \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{2}}_s}\Big) d\textbf{W}^{\textbf{2}}_s \\[4pt] =& \boldsymbol{\Lambda} \int_0^t (\textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu^1}_s +\textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu^2}_s) ds + \int_0^t \boldsymbol{\Lambda} \textbf{O}_{\textbf{11}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{1}}_s}\Big) d\textbf{W}^{\textbf{1}}_s \\[4pt] &+ \int_0^t \boldsymbol{\Lambda} \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{2}}_s}\Big) d\textbf{W}^{\textbf{2}}_s \\[4pt] &- \boldsymbol{\Lambda} \int_0^t \left[ \textbf{F}^{\alpha, \boldsymbol{\Lambda}}(t-s)\textbf{O}^{({-}1)}_{\textbf{11}} \boldsymbol{\mu^1}_s + \int_0^s \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(s-u) \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{1}}_u}\Big) d\textbf{W}^{\textbf{1}}_u\right] ds \\[4pt] &- \boldsymbol{\Lambda} \int_0^t \left[ \textbf{F}^{\alpha, \boldsymbol{\Lambda}}(t-s)\textbf{O}^{({-}1)}_{\textbf{12}} \boldsymbol{\mu^2}_s + \int_0^s \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(s-u) \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\textbf{V}^{\textbf{2}}_u}\Big)d\textbf{W}^{\textbf{2}}_u \right] ds.\end{align*}

The last two terms can be rewritten using the definition of ${\mathop {\bf{V}}\limits^ \sim }$ , so that

\begin{align*} \mathcal{{K}}_t =\, & \boldsymbol{\Lambda} \int_0^t \Big(\textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu^1}_s + \textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu^2}_s - {\mathop {\bf{V}}\limits^ \sim }_s\Big) ds + \boldsymbol{\Lambda} \int_0^t \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }_s}\Big)d\textbf{W}^{\textbf{1}}_s \\[4pt] &+ \boldsymbol{\Lambda} \int_0^t \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{2}}_s.\end{align*}

Thanks to the Hölder regularity of ${\mathop {\bf{V}}\limits^ \sim }$ , we can now apply the fractional differentiation operator of order $1-\alpha$ (see Definition 1 in Appendix A.2) together with the stochastic Fubini theorem to deduce that

\begin{align*}{\mathop {\bf{V}}\limits^ \sim }_t &= \dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1}\Big(\textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu^1}_s + \textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu^2}_s - {\mathop {\bf{V}}\limits^ \sim }_s\Big) ds\\[4pt] &+\dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1} \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{1}}_s\\[4pt] &+\dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1} \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{2}}_s.\end{align*}

This concludes the proof of the first implication. We now show the second implication. Suppose there exists ${\mathop {\bf{V}}\limits^ \sim }$ of Hölder regularity $\alpha - 1/2 - \epsilon$ for any $\epsilon > 0$ such that $\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }$ and $\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }$ are positive, solving the following stochastic Volterra equation:

\begin{align*}{\mathop {\bf{V}}\limits^ \sim }_t =\, & \dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1}\Big(\textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu^1}_s + \textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu^2}_s - {\mathop {\bf{V}}\limits^ \sim }_s\Big) ds \\[4pt] &+\dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1} \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{1}}_s\\[4pt] &+\dfrac{1}{\Gamma(\alpha)}\boldsymbol{\Lambda} \int_0^t (t-s)^{\alpha-1} \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{2}}_s.\end{align*}

For this proof, let us write

\begin{equation*}\boldsymbol{\theta} \,{:}\,{\raise-1.5pt{=}}\, \boldsymbol{\Lambda} \textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu^1} + \boldsymbol{\Lambda} \textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu^2}, \qquad \boldsymbol{\Lambda}_1 \,{:}\,{\raise-1.5pt{=}}\, \boldsymbol{\Lambda} \textbf{O}^{({-}1)}_{\textbf{11}}, \qquad \boldsymbol{\Lambda}_2 \,{:}\,{\raise-1.5pt{=}}\, \boldsymbol{\Lambda} \textbf{O}^{({-}1)}_{\textbf{12}},\end{equation*}

so that, for any t in [0,1],

\begin{align*}{\mathop {\bf{V}}\limits^ \sim }_t =\, & \dfrac{1}{\Gamma(\alpha)} \int_0^t (t-s)^{\alpha-1}\big(\boldsymbol{\theta}_s - \boldsymbol{\Lambda} {\mathop {\bf{V}}\limits^ \sim }_s\big) ds \\[4pt] &+\dfrac{1}{\Gamma(\alpha)} \int_0^t (t-s)^{\alpha-1}\boldsymbol{\Lambda_1} {diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{1}}_s\\[4pt] &+\dfrac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1} \boldsymbol{\Lambda_2}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{2}}_s.\end{align*}

Notice that the above can be written

\begin{equation*} {\mathop {\bf{V}}\limits^ \sim }_t = I^{\alpha}(\boldsymbol{\theta} - \boldsymbol{\Lambda} {\mathop {\bf{V}}\limits^ \sim })_t + I^{\alpha}_{\textbf{B}^1}\Big(\boldsymbol{\Lambda_1}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big)_t + I^{\alpha}_{\textbf{B}^2}(\boldsymbol{\Lambda_2}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big)_t,\end{equation*}

where $I^{\alpha}_{\textbf{B}}$ is the fractional integration operator with respect to $\textbf{B}$ (see Definition 2 in Appendix A.2). Iterating the application of $I^{\alpha}$ we find that, for any $N \geq 1$ , ${\mathop {\bf{V}}\limits^ \sim }$ satisfies

\begin{align*}{\mathop {\bf{V}}\limits^ \sim } =& \sum_{1 \leq k \leq N} \boldsymbol{\Lambda}^{k-1} ({-}1)^{k-1} I^{(k-1) \alpha} \Big[I^{\alpha} \boldsymbol{\theta} + I^{\alpha}_{\textbf{B}^1}\Big(\boldsymbol{\Lambda_1}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big) + I^{\alpha}_{\textbf{B}^2}\Big(\boldsymbol{\Lambda_2}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }})\Big)\Big] \\[5pt] &+ \boldsymbol{\Lambda}^{N} ({-}1)^{N} I^{(N+1) \alpha}{\mathop {\bf{V}}\limits^ \sim }.\end{align*}

Now, note that $\boldsymbol{\theta}$ , ${diag}(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim })}$ , ${diag}(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim })}$ , and ${\mathop {\bf{V}}\limits^ \sim }$ are square-integrable processes, and Lemma 8 in Appendix A.2 shows that the sum converges almost surely to the series, while $\boldsymbol{\Lambda}^{N} ({-}1)^{N} I^{(N+1) \alpha}{\mathop {\bf{V}}\limits^ \sim }$ converges almost surely to zero, as N tends to infinity. Thus we have

\begin{align*}{\mathop {\bf{V}}\limits^ \sim } = & \sum_{k \geq 0} \boldsymbol{\Lambda}^{k} ({-}1)^{k} I^{k \alpha} \Big[I^{\alpha} \boldsymbol{\theta} + I^{\alpha}_{\textbf{B}^1}\Big(\boldsymbol{\Lambda_1}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big) + I^{\alpha}_{\textbf{B}^2}\Big(\boldsymbol{\Lambda_2}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big)\Big] \\ = & \sum_{k \geq 0} \boldsymbol{\Lambda}^{k} ({-}1)^{k} I^{k \alpha} I^{\alpha} \boldsymbol{\theta} + \sum_{k \geq 0} \boldsymbol{\Lambda}^{k} ({-}1)^{k} I^{k \alpha} I^{\alpha}_{\textbf{B}^1}\Big(\boldsymbol{\Lambda_1}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big)\\ &+ I^{\alpha}_{\textbf{B}^2}\Big(\boldsymbol{\Lambda_2}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big)\Big] \\ = & \boldsymbol{\Lambda}^{-1} \sum_{k \geq 0} \boldsymbol{\Lambda}^{k+1} ({-}1)^{k} I^{(k+1) \alpha} \boldsymbol{\theta} + \sum_{k \geq 0} \boldsymbol{\Lambda}^{k} ({-}1)^{k} I^{k \alpha} I^{\alpha}_{\textbf{B}^1}\Big(\boldsymbol{\Lambda_1}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }}\Big)\Big)\\ &+ I^{\alpha}_{\textbf{B}^2}\Big(\boldsymbol{\Lambda_2}{diag}\Big(\sqrt{\boldsymbol{\Theta^2}{\mathop {\bf{V}}\limits^ \sim }}\Big)\Big)\Big].\end{align*}

