2.1. Procedure of the elementary approach

The main idea of our elementary approach for obtaining the moments of Hawkes processes is as follows. Given the history of Hawkes process ${\mathcal{F}_t} = \sigma (N(s),0 \leq s \lt t)$ and a small positive $\Delta \gt 0$ , we know that $N(t + \Delta ) - N(t)$ is approximately Bernoulli-distributed, so $\lambda (t + \Delta \mid (N(t + \Delta ) - N(t),{\mathcal{F}_t}))$ are two-point distributed random variables approximately. It is then easy to obtain $\mathbb{E}[g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta ) \mid {\mathcal{F}_t}]$ approximately based on these Bernoulli and two-point distributions. Furthermore, some differential or partial differential equations on $\mathbb{E}[g(N(t),\lambda (t),t)]$ can be given by letting $\Delta \to 0$ and using the double expectation theorem; these can be solved analytically or numerically to obtain $\mathbb{E}[g(N(t),\lambda (t),t)]$ . In the following, we present a procedure exemplified by application to a simple Hawkes process to show how our elementary approach works. Consider a simple Hawkes process N(t), $t \geq 0$ with an intensity function given by

\begin{equation*}\lambda (t) = v + \int_0^t {\alpha \,{{\mathrm{e}}^{ - \beta (t - u)}}\,{\mathrm{d}} N(u)} = v + \sum_{{T_r} < t} {\alpha \,{{\mathrm{e}}^{ - \beta (t - {T_r})}}} ,\end{equation*}

where ${T_r}$ are positive event times, $r = 1,2,\ldots, v,\alpha ,\beta \gt 0$ .

Procedure: the start.

Step 1. Set an objective function $f(t) = \mathbb{E}[g(N(t),\lambda (t),t)]$ that is to be calculated: for example, $\mathbb{E}[{N^m}(t){\lambda ^n}(t)]$ . (The required conditions on $g(n,\lambda ,t)$ will be given in Section 2.2.)

Step 2. Find the probabilities of the following related point process events.

For a simple (or orderly) point process, we need conditional probabilities

\begin{align*}\mathbb{P}\{N(t + \Delta ) - N(t) & = 0 \mid {\mathcal{F}_t}\},\\*\mathbb{P}\{N(t + \Delta ) - N(t) & = 1 \mid {\mathcal{F}_t}\},\end{align*}

where the filtration ${\mathcal{F}_t} = \sigma (N(s),0 \leq s \lt t)$ represents the history of the process up to time t.

For the simple Hawkes process, we have

\begin{align*}\mathbb{P}\{N(t + \Delta ) - N(t) = 0 \mid {\mathcal{F}_t}\} & =1 - \lambda (t)\Delta + \mathrm{o}(\Delta ),\\*\mathbb{P}\{N(t + \Delta ) - N(t) = 1 \mid {\mathcal{F}_t}\} & = \lambda (t)\Delta + \mathrm{o}(\Delta ).\end{align*}

Step 3. Calculate the intensity function values $\lambda (t + \Delta )$ given the following two events:

\begin{align*} \{N(t + \Delta ) - N(t) & = 0\},\\* \{N(t + \Delta ) - N(t) & = 1\}.\end{align*}

For the simple Hawkes process, we have the following.

1. (i) When $\{ N(t + \Delta ) - N(t) = 0\} $ occurs,

   \begin{equation*}{\lambda _0}(t + \Delta ) = v + \alpha \sum_{{T_i} < t } {\,{{\mathrm{e}}^{ - \beta (t - {T_i} + \Delta )}}} = (1 - \beta \Delta )\lambda (t) + \beta v\Delta + \mathrm{o}(\Delta ){.}\end{equation*}

2. (ii) When $\{ N(t + \Delta ) - N(t) = 1\} $ occurs,

   \begin{align*}{\lambda _1}(t + \Delta ) &= v + \alpha \sum_{{T_i} \lt t} {\,{{\mathrm{e}}^{ - \beta (t - {T_i} + \Delta )}}} + \alpha \,{{\mathrm{e}}^{ - \beta (t - {T_{N(t) + 1}} + \Delta )}} \\*&= {\lambda _0}(t + \Delta ) + \alpha (1 - \beta {\Delta _1}) + \mathrm{o}({\Delta _1}),\quad 0 \lt {\Delta _1} < \Delta.\end{align*}

Step 4. Calculate and simplify $\mathbb{E}[g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta ) \mid {\mathcal{F}_t}]$ by using the results obtained in Steps 2 and 3.

For the simple Hawkes process, we have

\begin{align*}&\mathbb{E}[{N^m}(t + \Delta ){\lambda ^n}(t + \Delta ) \mid {\mathcal{F}_t}] \\*&\quad = {N^m}(t)\lambda _0^n(t + \Delta )[1 - \lambda (t)\Delta ] + {[N(t) + 1]^m}\lambda _1^n(t + \Delta )\lambda (t)\Delta + \mathrm{o}(\Delta ) \\&\quad = {N^m}(t){\lambda ^n}(t) + {N^m}(t)\sum_{i = 0}^{n - 1} \dbinom n i {({-} {\nabla _1})^{n - i}}{\lambda ^i}(t)- {N^m}(t)\sum_{i = 0}^n \dbinom n i {({-} {\nabla _1})^{n - i}}{\lambda ^{i + 1}}(t)\Delta \\*&\quad \quad\, +\sum_{j = 0}^m \dbinom m j {N^{\kern0.5ptj}}(t)\sum_{i = 0}^n \dbinom n i \alpha^{n-i}{\lambda ^{i + 1}}(t)\Delta + \mathrm{o}(\Delta ),\end{align*}

where ${\nabla _1} = \beta [\lambda (t) - v]\Delta$ , $0 \lt {\Delta _1} \lt \Delta$ .

Step 5. Take expectations on both sides of the formula of $\mathbb{E}[g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta ) \mid {\mathcal{F}_t}]$ which was obtained in Step 4, then let $\Delta \to 0$ to obtain the differential or partial differential equation on $\mathbb{E}[g(N(t),\lambda (t),t)]$ .

For the simple Hawkes process, we have

\begin{align*} \dfrac{{\mathrm{d}}}{{\mathrm{d}} t}{\mathbb{E}[{N^m}(t){\lambda ^n}(t)]} &= \mathop {\lim }_{\Delta \downarrow 0} \dfrac{{\mathbb{E}[{N^m}(t + \Delta ){\lambda ^n}(t + \Delta )] - \mathbb{E}[{N^m}(t){\lambda ^n}(t)]}}{\Delta }\notag\\*&=n\beta v\mathbb{E}[{N^m}(t){\lambda ^{n - 1}}(t)] - n\beta \mathbb{E}[{N^m}(t){\lambda ^n}(t)]\notag\\*&\quad\, + \sum_{j = 0}^{m - 1}\binom m j \mathbb{E}[{N^{\kern0.5ptj}}(t){\lambda ^{n + 1}}(t)]+ \sum_{j = 0}^m {\sum_{i = 0}^{n - 1} \binom m j \binom n i {\alpha ^{n - i}}\mathbb{E}[{N^{\kern0.5ptj}}(t){\lambda ^{i + 1}}(t)]}.\end{align*}

Note that $\sum_{i = 0}^w [\! \bullet \!] = 0$ when $w \lt 0$ .

Step 6. Solve the established differential equations based on the boundary conditions.

For the simple Hawkes process, this differential equation (obtained in Step 5) can be solved, subject to initial conditions $\mathbb{E}[{N^i}(0)] = 0$ , $\mathbb{E}[{\lambda ^i}(0)] = {v^i}$ $(i = 1,2, \ldots )$ .

Procedure: the end.

This procedure provides the main result for our elementary method. The quantities ${\lambda _0}(t + \Delta )$ and ${\lambda _1}(t + \Delta )$ are required, and the double expectation theorem is used, from which we obtain differential equations which provide a solution for $\mathbb{E}[g(N(t),\lambda (t),t)]$ .