Lemmas 5 and 7 in Appendix A.2 enable us to rewrite the above using the matrix Mittag-Leffler function. This yields, for any t in [0,1] and almost surely,

\begin{align*} {\mathop {\bf{V}}\limits^ \sim }_t &= \boldsymbol{\Lambda}^{-1} \int_0^t \textbf{f}^{\alpha, \Lambda}(t-s) \boldsymbol{\theta_s} ds + \boldsymbol{\Lambda}^{-1} \int_0^{t} \textbf{f}^{\alpha, \Lambda}(t-s) \boldsymbol{\Lambda_1}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{1}}_s \\[5pt] &+ \boldsymbol{\Lambda}^{-1} \int_0^{t} \textbf{f}^{\alpha, \Lambda}(t-s) \boldsymbol{\Lambda_2}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{2}}_s .\end{align*}

Replacing $\boldsymbol{\theta}, \boldsymbol{\Lambda}_1, \boldsymbol{\Lambda}_2$ by their expressions, almost surely and for any t in [0,1], we have

\begin{align*} {\mathop {\bf{V}}\limits^ \sim }_t &= \int_0^t \textbf{f}^{\alpha, \Lambda}(t-s) (\textbf{O}^{({-}1)}_{\textbf{11}}\boldsymbol{\mu^1}_s + \boldsymbol{\Lambda} \textbf{O}^{({-}1)}_{\textbf{12}}\boldsymbol{\mu^2}_s)ds \\[3pt] &\quad + \int_0^{t} \textbf{f}^{\alpha, \Lambda}(t-s) \textbf{O}^{({-}1)}_{\textbf{11}}{diag}\Big(\sqrt{\boldsymbol{\Theta^1} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{1}}_s \\[3pt] &\quad + \int_0^{t} \textbf{f}^{\alpha, \Lambda}(t-s) \textbf{O}^{({-}1)}_{\textbf{12}}{diag}\Big(\sqrt{\boldsymbol{\Theta^2} {\mathop {\bf{V}}\limits^ \sim }_s}\Big) d\textbf{W}^{\textbf{2}}_s.\end{align*}

This concludes the second implication and the proof.

5.4. Step 4: Equation satisfied by the limiting price process

The previous results on the convergence of the intensity process enable us to now turn to the question of the limiting price dynamics. Recall that the sequence of rescaled price processes $\textbf{P}^T$ is defined as

\begin{equation*}\textbf{P}^T \,{:}\,{\raise-1.5pt{=}}\, {{}^ \top {{\textbf{Q}}}} \textbf{X}^T,\end{equation*}

where $\textbf{Q} = \begin{pmatrix}\textbf{e}_1 - \textbf{e}_2 \mid & \cdots &\mid \textbf{e}_{2m-1} - \textbf{e}_{2m}\end{pmatrix}.$ We have the following result.

Proposition 5. Let $(\textbf{X},\textbf{Z})$ be a limit point of $(\textbf{X}^T, \textbf{Z}^T)$ and $\textbf{P} = {{}^ \top {{\textbf{Q}}}} \textbf{X}$ . Then

\begin{equation*}\textbf{P}_t = (\textbf{I} + \boldsymbol{\Delta}) {{}^ \top {{\textbf{Q}}}} \bigg(\textbf{Z}_t + \int_0^t \boldsymbol{\mu}_s ds\bigg),\end{equation*}

where $\boldsymbol{\Delta} = ({\int_{0}^{\infty}{{\delta^T_{ij}}}})_{1 \leq i,j \leq m}$ .

Proof. Let $(\textbf{X}, \textbf{Z})$ be a limit point of $(\textbf{X}^T, \textbf{Z}^T)$ . For any $1 \leq i \leq m$ we can compute the difference between upward and downard jumps on Asset i as

\begin{align*} \textbf{v}_i \cdot \textbf{N}^T_t = \textbf{v}_i \cdot \textbf{M}^T_t + \textbf{v}_i \cdot \int_0^t \boldsymbol{\lambda}_s^T ds,\end{align*}

with the following expression for the integrated intensity:

\begin{align*} \int_0^{tT} \boldsymbol{\lambda}^T_s ds & = T \int_0^{t} \boldsymbol{\mu}^T_{sT} ds + T \int_0^t \int_0^{T(t-s)} \boldsymbol{\psi}^T(u) du \boldsymbol{\mu}^T_{sT} ds + {{\left\lVert{\psi^{\textbf{T}}}\right\rVert_{1}}} \textbf{M}^{\textbf{T}}_{\textbf{tT}} \\[5pt] &- \int_0^{tT} \int_{tT-s}^{\infty} \boldsymbol{\psi}^T(u) du d\textbf{M}^T_s.\end{align*}

Thus the microscopic price for Asset i satisfies

\begin{align*} T^{-\alpha} \textbf{v}_i \cdot \textbf{N}^T_{tT} \,=\, & T^{1-\alpha} \int_0^{t} \textbf{v}_i \cdot \boldsymbol{\mu}^T_{sT} ds + T^{1-\alpha} {{}^ \top {{{\left\lVert{\boldsymbol{\psi^T}}\right\rVert_{1}}}}}\textbf{v}_i \cdot \int_0^t \boldsymbol{\mu}^T_{sT} ds + \textbf{v}_i \cdot \textbf{Z}^{\textbf{T}}_{\textbf{t}} + {{}^ \top {{{\left\lVert{\psi^{\textbf{T}}}\right\rVert_{1}}}}} \textbf{v}_i \cdot \textbf{Z}^{\textbf{T}}_{\textbf{t}} \\[5pt] & - T^{-\alpha} \int_0^t \int_{T(t-s)}^{\infty} {{}^ \top {{\boldsymbol{\psi}^T}}}(u)\textbf{v}_i \cdot \boldsymbol{\mu}^T_{sT} du ds - T^{-\alpha} \int_0^{tT} \int_{tT-s}^{\infty} \boldsymbol{\psi}^T(u) du d\textbf{M}^T_s \\ =\,& \sum_{1 \leq k \leq m} \bigg(\mathbb{1}_{ik} + {\int_{0}^{\infty}{{\delta^T_{ik}}}}\bigg) \textbf{v}_k \cdot \textbf{Z}^{\textbf{T}}_{\textbf{t}} + \sum_{1 \leq k \leq m} \bigg(\mathbb{1}_{ik} + {\int_{0}^{\infty}{{\delta^T_{ik}}}}\bigg) T^{1-\alpha} \int_0^{t} \textbf{v}_k \cdot \boldsymbol{\mu}^T_{sT} ds \\[5pt] & - \int_0^{t} \int_{tT-s}^{\infty} {{}^ \top {{\boldsymbol{\psi}^T}}}(u) \textbf{v}_i du \cdot d\textbf{Z}^{\textbf{T}}_{\textbf{s}} - T^{-\alpha} \int_0^t \int_{T(t-s)}^{\infty} {{}^ \top {{\boldsymbol{\psi}^T}}}(u)\textbf{v}_i \cdot \boldsymbol{\mu}^T_{sT} du ds.\end{align*}

It is straightforward to show that the last two terms converge to zero, and thus any limit point $\textbf{P}$ of $\textbf{P}^T = {{}^ \top {{\textbf{Q}}}} \textbf{X}^T$ is such that

$${{\bf{P}}_t} = {({\bf{I}} + \Delta )^ \top }{\bf{Q}}({{\bf{Z}}_t} + \int_0^t {{\mu _s}} ds).$$

Replacing $\textbf{Z}$ by the expression obtained in Proposition 3 concludes the proof of Theorem 1, since

\begin{equation*}\textbf{P}_t = (\textbf{I} + \boldsymbol{\Delta}) {{}^ \top {{\textbf{Q}}}} \bigg( \int_0^{t} {diag}\Big(\sqrt{\textbf{V}_s}\Big) d\textbf{B}_s + \int_0^t \boldsymbol{\mu}_s ds \bigg).\end{equation*}