Remark 1. In Steps 3 and 4 the symbol $\mathrm{o}(\Delta )$ is used: $X = \mathrm{o}(\Delta )$ means that

\[\mathop {\lim }_{\Delta \downarrow 0} \dfrac{X}{\Delta } = 0\ \ \text{a.e.},\]

where X is a random variable. The $\mathrm{o}(\Delta )$ notation is used both in the deterministic sense and for the random case, as mentioned in Step 3. However, this does not affect the calculations, as explained in Remark 2.

Remark 2. We have used the conclusion $\mathop {\lim }_{\Delta \downarrow 0} \mathbb{E}[X/\Delta ] = 0$ such that $X = \mathrm{o}(\Delta )$ (of course, it is not a general infinitesimal, as mentioned in our paper). This operation means that

\[ \mathop {\lim }_{\Delta \downarrow 0} \mathbb{E}[X/\Delta ] = \mathbb{E}\bigg[\mathop {\lim }_{\Delta \downarrow 0} X/\Delta \bigg] = 0. \]

The above equality generally needs the sequence

\[ {\bigg\{ \dfrac{{\mathrm{o}(\Delta )}}{\Delta }\bigg\}_{\Delta \gt 0}} = {\bigg\{ \dfrac{X}{\Delta }\bigg\}_{\Delta \gt 0}} \]

to be uniformly integrable. In fact we have

\[ \mathop {\sup }_{0 \lt \Delta \lt \delta } \mathbb{E}[|X/\Delta |] < \infty , \]

where $\delta$ is a given very small positive value; it is well known that this condition can guarantee that the sequence ${\{ {X}/{\Delta }\} _{\Delta \gt 0}}$ is uniformly integrable. Thus the interchange of expectation and limit is a reasonable operation in our procedure. In the following, we shall show some facts that appear in our paper. Note that we assume

\[ \mathbb{E}[|g(N(t),\lambda (t),t)|] < \infty . \]

Case 1. In Step 3 above, we have

\[ {\lambda _0}(t + \Delta ) = v + \alpha \sum_{{T_i} < t + \Delta } {\,{{\mathrm{e}}^{ - \beta (t - {T_i} + \Delta )}}} = (1 - \beta \Delta )\lambda (t) + \beta v\Delta + \mathrm{o}(\Delta ), \]

where

\[ \mathrm{o}(\Delta ) = \bigg[\dfrac{{{{(\beta \Delta )}^2}}}{{2!}} - \dfrac{{{{(\beta \Delta )}^3}}}{{3!}} + \dfrac{{{{(\beta \Delta )}^4}}}{{4!}} - \cdots \bigg] [\lambda (t) - v] = [{{\mathrm{e}}^{ - \beta \Delta }} - 1 + \beta \Delta ][\lambda (t) - v]. \]

On the other hand, we know that

\[\mathbb{E}\bigg[\bigg|\dfrac{{\mathrm{o}(\Delta )}}{\Delta }\bigg|\bigg] \le \bigg| \dfrac{{{{\mathrm{e}}^{ - \beta \Delta }} - 1 + \beta \Delta }}{\Delta }\bigg|\mathbb{E}[|\lambda (t) - v|],\]

and

\[\bigg|\dfrac{{{{\mathrm{e}}^{ - \beta \Delta }} - 1 + \beta \Delta }}{\Delta }\bigg| \le \dfrac{1}{\Delta }\sum_{n = 2}^\infty {{{( - 1)}^n}} \dfrac{{{{(\beta \Delta )}^n}}}{{n!}} \le \dfrac{1}{\Delta }\sum_{n = 2}^\infty {\dfrac{{{{(\beta \Delta )}^n}}}{{n!}}} \le \beta \sum_{n = 1}^\infty {\dfrac{{{{(\beta \Delta )}^n}}}{{n!}}} = \beta ({{\mathrm{e}}^{\beta \Delta }} - 1).\]

If we choose $\Delta \beta \lt 1$ , i.e. $\Delta \lt 1/\beta = \delta $ , then we have

\[\bigg|\dfrac{{{{\mathrm{e}}^{ - \beta \Delta }} - 1 + \beta \Delta }}{\Delta }\bigg| \le \beta (e - 1).\]