A.1 Independence of Equation (11) from chosen basis

We consider two representations which satisfy Assumption 1. Let $\textbf{P}, {\mathop {\bf{P}}\limits^ \sim }$ be invertible matrices, $0 \leq n_c, n_{c^{\prime}} \leq n$ , and let

\begin{equation*}\textbf{A}^T \in {\mathcal{F}(\mathcal{M}_{n_c}({\mathbb{R}}))}, \qquad \textbf{C}^T \in {\mathcal{F}(\mathcal{M}_{{n-n_c}}({\mathbb{R}}))}, \qquad \textbf{B}^T \in {\mathcal{F}(\mathcal{M}_{{n-n_c,n_c}}({\mathbb{R}}))},\end{equation*}

$${\mathop {{{\bf{A}}^T}}\limits^ \sim } \in {\cal F}({{\cal M}_{{n_{{c^\prime }}}}}()),\qquad {\mathop {{{\bf{C}}^T}}\limits^ \sim } \in {\cal F}({{\cal M}_{n - {n_{{c^\prime }}}}}()),\qquad {\mathop {{{\bf{B}}^T}}\limits^ \sim } \in {\cal F}({{\cal M}_{n - {n_{{c^\prime }}},{n_{{c^\prime }}}}}())$$

be such that

\begin{align*} \boldsymbol{\phi}^T &= \textbf{P} \begin{pmatrix} \textbf{A}^T & \quad \textbf{0} \\[5pt] \textbf{B}^T & \quad \textbf{C}^T \end{pmatrix} \textbf{P}^{-1} = {\mathop {\bf{P}}\limits^ \sim } \begin{pmatrix} {\mathop {{{\bf{A}}^T}}\limits^ \sim } & \quad \textbf{0} \\[5pt] {\mathop {{{\bf{B}}^T}}\limits^ \sim } & \quad {\mathop {{{\bf{C}}^T}}\limits^ \sim }\end{pmatrix} {\mathop {\bf{P}}\limits^ \sim }^{-1}.\end{align*}

We write $\textbf{A}$ for the limit of $\textbf{A}^T$ (and similarly for $\textbf{B}^T, \textbf{C}^T$ , etc.). First, notice that we must have $n_c = n_{c^{\prime}}$ . Indeed, since ${\rho{({{\int_{0}^{\infty}{{\textbf{C}}}}})}}<1$ and $$\rho (\int_{0}^{\infty} {\mathop {\bf{C}}\limits^ \sim }) < 1$$ , 1 is neither an eigenvalue of ${\int_{0}^{\infty}{{\textbf{C}}}}$ nor of ${\int_{0}^{\infty}{{\boldsymbol{\mathop {\bf{C}}\limits^ \sim }}}}$ . Yet, since $\textbf{A} = \textbf{I}$ and $\boldsymbol{\mathop {\bf{A}}\limits^ \sim } = \textbf{I}$ , 1 is an eigenvalue of $\boldsymbol{\phi}$ with multiplicity $n_{c}$ and $n_{c^{\prime}}$ . Therefore $n_{c} = n_{c^{\prime}}$ .

Writing $\textbf{L}=\textbf{P}^{-1} {\mathop {\bf{P}}\limits^ \sim }$ , we have

\begin{align*} \begin{pmatrix} \textbf{A} & \quad \textbf{0} \\[4pt] \textbf{B} & \quad\textbf{C} \end{pmatrix} &= \textbf{L} \begin{pmatrix} \boldsymbol{\mathop {\bf{A}}\limits^ \sim } & \quad\textbf{0} \\[4pt] \boldsymbol{\mathop {\bf{B}}\limits^ \sim } & \quad\boldsymbol{\mathop {\bf{C}}\limits^ \sim } \end{pmatrix} \textbf{L}^{-1}.\end{align*}

Since $\textbf{A} = \boldsymbol{\mathop {\bf{A}}\limits^ \sim } = \textbf{I}$ because of Equation (5), writing the blockwise matrix product and using the assumption that $\textbf{I}-\textbf{C}$ is invertible, we get

\begin{align*} \textbf{L}_{\textbf{12}} &= \textbf{0}, \\ (\textbf{I}-\textbf{C}) \textbf{L}_{\textbf{21}} &= \textbf{B} \textbf{L}_{\textbf{11}} - \textbf{L}_{\textbf{22}} \boldsymbol{\mathop {\bf{B}}\limits^ \sim }, \\ \textbf{C} \textbf{L}_{\textbf{22}} &= \textbf{L}_{\textbf{22}} \boldsymbol{\mathop {\bf{C}}\limits^ \sim }.\end{align*}

Since $\textbf{L} \textbf{L}^{-1} = \textbf{I}$ , $\textbf{L}_{\textbf{11}} = \textbf{I}$ , $\textbf{L}_{\textbf{22}} = \textbf{I}$ , and $\textbf{L}_{\textbf{21}} = - \textbf{L}^{(\textbf{-1})}_{\textbf{21}}$ , we deduce that

\begin{align*} \textbf{L}_{\textbf{11}} = \textbf{I}, \qquad \textbf{L}_{\textbf{22}} = \textbf{I}, \qquad \textbf{L}_{\textbf{12}} = \textbf{0}, \qquad (\textbf{I}-\textbf{C}) \textbf{L}_{\textbf{21}} &= \textbf{B} - \boldsymbol{\mathop {\bf{B}}\limits^ \sim }, \qquad \textbf{C} = {\mathop {\bf{C}}\limits^ \sim }.\end{align*}

As $\textbf{L} = \textbf{P}^{-1} {\mathop {\bf{P}}\limits^ \sim }$ , we have

\begin{align*} \textbf{P}^{-1} =& \begin{pmatrix} \textbf{I} & \quad\textbf{0} \\[4pt] (\textbf{I}-\textbf{C})^{-1} (\textbf{B}-{\mathop {\bf{B}}\limits^ \sim }) & \quad \textbf{I} \end{pmatrix} {\mathop {\bf{P}}\limits^ \sim }^{-1} \\[4pt] =& \begin{pmatrix} {\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{11} & \quad {\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{12} \\[4pt] (\textbf{I}-\textbf{C})^{-1} (\textbf{B}-{\mathop {\bf{B}}\limits^ \sim }){\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{11} + {\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{21} & \quad (\textbf{I}-\textbf{C})^{-1} (\textbf{B}-{\mathop {\bf{B}}\limits^ \sim }){\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{12} + {\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{22} \end{pmatrix}.\end{align*}

Computing the matrix product ${\mathop {\bf{P}}\limits^ \sim } = \textbf{P} \textbf{L}$ blockwise and using the above, we find

\begin{align*} &{\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{11} = \textbf{P}^{({-}1)}_{11}, \qquad {\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{12} = \textbf{P}^{({-}1)}_{12}, \qquad {\mathop {\bf{P}}\limits^ \sim }_{12} = \textbf{P}_{\textbf{12}}, \qquad {\mathop {\bf{P}}\limits^ \sim }_{22} = \textbf{P}_{\textbf{22}}, \\ &{\mathop {\bf{P}}\limits^ \sim }_{11} = \textbf{P}_{\textbf{11}} + \textbf{P}_{\textbf{12}} (\textbf{I}-\textbf{C})^{-1} (\textbf{B}-{\mathop {\bf{B}}\limits^ \sim }), \\ &{\mathop {\bf{P}}\limits^ \sim }_{21} = \textbf{P}_{\textbf{21}} + \textbf{P}_{\textbf{22}}(\textbf{I}-\textbf{C})^{-1} (\textbf{B}-{\mathop {\bf{B}}\limits^ \sim }).\end{align*}

Thus

\begin{align*} & {\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{11} = \textbf{P}^{({-}1)}_{11}, \qquad {\mathop {\bf{P}}\limits^ \sim }^{({-}1)}_{12} = \textbf{P}^{({-}1)}_{12}, \\ & {\mathop {\bf{P}}\limits^ \sim }_{11} + {\mathop {\bf{P}}\limits^ \sim }_{12}(\textbf{I}-\textbf{C})^{-1}{\mathop {\bf{B}}\limits^ \sim } = \textbf{P}_{\textbf{11}} + \textbf{P}_{\textbf{12}} (\textbf{I}-\textbf{C})^{-1} \textbf{B}, \\ & \mathop {\bf{P}}\limits^ \sim _{21} + {\mathop {\bf{P}}\limits^ \sim }_{22}(\textbf{I}-\textbf{C})^{-1} {\mathop {\bf{B}}\limits^ \sim } = \textbf{P}_{\textbf{21}} + \textbf{P}_{\textbf{22}}(\textbf{I}-\textbf{C})^{-1} \textbf{B}.\end{align*}

Therefore, regardless of the chosen basis, Equation (11) is the same, which concludes the proof.

A.2 Fractional operators

This section is a brief reminder about fractional operators, which are used in the proofs. We also introduce the matrix-extended Mittag-Leffler function.

Definition 1. (Fractional differentiation and integration operators.) For $\alpha \in (0,1)$ , the fractional differentiation operator, denoted by $D^{\alpha}$ is defined as

\begin{align*} D^{\alpha}f(t) \,{:}\,{\raise-1.5pt{=}}\, \dfrac{1}{\Gamma(1 - \alpha)}\dfrac{d}{dt} \int_0^t (t-s)^{-\alpha} f(s) ds,\end{align*}

where f is a measurable, Hölder continuous function of order strictly greater than $\alpha$ . The fractional integration operator, denoted by $I^{\alpha}$ , is defined as

\begin{equation*}I^{\alpha}f(t) \,{:}\,{\raise-1.5pt{=}}\, \dfrac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1} f(s) ds,\end{equation*}

where f is a measurable function.

It will be convenient for us to define fractional integration with respect to a Brownian motion.

Definition 2. (Fractional integration operator with respect to a Brownian motion.) Given a Brownian motion B and $\alpha \in (1/2,1)$ , the fractional integration operator with respect to B, denoted by $I^{\alpha}_{B}$ , is defined as

\begin{equation*}I^{\alpha}_{B}f(t) = \dfrac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{1-\alpha} f(s) dB_s,\end{equation*}

for f a measurable, square-integrable stochastic process.

Remark 2. The fractional integration of a matrix-valued stochastic process $\textbf{f}$ with respect to a multivariate Brownian motion $\textbf{B}$ is

\begin{equation*}I^{\alpha}_{\textbf{B}}\textbf{f}(t) = \dfrac{1}{\Gamma(\alpha)}\int_0^t (t-s)^{1-\alpha} \textbf{f}(s) d\textbf{B}_s.\end{equation*}

We now extend the Mittag-Leffler function to matrices (for a theory of matrix-valued functions, see for example [Reference Higham16]). We have the following definition.

Definition 3. (Matrix-extended Mittag-Leffler function.) Let $\alpha, \beta \in \mathbb{C}$ such that ${Re}(\alpha), {Re}(\beta) > 0$ , and let $\boldsymbol{\Lambda} \in \mathcal{M}_{n}(\mathbb{R})$ . Then the matrix Mittag-Leffler function is defined as

\begin{align*} \textbf{E}_{\alpha, \beta}(\boldsymbol{\Lambda}) \,{:}\,{\raise-1.5pt{=}}\, \sum_{n \geq 0} \dfrac{\boldsymbol{\Lambda}^n}{\Gamma(\alpha n + \beta)}.\end{align*}

We also extend the Mittag-Leffler density function for matrices.

Definition 4. (Mittag-Leffler density for matrices.) Let $\alpha \in \mathbb{C}$ such that ${Re}(\alpha) > 0$ , $\boldsymbol{\Lambda} \in \mathcal{M}_{n}(\mathbb{R})$ . Then the matrix Mittag-Leffler density function $\textbf{f}^{\alpha, \boldsymbol{\Lambda}}$ is defined as

\begin{equation*} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t) \,{:}\,{\raise-1.5pt{=}}\, \boldsymbol{\Lambda} t^{\alpha - 1} \textbf{E}_{\alpha, \alpha}({-}\boldsymbol{\Lambda} t^{\alpha}).\end{equation*}

We write $\textbf{F}^{\alpha, \Lambda}$ for the cumulative matrix Mittag-Leffler density function,

\begin{equation*} \textbf{F}^{\alpha, \Lambda}(t) \,{:}\,{\raise-1.5pt{=}}\, \int_0^t \textbf{f}^{\alpha, \Lambda}(s)ds.\end{equation*}

Using Definition 3, it is easy to prove the following lemma.

Lemma 4. Let $\alpha\in \mathbb{C}$ such that ${Re}(\alpha)> 0$ , and let $\boldsymbol{\Lambda} \in \mathcal{M}_{n}(\mathbb{R})$ . Then

\begin{equation*} I^{1-\alpha} \textbf{f}^{\alpha, \boldsymbol{\Lambda}} = \boldsymbol{\Lambda} (\textbf{I} - \textbf{F}^{\alpha, \boldsymbol{\Lambda}}).\end{equation*}

Furthermore, if $\alpha \in (1/2,1)$ , then

\begin{equation*}\widehat{\textbf{f}^{\alpha, \boldsymbol{\Lambda}}}(z) = \boldsymbol{\Lambda} (\textbf{I} z^{\alpha} + \boldsymbol{\Lambda})^{-1}.\end{equation*}

We need another important property relating Mittag-Leffler functions to fractional integration operators.

Lemma 5. Let $\alpha > 0$ and $\boldsymbol{\Lambda} \in \mathcal{M}_{m}(\mathbb{R})$ . Then

\begin{align*} I^1 \textbf{f}^{\alpha, \boldsymbol{\Lambda}} &= \sum_{n \geq 1} ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} I^{n \alpha}(1).\end{align*}

Proof. Using Lemma 4 and repeated applications of $I^{\alpha}$ , for all $N \geq 1$ we have

\begin{align*} I \textbf{f}^{\alpha, \boldsymbol{\Lambda}} &= \sum_{1 \leq n \leq N} ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} I^{n \alpha}(1) + ({-}1)^{N-1}\boldsymbol{\Lambda}^{N} I^{N \alpha} I \textbf{f}^{\alpha, \boldsymbol{\Lambda}}.\end{align*}

Therefore, if we show that

\begin{equation*}({-}1)^{N-1}\boldsymbol{\Lambda}^{N} I^{N \alpha} I \textbf{f}^{\alpha, \boldsymbol{\Lambda}} \underset{N \to \infty}{\to} 0,\end{equation*}

the result will follow. To prove this we make use of the series expansion of $I^{N \alpha} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}$ to deduce bounds which will converge to zero. Writing C for a constant independent of t and N which may change from line to line, $N_{\alpha} = \lfloor \dfrac{1}{\alpha} \rfloor$ , and ${\left\lVert{\cdot}\right\rVert_{{op}}}$ for the operator norm, we have

\begin{align*} {\left\lVert{\boldsymbol{\Lambda}^{N} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t)}\right\rVert_{{op}}} &= {\left\lVert{\boldsymbol{\Lambda}^{N+1} \sum_{n \geq 0} ({-}1)^{n} \dfrac{t^{(n+1)\alpha - 1}}{\Gamma((n+1)\alpha)}}\right\rVert_{{op}}} \\ & \leq {\left\lVert{\boldsymbol{\Lambda}^{N+1} \sum_{0 \leq n \leq N_{\alpha}} ({-}1)^n \dfrac{t^{(n+1)\alpha - 1}}{\Gamma((n+1)\alpha)} + \boldsymbol{\Lambda}^{N+1} C}\right\rVert_{{op}}}.\end{align*}

Therefore, applying the fractional integration operator of order $N\alpha$ , and writing $g_{n}: t \mapsto t^{(n+1)\alpha-1}$ , we have

\begin{align*} I^{N \alpha} {\left\lVert{\boldsymbol{\Lambda}^{N} \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t)}\right\rVert_{{op}}} &\leq {\left\lVert{\boldsymbol{\Lambda}^{N+1}I^{N \alpha} \bigg(\sum_{0 \leq n \leq N_{\alpha}} ({-}1)^n \dfrac{g_n}{\Gamma((n+1)\alpha)}\bigg) + \boldsymbol{\Lambda}^{N+1}I^{N \alpha}(C)}\right\rVert_{{op}}} \\ & \leq \sum_{0 \leq n \leq N_{\alpha}} \dfrac{1}{\Gamma((n+1)\alpha)} {\left\lVert{\boldsymbol{\Lambda}^{N+1}I^{N \alpha} (g_{n})}\right\rVert_{{op}}} + {\left\lVert{\boldsymbol{\Lambda}^{N+1}I^{N \alpha}(C)}\right\rVert_{{op}}}.\end{align*}

An explicit computation of $I^{N\alpha}(g_n)$ shows the convergence to zero of the right-hand side as N tends to infinity, which concludes the proof.

Finally, we need to combine fractional integration $I^{\alpha}$ with $I^{\alpha}_{B}$ . We have the following lemma.

Lemma 6. Let $m \geq 1$ , $\textbf{B}$ an m-dimensional Brownian motion, $\textbf{X}$ an $m \times m$ matrix-valued adapted square-integrable stochastic process, and $\alpha, \beta > 0$ . Then we have

\begin{equation*}I^{\alpha} I^{\beta}_{\textbf{B}}(\textbf{X}) = I^{\alpha + \beta}_{\textbf{B}}(\textbf{X}).\end{equation*}

Proof. The proof is a straightforward application of the definitions of the operators together with the stochastic Fubini theorem.