For any given time t, we know that $\mathbb{E}[|\lambda (t) - v|] \le C$ , that is, it has a bound (where C is not related to $\Delta$ ). Thus we obtain

\[\mathbb{E}\bigg[\bigg|\dfrac{{\mathrm{o}(\Delta )}}{\Delta }\bigg|\bigg] \le \beta (e - 1)C\]

provided that $\Delta \lt 1/\beta = \delta $ . This proves $\mathop {\sup }_{0 \lt \Delta \lt \delta } \mathbb{E}[|X/\Delta |] \lt \infty .$

Case 2. In Step 3 above, we have

\begin{align*}{\lambda _1}(t + \Delta ) & = v + \alpha \sum_{{T_i} \lt t} {\,{{\mathrm{e}}^{ - \beta (t - {T_i} + \Delta )}}} + \alpha \,{{\mathrm{e}}^{ - \beta (t - {T_{N(t) + 1}} + \Delta )}}\\* & = {\lambda _0}(t + \Delta ) + \alpha (1 - \beta {\Delta _1}) + \mathrm{o}({\Delta _1}), \quad 0 \lt {\Delta _1} < \Delta ,\end{align*}

where $\Delta_1$ is a random variable, but it has the lower and upper bounds 0 and $\Delta$ , respectively. Similarly, we can prove that $\mathop {\sup }_{0 \lt \Delta \lt \delta } \mathbb{E}[|X/\Delta |] \lt \infty , \Delta \lt 1/\beta = \delta. $

Case 3. In Step 3 above, in $\mathbb{E}[g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta ) \mid {\mathcal{F}_t}]$ , we used

\[\mathrm{o}(\Delta ) = g(N(t + \Delta ),{\lambda _i}(t + \Delta ),t + \Delta )P\{ N(t + \Delta ) - N(t) \ge 2\} \quad ( i \ge 2){.}\]

It was assumed that

\[P\{ N(t + \Delta ) - N(t) \ge 2\} = {o_D}(\Delta )\]

which is a deterministic case (denoted by D). On the other hand, based on the boundedness of $\mathbb{E}[|g(N(t),\lambda (t),t)|]$ for any time t, we have

\[\mathbb{E}[\mathrm{o}(\Delta )/\Delta ] = [{o_D}(\Delta )/\Delta ]\mathbb{E}[g(N(t + \Delta ),{\lambda _i}(t + \Delta ),t + \Delta )] = {o_D}(\Delta )/\Delta ,\]

which proves that $\mathop {\sup }_{0 \lt \Delta \lt \delta } \mathbb{E}[|X/\Delta |] \lt \infty .$

Remark 3. The double expectation theorem is used in our elementary method, that is,

\[\mathbb{E}[ {g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta )} ] = \mathbb{E}[ {\mathbb{E}[g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta ) \mid \mathcal{F}_t}] ].\]

2.2. Relation of our method to use of infinitesimal generators and Dynkin’s formula

From the procedure provided above, we know that the key issue is to obtain the differential equation. To embody this key point, we summarize and provide the conditions that a general function $g(N(t),\lambda (t),t)$ and its related expectations must satisfy:

1. (i) $({\partial }/{{\partial t}})g(N,\lambda ,t)$ is uniformly continuous in t;

2. (ii) $({\partial }/{{\partial \lambda}})g(N,\lambda ,t)$ and $({\partial }/{{\partial \lambda }})g(N,\lambda ,t) {{{\mathrm{d}} \lambda }}/{{{\mathrm{d}} t}}$ are uniformly continuous in $\lambda$ ;

3. (iii) all expectations exist and are finite.