The next lemma is useful for transforming stochastic convolutions of stochastic processes with the Mittag-Leffler density function into series of repeated applications of $I^{\alpha}_{\textbf{B}}$ .

Lemma 7. Let $m \geq 1$ , $\textbf{B}$ an m-dimensional Brownian motion, $\textbf{X}$ an $m \times m$ matrix-valued adapted and square-integrable stochastic process, $\alpha > 0$ , and $\boldsymbol{\Lambda} \in \mathcal{M}_{m}(\mathbb{R})$ . Then, for all $t \geq 0$ and almost surely,

\begin{equation*} \int_0^t \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{X}_s d\textbf{B}_s = \sum_{n \geq 1} ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} I^{n \alpha}_{\textbf{B}}(\textbf{X}),\end{equation*}

where the series converges almost surely.

Proof. Using Lemma 5, we can write the integral using a series of fractional integration operators and apply the stochastic Fubini theorem (as $\textbf{X}$ is square-integrable) to obtain

\begin{align*} \int_0^t \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{X}_s d\textbf{B}_s &= \int_0^t \sum_{n \geq 1} ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} I^{n\alpha-1}(1)_{t-s} \textbf{X}_s d\textbf{B}_s \\ &= \sum_{n \geq 1} \int_0^t ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} I^{n\alpha-1}(1)_{t-s} \textbf{X}_s d\textbf{B}_s \\ &= \sum_{n \geq 1} ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} \int_0^t I^{n\alpha-1}(1)_{t-s} \textbf{X}_s d\textbf{B}_s \\ &= \sum_{n \geq 1} \dfrac{({-}1)^{n-1}}{\Gamma(n\alpha-1)} \boldsymbol{\Lambda}^{n} \int_0^t \int_0^{t-s} (t-s-\tau)^{n\alpha - 2} d \tau \textbf{X}_s d\textbf{B}_s.\end{align*}

After a change of variables and using the stochastic Fubini theorem (see for example [Reference Veraar27]), we deduce the simpler expression

\begin{align*} \int_0^t \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{X}_s d\textbf{B}_s &= \sum_{n \geq 1} \dfrac{({-}1)^{n-1}}{\Gamma(n\alpha-1)} \boldsymbol{\Lambda}^{n} \int_0^t (t-\tau)^{n\alpha - 2} \int_0^{\tau} \textbf{X}_s d\textbf{B}_s d\tau.\end{align*}

Integrating by parts, we finally obtain the result:

\begin{align*} \int_0^t \textbf{f}^{\alpha, \boldsymbol{\Lambda}}(t-s) \textbf{X}_s d\textbf{B}_s &= \sum_{n \geq 1} \dfrac{({-}1)^{n-1}}{\Gamma(n\alpha-1) (n\alpha-1)} \boldsymbol{\Lambda}^{n} \int_0^t (t-\tau)^{n\alpha-1} \textbf{X}_{\tau} d\textbf{B}_{\tau}, \\ &= \sum_{n \geq 1} \dfrac{({-}1)^{n-1}}{\Gamma(n\alpha)} \boldsymbol{\Lambda}^{n} \int_0^t (t-\tau)^{n\alpha-1} \textbf{X}_{\tau} d\textbf{B}_{\tau}, \\ &= \sum_{n \geq 1} ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} I^{n\alpha}_{\textbf{B}}(\textbf{X}).\end{align*}

The last lemma gives convergence for terms of a series of repeated iterations of $I^{\alpha}$ .

Lemma 8. Let $\alpha > 0$ , $\boldsymbol{\Lambda} \in \mathcal{M}_{m}(\mathbb{R})$ , $\textbf{B}$ an m-dimensional Brownian motion, and $\textbf{X}$ an m-dimensional vector-valued square-integrable stochastic process. Then, almost surely and for all $t \in [0,1]$ ,

\begin{align*} & ({-}1)^{N-1} \boldsymbol{\Lambda}^{N} I^{N\alpha}(\textbf{X})_t \underset{N \to \infty}{\to} 0, \\ &\sum_{n \geq N} ({-}1)^{n-1} \boldsymbol{\Lambda}^{n} I^{n\alpha}_{\textbf{B}}({diag}(\textbf{X}))_t \underset{N \to \infty}{\to} 0.\end{align*}

Proof. Let $N^* > N_{\alpha} \,{:}\,{\raise-1.5pt{=}}\, \lfloor \dfrac{1}{\alpha} \rfloor$ . Since $\textbf{X}$ is square-integrable, we have

\begin{align*} \mathbb{E}\Bigg[ \Big \lVert & \sum_{N > N_{*}} \boldsymbol{\Lambda}^{N} I^{(N+1)\alpha}_{\textbf{B}}({diag}(\textbf{X}))_t \Big \rVert^2 \Bigg] \\ \leq & \sum_{N_1, N_2 > N_{*}} \mathbb{E}\bigg[ {{}^ \top {{\bigg(\boldsymbol{\Lambda}^{N_1} I^{(N_1+1)\alpha}_{\textbf{B}}({diag}(\textbf{X}))_t}\bigg) (\boldsymbol{\Lambda}^{N_2} I^{(N_2+1)\alpha}_{\textbf{B}}({diag}(\textbf{X}))_t)}}\bigg].\end{align*}

Using the Cauchy–Schwartz inequality and writing ${\left\lVert{\cdot}\right\rVert_{{op}}}$ for the operator norm associated to the Euclidean norm, we find

\begin{align*} \mathbb{E}\Bigg[ \Bigg \lVert & \sum_{N > N_{*}} \boldsymbol{\Lambda}^{N} I^{(N+1)\alpha}_{\textbf{B}}({diag}(\textbf{X}))_t \Bigg \rVert^2 \Bigg] \\ \leq & \sum_{N_1, N_2 > N_{*}} \left\lVert{\boldsymbol{\Lambda}}\right\rVert_{{op}}^{N_1+N_2} \sum_{1 \leq k,l \leq m} \mathbb{E}\Big[ I^{(N_1+1)\alpha}_{B^k}(X^{k})_t I^{(N_2+1)\alpha}_{B^l}(X^{l})_t\Big] \\ \leq & \sum_{N_1, N_2 > N_{*}} \dfrac{\left\lVert{\boldsymbol{\Lambda}}\right\rVert_{{op}}^{N_1+N_2} }{\Gamma((N_1+1)\alpha)\Gamma((N_2+1)\alpha)} \sum_{1 \leq i \leq m} \int_0^t (t-s)^{(N_1+N_2)\alpha - 2} \mathbb{E}\Big[(X^i_s)^2\Big] ds \\ \leq & c \sum_{N_1, N_2 > N_{*}} \dfrac{\left\lVert{\boldsymbol{\Lambda}}\right\rVert_{{op}}^{N_1+N_2}}{\Gamma((N_1+1)\alpha)\Gamma((N_2+1)\alpha)} \\ \leq & c \Bigg( \sum_{N > N_{*}} \dfrac{\left\lVert{\boldsymbol{\Lambda}}\right\rVert_{{op}}^{N}}{\Gamma((N+1)\alpha)} \Bigg)^2.\end{align*}

Thus, by comparison of functions (for example by application of Stirling’s formula), for all $\epsilon > 0$ ,

\begin{align*}\sum_{N > N_{\alpha}} & \mathbb{P} \left( \Big \lVert \sum_{N > N_{*}} \boldsymbol{\Lambda}^{N} I^{(N+1)\alpha}_{\textbf{B}}({diag}(\textbf{X}))_t \Big \rVert > \epsilon \right)\\[5pt] \leq & \dfrac{1}{\epsilon^2} \sum_{N_* \geq N_{\alpha}} \mathbb{E}\Bigg[ \Big \lVert \sum_{N > N_{*}} \boldsymbol{\Lambda}^{N} I^{(N+1)\alpha}_{\textbf{B}}({diag}(\textbf{X}))_t \Big \rVert^2 \Bigg] < \infty.\end{align*}

The Borel–Cantelli lemma yields the almost sure convergence to zero of $\boldsymbol{\Lambda}^{N} I^{(N+1)\alpha}_{\textbf{B}}({diag}(\textbf{X}))$ as $N \to \infty$ . The same approach yields the almost sure convergence to zero of $({-}1)^{N-1} \boldsymbol{\Lambda}^{N} I^{N\alpha}(\textbf{X})$ as $N \to \infty$ .

A.3 Proof of Corollary 1

We split the proof into two steps. First, we show that the structure of the kernel satisfies the assumptions of Section 2. Then, we compute the equations satisfied by variance and prices.

Checking for the assumptions of Theorem 1. We write

\begin{equation*}\textbf{O}_1 \,{:}\,{\raise-1.5pt{=}}\, \begin{pmatrix}1 \\[3pt] 1 \\[3pt] 0 \\[3pt] 0\end{pmatrix}, \qquad\textbf{O}_2 \,{:}\,{\raise-1.5pt{=}}\, \begin{pmatrix}0 \\[3pt] 0 \\[3pt] 1 \\[3pt] 1\end{pmatrix}, \qquad\textbf{O}_3 \,{:}\,{\raise-1.5pt{=}}\, \begin{pmatrix}1 \\[3pt] -1 \\[3pt] 0 \\[3pt] 0\end{pmatrix}, \qquad\textbf{O}_4 \,{:}\,{\raise-1.5pt{=}}\, \begin{pmatrix}0 \\[3pt] 0 \\[3pt] 1 \\[3pt] -1\end{pmatrix}\!.\end{equation*}

Then, setting $\textbf{O} \,{:}\,{\raise-1.5pt{=}}\, \begin{pmatrix}\textbf{O}_1 \mid \textbf{O}_2 \mid \textbf{O}_3 \mid \textbf{O}_4\end{pmatrix}$ , we have

\begin{align*} \boldsymbol{\phi}^T &= \textbf{O} \begin{pmatrix} \phi_1^T + \phi_2^T & \quad \phi^{T,c}_{12} + \phi^{T,a}_{12} & \quad 0 & \quad 0 \\[8pt] \phi^b_{21} + \phi^s_{21} & \quad \tilde{\phi}^T_1 + \tilde{\phi}^T_2 & \quad 0 & \quad 0 \\[8pt] 0 & \quad 0 & \quad \phi_1^T - \phi_2^T & \quad \phi^{T,c}_{12} - \phi^{T,a}_{12} \\[8pt] 0 & \quad 0 & \quad \phi^b_{21} - \phi^s_{21} & \quad \tilde{\phi}^T_1 - \tilde{\phi}^T_2 \end{pmatrix} \textbf{O}^{-1}.\end{align*}

It is straightforward to check that the assumptions are satisfied if

\begin{align*} 0 &\leq (H^c_{12}+H^a_{12})(H^c_{21}+H^a_{21}) < 1, \\ 0 &\leq {\mid{{1 - (\gamma_1 + \gamma_2) - \sqrt{(H^c_{12}-H^a_{12})(H^c_{21}-H^a_{21}) + (\gamma_1-\gamma_2)^2}}}\mid} < 1, \\ 0 &\leq {\mid{{1 - (\gamma_1 + \gamma_2) + \sqrt{(H^c_{12}-H^a_{12})(H^c_{21}-H^a_{21}) + (\gamma_1-\gamma_2)^2}}}\mid} < 1.\end{align*}

Under those conditions, $\textbf{K}=\textbf{I}-\textbf{H}$ has positive eigenvalues, and therefore $\textbf{KM}^{-1} = \dfrac{1}{\alpha}\textbf{K}$ has positive eigenvalues. Therefore all the assumptions of Theorem 1 are satisfied.

Limiting variance process. Since we can apply Theorem 1, we now compute the relevant quantities. As the blockwise matrix $\textbf{B}$ is equal to zero, writing $H^{12} \,{:}\,{\raise-1.5pt{=}}\, H^{a}_{12} + H^{c}_{12}$ and $H^{21} \,{:}\,{\raise-1.5pt{=}}\, H^{a}_{21} + H^{c}_{21}$ , we have

\begin{align*} &\textbf{O}^{-1} = \dfrac{1}{2}\begin{pmatrix} 1 & \quad 1 & \quad 0 & \quad 0 \\[4pt] 0 & \quad 0 & \quad 1 & \quad 1 \\[4pt] 1 & \quad -1 & \quad 0 & \quad 0 \\[4pt] 0 & \quad 0 & \quad 1 & \quad -1 \end{pmatrix}, \qquad \textbf{K}^{-1} = \dfrac{1}{1 - H_{12}H_{21}} \begin{pmatrix} 1 & \quad H_{12} \\[4pt] H_{21} & \quad 1 \end{pmatrix}, \\[4pt] &\boldsymbol{\Theta^1} = \dfrac{1}{1 - H_{12}H_{21}} \begin{pmatrix} 1 & \quad H_{12} \\[4pt] 1 & \quad H_{12} \end{pmatrix}, \qquad \boldsymbol{\Theta^2} = \dfrac{1}{1 - H_{12}H_{21}} \begin{pmatrix} H_{21} & \quad 1 \\[4pt] H_{21} & \quad 1 \end{pmatrix}.\end{align*}

One can check that the equations satisfied by $\boldsymbol{\Theta^1} \boldsymbol{\tilde{V}}$ and $\boldsymbol{\Theta^2} \boldsymbol{\tilde{V}}$ are the following, where $\textbf{B}$ is a Brownian motion:

\begin{align*} \boldsymbol{\Theta^1} \boldsymbol{\tilde{V}}_t &= \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} \left[ \begin{pmatrix} \mu_1 \\[2pt] \mu_1 \end{pmatrix} - \begin{pmatrix} \tilde{V}^1_s \\[5pt] \tilde{V}^1_s \end{pmatrix} \right]ds \\[2pt] &+ \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} \sqrt{\tilde{V}^1_s + H_{12}\tilde{V}^2_s} \begin{pmatrix} dB^{1}_s+dB^{2}_s \\[7pt] dB^{1}_s+dB^{2}_s \end{pmatrix}, \\[2pt] \boldsymbol{\Theta^2} \boldsymbol{\tilde{V}}_t &= \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} \left[ \begin{pmatrix} \mu_2 \\[2pt] \mu_2 \end{pmatrix} - \begin{pmatrix} \tilde{V}^2_s \\[5pt] \tilde{V}^2_s \end{pmatrix} \right]ds \\[2pt] &+ \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} \sqrt{\tilde{V}^2_s + H_{21}\tilde{V}^1_s} \begin{pmatrix} dB^{3}_s+dB^{4}_s \\[6pt] dB^{3}_s+dB^{4}_s \end{pmatrix}.\end{align*}

Note that the above implies that $V^{1+} = V^{1-}$ and $V^{2+} = V^{2-}$ . This property is due to the symmetric structure of the baselines and kernels. Therefore, the joint dynamics can be fully captured by considering the joint dynamics of $(V^{1+}, V^{2+})$ . Thus, writing $V^{1} \,{:}\,{\raise-1.5pt{=}}\, V^{1+}=V^{1-}$ and $V^{2} \,{:}\,{\raise-1.5pt{=}}\, V^{2+} = V^{2-}$ , we have

\begin{align*} \Gamma(\alpha) \dfrac{\Gamma(1 - \alpha)}{\alpha} V^{1}_t &= \int_0^t (t-s)^{\alpha - 1} \big(\mu_1 - \tilde{V}^{1}_s\big)ds + \int_0^t \sqrt{V^{1}_t} \big(dB^{1}_s+dB^{2}_s\big), \\[5pt] \Gamma(\alpha) \dfrac{\Gamma(1 - \alpha)}{\alpha}V^{2}_t &= \int_0^t (t-s)^{\alpha - 1} \big(\mu_2 - \tilde{V}^{2}_s\big)ds + \int_0^t \sqrt{V^{2}_t} \big(dB^{3}_s+dB^{4}_s\big).\end{align*}

We can write the above without $\boldsymbol{\tilde{V}}$ as

\begin{align*} \Gamma(\alpha) \dfrac{\Gamma(1 - \alpha)}{\alpha} \begin{pmatrix} V^{1}_t \\[6pt] V^{2}_t \end{pmatrix} &= \int_0^t (t-s)^{\alpha - 1} \left( \begin{pmatrix} \mu_1 \\[2pt] \mu_2 \end{pmatrix} - \textbf{K}^{-1} \begin{pmatrix} V^{1}_s \\[5pt] V^{2}_s \end{pmatrix} \right)ds \\[7pt] &+ \int_0^t (t-s)^{\alpha - 1} \begin{pmatrix} \sqrt{V^1_s} \big(dB^{1}_s + dB^{2}_s\big) \\[6pt] \sqrt{V^2_s} \big(dB^{3}_s + dB^{4}_s\big) \end{pmatrix}.\end{align*}