Given the filtration ${\mathcal{F}_t} = \sigma (N(s),0 \leq s \lt t)$ , and noting that $\lambda (t) = \lambda (t,N(t),Z)$ , where the random variable Z is a mark in the Hawkes process N(t) with distribution function ${G_Z}(z)$ , we have

\begin{align*} \notag& \mathbb{E}[g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta ) \mid \mathcal{F}_t]\\*& \quad = g(N(t),{\lambda _0}(t + \Delta ),t + \Delta )[1 - \lambda (t)\Delta ] \\*& \quad \quad\, + g(N(t) + 1,{\lambda _1}(t + \Delta ),t + \Delta )[\lambda (t)\Delta ] + \mathrm{o}(\Delta )\\& \quad = g(N(t),{\lambda _0}(t + \Delta ),t + \Delta ) - g(N(t),{\lambda _0}(t + \Delta ),t + \Delta )\lambda (t)\Delta \\*& \quad \quad\, + g(N(t) + 1,{\lambda _1}(t + \Delta ),t + \Delta )\lambda (t)\Delta + \mathrm{o}(\Delta ),\end{align*}

where ${\lambda _0}(t + \Delta ) = \lambda (t + \Delta ,N(t),{Z_1},{Z_2}, \ldots ,{Z_{N(t)}})$ . Then

\begin{align*}{\lambda _1}(t + \Delta ) & = \lambda (t + \Delta ,N(t) + 1,{Z_1},{Z_2}, \ldots ,{Z_{N(t) + 1}}),\\*\mathop {\lim }_{\Delta \downarrow 0} {\lambda _0}(t + \Delta ) & = \lambda (t,N(t),{Z_1},{Z_2}, \ldots ,{Z_{N(t)}}){,}\end{align*}

and

\[\mathop {\lim }_{\Delta \downarrow 0} {\lambda _1}(t + \Delta ) = \lambda (t,N(t) + 1,{Z_1},{Z_2}, \ldots ,{Z_{N(t) + 1}}).\]

Thus

\begin{align*}& \lim _{\Delta \downarrow 0} \dfrac{{\mathbb{E}\{ \mathbb{E}[g(N(t + \Delta ),\lambda (t + \Delta ),t + \Delta ) \mid \mathcal{F}_t] - g(N(t),\lambda (t),t)\} }}{\Delta }\\*& \quad = \lim _{\Delta \downarrow 0} \dfrac{{\mathbb{E}[g(N(t),{\lambda _0}(t + \Delta ),t + \Delta )] - \mathbb{E}[g(N(t),{\lambda _0}(t + \Delta ),t)]}}{\Delta }\\*& \quad \quad\, + \lim_{\Delta \downarrow 0} \dfrac{{\mathbb{E}[g(N(t),{\lambda _0}(t + \Delta ),t)] - \mathbb{E}[g(N(t),\lambda (t),t)]}}{\Delta }\\& \quad \quad\, + \mathop {\lim }_{\Delta \downarrow 0} \{\mathbb{E}[\lambda (t)g(N(t) + 1,{\lambda _1}(t + \Delta ),t + \Delta )]\\*& \quad \quad\, - \mathbb{E}[\lambda (t)g(N(t),{\lambda _0}(t + \Delta ),t + \Delta )]\}.\end{align*}

Hence

(4) \begin{align}\dfrac{{\mathrm{d}}} {{\mathrm{d}} t}\mathbb{E}[g(N(t),\lambda (t),t)]& = \mathbb{E}\bigg[\dfrac{\partial }{{\partial t}}g(N(t),\lambda (t),t)\bigg] + \mathbb{E}\bigg[\dfrac{\partial }{{\partial \lambda }}g(N(t),\lambda (t),t)\dfrac{{\mathrm{d}}} {{\mathrm{d}} t}\lambda (t)\bigg] \notag \\*&\quad\, + \mathbb{E}[\lambda (t)g(N(t) + 1,\lambda (t) + Z,t)] - \mathbb{E}[\lambda (t)g(N(t),\lambda (t),t)]{,}\end{align}

provided that

\[\mathop {\lim }_{\Delta \downarrow 0} \dfrac{{\mathbb{E}[g(N(t),{\lambda _0}(t + \Delta ),t)] - \mathbb{E}[g(N(t),\lambda (t),t)]}}{\Delta } =\mathbb{E}\bigg[\dfrac{\partial }{{\partial \lambda }}g(N(t),\lambda (t),t)\dfrac{{\mathrm{d}} }{{{\mathrm{d}} t}}\lambda (t)\bigg]\]

and

\[|\mathbb{E}[\lambda (t)g(N(t) + 1,\lambda (t) + Z,t)] - \mathbb{E}[\lambda (t)g(N(t),\lambda (t),t)]| < \infty {.}\]