Limiting price process. Turning now to the price process, we compute $\boldsymbol{\Delta}$ (see Equation (10)) using the definition. We have

\begin{align*} {{}^ \top {{{\left\lVert{\boldsymbol{\psi}}\right\rVert_{1}}}}}\textbf{O}_3 &= \sum_{k \geq 1} {{}^ \top {{{\left\lVert{\boldsymbol{\phi}}\right\rVert_{1}}}}}^k \textbf{O}_3 \\ &= \textbf{O} \sum_{k \geq 1} \bigg[ \bigg( {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds \bigg)^{k}_{11} \textbf{e}_3 + \bigg( {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds \bigg)^{k}_{12} \textbf{e}_4 \bigg] \\ &= \sum_{k \geq 1} \bigg[ \bigg( {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds \bigg)^{k}_{11} \textbf{O}_3 + \bigg( {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds \bigg)^{k}_{12} \textbf{O}_4 \bigg] \\ &= \bigg[(\textbf{I} - {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds)^{-1} - \textbf{I}\bigg]_{11} \textbf{O}_3 + \bigg[(\textbf{I} - {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds)^{-1} - \textbf{I}\bigg]_{12} \textbf{O}_4,\end{align*}

which, by definition of $\boldsymbol{\Delta}$ , yields

\begin{align*} \Delta_{11} &= \bigg[ \bigg(\textbf{I} - {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds \bigg)^{-1} - \textbf{I} \bigg]_{11} = \dfrac{2 \gamma_2}{4 \gamma_1 \gamma_2 - \big(H_{12}^c - H_{12}^a\big)\big(H_{21}^c - H_{21}^a\big)} - 1, \\[5pt] \Delta_{12} &= \bigg[ \bigg(\textbf{I} - {\int_{0}^{\infty}{{\textbf{C}}}}(s)ds \bigg)^{-1} - \textbf{I} \bigg]_{12} =\dfrac{H_{21}^c - H_{21}^a}{4 \gamma_1 \gamma_2 - \big(H_{12}^c - H_{12}^a\big)\big(H_{21}^c - H_{21}^a\big)}.\end{align*}

Therefore,

\begin{align*} \boldsymbol{\Delta} = \dfrac{1}{4 \gamma_1 \gamma_2 - \big(H_{12}^c - H_{12}^a\big)\big(H_{21}^c - H_{21}^a\big)}\begin{pmatrix} 2 \gamma_2 & \quad H_{21}^c - H_{21}^a \\[8pt] H_{12}^c - H_{12}^a & \quad 2 \gamma_1 \end{pmatrix} - \textbf{I}.\end{align*}

Finally, by application of Theorem 1, any limit point $\textbf{P}$ of the sequence of microscopic price processes satisfies the following equation:

\begin{align*} \textbf{P}_t &= \dfrac{1}{4 \gamma_1 \gamma_2 - \big(H_{12}^c - H_{12}^a\big)\big(H_{21}^c - H_{21}^a\big)}\begin{pmatrix} 2 \gamma_2 & \quad H_{21}^c - H_{21}^a \\[4pt] H_{12}^c - H_{12}^a & \quad 2 \gamma_1 \end{pmatrix}\\[5pt] &\qquad\qquad\qquad \begin{pmatrix} 1 & \quad -1 & \quad 0 & \quad 0 \\[4pt] 0 & \quad 0 & \quad 1 & \quad -1 \end{pmatrix} \int_0^t \begin{pmatrix} \sqrt{V^1_s}dB^{1}_s \\[7pt] \sqrt{V^1_s}dB^{2}_s \\[7pt] \sqrt{V^2_s}dB^{3}_s \\[7pt] \sqrt{V^2_s}dB^{4}_s \end{pmatrix} \\[8pt] &= \dfrac{1}{4 \gamma_1 \gamma_2 - \big(H_{12}^c - H_{12}^a\big)\big(H_{21}^c - H_{21}^a\big)}\begin{pmatrix} 2 \gamma_2 & \quad H_{21}^c - H_{21}^a \\[6pt] H_{12}^c - H_{12}^a & \quad 2 \gamma_1 \end{pmatrix} \int_0^t \begin{pmatrix} \sqrt{V^1_s} \big(dB^{1}_s - dB^{2}_s\big) \\[7pt] \sqrt{V^2_s} \big(dB^{3}_s - dB^{4}_s\big) \end{pmatrix}.\end{align*}

Introducing the independent bi-dimensional Brownian motions

\begin{equation*}\textbf{Z} \,{:}\,{\raise-1.5pt{=}}\, \frac{1}{\sqrt{2}}\begin{pmatrix}B^1 + B^2 \\[4pt] B^3 + B^4\end{pmatrix}, \qquad \textbf{W} \,{:}\,{\raise-1.5pt{=}}\, \frac{1}{\sqrt{2}}\begin{pmatrix}B^1 - B^2 \\[4pt] B^3 - B^4\end{pmatrix},\end{equation*}

this concludes the proof of Corollary 1.

A.4. Proof of Corollary 2

We define the interaction kernel between Asset i and Asset j. For $1 \leq i,j \leq m$ , define

\begin{equation*}\boldsymbol{\phi}_{ij}^T(t) \,{:}\,{\raise-1.5pt{=}}\, \left\{ \begin{array}{l@{\quad}l} \alpha (1-T^{-\alpha}) \mathbb{1}_{t \geq 1} t^{-(\alpha + 1)} \begin{pmatrix} (1-\gamma) & \quad\gamma \\[10pt] \gamma& \quad(1-\gamma) \end{pmatrix} & \mbox{if } i=j,\\[18pt] \alpha T^{-\alpha} \mathbb{1}_{t \geq 1} t^{-(\alpha + 1)} \begin{pmatrix} H^{c} & \quad H^{a} \\[10pt] H^{a} & \quad H^{c} \end{pmatrix} & \mbox{if Asset } i \mbox{ and Asset } j \mbox{ belong to the same sector}, \\[18pt] \alpha T^{-\alpha} \mathbb{1}_{t \geq 1} t^{-(\alpha + 1)} \begin{pmatrix} H^{c} + H_r^{c} & \quad H^{a} + H_r^{a} \\[10pt] H^{a} + H_r^{a} & \quad H^{c} + H_r^{c} \end{pmatrix} & \mbox{otherwise.} \end{array}\right.\end{equation*}

Finally, the complete Hawkes baseline and kernel structure is

\begin{equation*}\boldsymbol{\mu}^T = T^{\alpha - 1}\begin{pmatrix}\mu^{1} \\[4pt] \mu^{1} \\[4pt] \vdots \\[4pt] \mu^{m} \\[4pt] \mu^{m} \\ \end{pmatrix}, \qquad\boldsymbol{\phi}^T = \begin{pmatrix}\boldsymbol{\phi_{11}}^T &\quad \boldsymbol{\phi_{12}}^T &\quad \dots &\quad \boldsymbol{\phi_{1m}}^T \\[4pt] \boldsymbol{\phi_{21}}^T &\quad \boldsymbol{\phi_{22}}^T &\quad \dots &\quad \boldsymbol{\phi_{2m}}^T \\[4pt] \vdots &\quad \dots &\quad \ddots &\quad \vdots \\[4pt] \boldsymbol{\phi_{m1}}^T &\quad \dots &\quad \dots &\quad \boldsymbol{\phi_{mm}}^T \\[4pt] \end{pmatrix}.\end{equation*}

As in the previous example, the proof is split into two steps. First, we show that the kernel satisfies the assumptions required to apply Theorem 1. Then, we compute the equations satisfied by the limiting variance and price processes.

Checking for the assumptions of Theorem 3. We can examine the structure of the kernel as in the two-asset example. Define the following basis:

\begin{align*} \textbf{O}_{i} \,{:}\,{\raise-1.5pt{=}}\, \left\{ \begin{array}{ll} \textbf{e}_{2i}+\textbf{e}_{2i+1} & \mbox{if } 1 \leq i \leq m, \\[5pt] \textbf{e}_{2i}-\textbf{e}_{2i} & \mbox{if } m+1 \leq i \leq 2m. \end{array}\right.\end{align*}

Using the notation of Section 2, straightforward computations allow us to write

\begin{align*} \boldsymbol{\phi}^T &= \textbf{O} \begin{pmatrix} \textbf{A}^T & \quad \textbf{0} \\[5pt] \textbf{B}^T & \quad \textbf{C}^T \end{pmatrix} \textbf{O}^{-1} = \textbf{O} \begin{pmatrix} \textbf{A}^T & \quad \textbf{0} \\[5pt] \textbf{0} & \quad \textbf{C}^T \end{pmatrix} \textbf{O}^{-1},\end{align*}

where we can compute $\textbf{A}^T$ and $\textbf{C}^T$ . Checking the assumptions is done as in the two-asset case, though the conditions have changed here because of the new structure of the kernel. For example, since

\begin{align*} \underset{T \to \infty}{\lim} {\int_{0}^{\infty}{{\boldsymbol{\phi}^T}}}(s)ds \textbf{O}_{m+i} \,=\, & (1-2\gamma) \textbf{O}_{n+i} + (H^c - H^a) \sum_{1 \leq j \neq i \leq m} \textbf{O}_{m+j} \\ &+ \sum_{1 \leq j \neq i \leq m} \sum_{1 \leq r \leq R} \big(H^{c}_r - H^{a}_r\big) \textbf{O}_{m+j},\end{align*}

we have, writing $\textbf{J} \,{:}\,{\raise-1.5pt{=}}\, \textbf{e}_1 {{}^ \top {{\textbf{e}_1}}} + \cdots + \textbf{e}_m {{}^ \top {{\textbf{e}_m}}}$ and for any $1 \leq r \leq R$ , $\textbf{J}_r \,{:}\,{\raise-1.5pt{=}}\, \textbf{e}_{i_r}{{}^ \top {{\textbf{e}_{i_r}}}} + \cdots + \textbf{e}_{i_r + m_r} {{}^ \top {{\textbf{e}_{i_r + m_r}}}}$ ,

\begin{equation*}{\int_{0}^{\infty}{{\textbf{C}}}}(s)ds = (1-2\gamma) \textbf{I} + \big(H^c - H^a\big) \textbf{J} + \sum_{1 \leq r \leq R} \big(H^{c}_{r} - H^{a}_{r}\big) \textbf{J}_{\textbf{r}}.\end{equation*}

Therefore, as the eigenvalues of ${\int_{0}^{\infty}{{\textbf{C}}}}(s)ds$ can be made explicit, if

\begin{align*} {\mid{{\lambda^{-} + \sum_{1 \leq r \leq R} \lambda^{-}_r}}\mid} < 2 \gamma,\end{align*}

then ${\rho{({{\int_{0}^{\infty}{{\textbf{C}^T}}}(s)ds})}}< 1$ and ${\rho{({{\int_{0}^{\infty}{{\textbf{C}}}}(s)ds})}} < 1$ . Similarly, we can easily check that a necessary condition for ${\rho{({{\int_{0}^{\infty}{{\textbf{A}^T}}}})}}< 1$ for T large enough is

\begin{align*} {\mid{{H^c + H^a + \sum_{1 \leq r \leq R} \dfrac{m_{r}-1}{m-1}\big(H^{c}_{r} + H^{a}_{r}\big)}}\mid} < \dfrac{1}{m-1}.\end{align*}

Since we are interested in the limit where the number of assets grows to infinity, we also impose

\begin{align*} {\mid{{\lambda^{-} + \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r}}\mid} &< 2 \gamma, \\[5pt] {\mid{{\lambda^{+} + \sum_{1 \leq r \leq R} \eta_r \lambda^{+}_r}}\mid}& < 1.\end{align*}

Combined, we have verified all the assumptions on the structure of the kernel that are needed to apply Theorem 1. We thus move to assumptions on $\textbf{K}$ and $\boldsymbol{\Lambda} = \textbf{K} \textbf{M}^{-1}$ . As in the two-asset example, we have here $\textbf{M} = \alpha \textbf{I}$ . Since $\textbf{K} = \textbf{I} - (H^{c}+H^{a})\textbf{J} - \sum_{1 \leq r \leq R} (H^{c}_{r}+H^{a}_{r})\textbf{J}_r$ , the eigenvalues of $\textbf{K}$ (and therefore those of $\boldsymbol{\Lambda}$ ) are all strictly positive. Thus we have checked all conditions necessary to apply Theorem 1. We can now state the equations satisfied by the limiting variance and price processes.

Limiting variance process. As in the previous example, we have $V^{i+} = V^{i-}$ . Thus, we write the underlying variance of asset i as $V^i$ and, using a (slight) abuse of notation, define $\textbf{V} \,{:}\,{\raise-1.5pt{=}}\, (V^1, V^2, \cdots, V^{m})$ . Then $\textbf{V}$ satisfies

\begin{align*} \textbf{V}_t &= \dfrac{\alpha}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} ( \boldsymbol{\theta} - \textbf{K}^{-1} \textbf{V}_s )ds \\[4pt] &+ \dfrac{\alpha \sqrt{2}}{\Gamma(\alpha)\Gamma(1 - \alpha)} \int_0^t (t-s)^{\alpha-1} {diag}\Big(\sqrt{\textbf{V}_s}\Big) d\textbf{B}_s,\end{align*}

where $\textbf{B}$ is a Brownian motion. We can rewrite $\textbf{K}^{-1}$ as

\begin{align*} \textbf{K}^{-1} = & \bigg( \textbf{I} - (H^c + H^a) \textbf{J} - \sum_{1 \leq r \leq R} (H^c_r + H^a_r) \textbf{J}_r \bigg)^{-1}\\ &= \bigg( \textbf{I} - (H^c + H^a) (m-1) \textbf{w} {{}^ \top {{\textbf{w}}}} - \sum_{1 \leq r \leq R} (H^c_r + H^a_r) (m_r - 1) \textbf{w}_r {{}^ \top {{\textbf{w}_r}}} - \epsilon \bigg) ^{-1},\end{align*}

with the small term $\boldsymbol{\epsilon}$ given by

\begin{equation*}\boldsymbol{\epsilon} \,{:}\,{\raise-1.5pt{=}}\, (H^c + H^a)(\textbf{J}-(m-1)\textbf{w} {{}^ \top {{\textbf{w}}}}) + \sum_{1 \leq r \leq R} \big(H^c_r + H^a_r\big) \bigg(\textbf{J}_r-(m_r - 1)\textbf{w}_r {{}^ \top {{\textbf{w}_r}}}\bigg).\end{equation*}

It is easy to check that

\begin{equation*}{\rho{({\boldsymbol{\epsilon}})}}\underset{m \to \infty}{=} o\bigg(\dfrac{1}{m}\bigg),\end{equation*}

which concludes our study of the variance process. We now turn to the equation satisfied by the limiting price process.

Limiting price process. Using the same approach as in the two-asset case, computing $\boldsymbol{\Delta}$ boils down to computing $(I-{\int_{0}^{\infty}{{\textbf{C}}}}(s)ds)^{-1}$ . Using the expression for ${\int_{0}^{\infty}{{\textbf{C}}}}(s)ds$ derived previously, we have

\begin{equation*}(\textbf{I} - \textbf{C})^{-1} = \dfrac{1}{2\gamma} \bigg(\textbf{I} - \dfrac{H^c - H^a}{2\gamma} \textbf{J} - \sum_{1 \leq r \leq R} \dfrac{H^c_r - H^a_r}{2\gamma} \textbf{J}_r\bigg)^{-1}.\end{equation*}

Therefore, repeating the same approach we used for $\textbf{K}^{-1}$ yields

\begin{equation*}(\textbf{I} - \textbf{C})^{-1} = \bigg(2 \gamma \textbf{I} - \lambda^{-} \textbf{w} {{}^ \top {{\textbf{w}}}} - \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r \textbf{w}_r {{}^ \top {{\textbf{w}_r}}}-\boldsymbol{\epsilon}\bigg)^{-1},\end{equation*}

with

\begin{equation*}{\rho{({\boldsymbol{\epsilon}})}}= o\Big(\dfrac{1}{m}\Big).\end{equation*}

Thus, we have the following expression for $\boldsymbol{\Delta}$ :

\begin{align*} \boldsymbol{\Delta} & = \bigg(2 \gamma \textbf{I} - \lambda^{-} \textbf{w} {{}^ \top {{\textbf{w}}}} - \sum_{1 \leq r \leq R} \eta_r \lambda^{-}_r \textbf{w}_r {{}^ \top {{\textbf{w}_r}}} - \boldsymbol{\epsilon}\bigg)^{-1} - \textbf{I}.\end{align*}

Plugging this into Theorem 1, we have the equation satisfied by any limit point $\textbf{P}$ of the sequence $\textbf{P}^T$ , which concludes the proof of Corollary 2.