Equation (4) can be used for any function $g(N,\lambda ,t)$ , for which the expectations of

\[\mathbb{E}\bigg[\dfrac{\partial }{{\partial \lambda }}g(N(t),\lambda (t),t)\dfrac{{\mathrm{d}}} {{\mathrm{d}} t} \lambda (t)\bigg]\quad\text{and}\quad\mathbb{E}[\lambda (t)g(N(t) + 1,\lambda (t) + Z,t)]\]

can be calculated, or it can be expressed as functions of $\mathbb{E}[g(N(t),\lambda (t),t)]$ . Equation (4) can be solved analytically or numerically.

For completeness, we recall that the definition of an infinitesimal generator $\mathcal{A}$ of the process $\emph{\textbf{X}}(t)$ is given for any function $g \colon {\Re ^n} \to \Re $ by

\[\mathcal{A}g(\emph{\textbf{x}}) = \mathop {\lim }_{t \downarrow 0} \dfrac{{\mathbb{E}[g(\emph{\textbf{X}}(t))] - g(\emph{\textbf{x}})}}{t},\quad \emph{\textbf{x}} \in {\Re ^n},\quad \emph{\textbf{X}}(0) = \emph{\textbf{x}},\quad \mathop {\lim }_{t \downarrow 0} g(\emph{\textbf{X}}(t)) = g(\emph{\textbf{x}}).\]

Furthermore, we can integrate from time t to $T(t \lt T)$ on both sides of (4), obtaining

\begin{align*}& \mathbb{E}[g(N(T),\lambda (T),T) \mid {\mathcal{F}_t}] - g(N(t),\lambda (t),t)\\*& \quad = \int_t^T {\mathbb{E}\bigg\{\dfrac{\partial }{{\partial u}}g(N(u),\lambda (u),u) \mid \mathcal{F}_t}\bigg\} \,{\mathrm{d}} u+ \int_t^T \mathbb{E}\bigg\{\dfrac{\partial }{\partial\lambda}g(N(u),\lambda(u),u)\dfrac{{\mathrm{d}}} {{\mathrm{d}} u}\lambda (u) \mid \mathcal{F}_t\bigg\}\,{\mathrm{d}} u\\*& \quad \quad\, +\int_t^T {\mathbb{E}[\lambda (u)g(N(u) + 1,\lambda (u) + Z,u) \mid {\mathcal{F}_t}]\,{\mathrm{d}} u - \mathbb{E}[\lambda (t)g(N(t),\lambda (t),t)]}\\& \quad = \mathbb{E}\bigg\{ \int_t^T \bigg[\lambda (u)\int_0^\infty {[g(N(u) + 1,\lambda (u) + z,u)- g(N(u),\lambda (u),u)]} \,{\mathrm{d}} {G_Z}(z)\\*& \quad \quad\, +{\dfrac{{\partial g}}{{\partial u}} + \dfrac{{\partial g}}{{\partial \lambda }}}\dfrac{{{\mathrm{d}} \lambda }}{{{\mathrm{d}} u}} \mid {\mathcal{F}_t}\bigg]\,{\mathrm{d}} u\bigg\}.\end{align*}

That is to say, we have

\begin{equation*} \mathbb{E}[g(N(T),\lambda (T),T) \mid {\mathcal{F}_t}] - g(N(t),\lambda (t),t) = \mathbb{E}\bigg[\int_t^T {\mathcal{A} g(N(u),\lambda (u),u)} \,{\mathrm{d}} u \mid \mathcal{F}_t\bigg],\end{equation*}

which is the Dynkin formula. The infinitesimal generator $\mathcal{A}$ is given for a function $g \colon \emph{\textbf{N}} \times {\Re ^ + } \times {\Re ^ + } \to \Re $ with continuous partial derivatives by

\begin{equation*} \mathcal{A}g(N,\lambda ,t) = \dfrac{{\partial g}}{{\partial t}} + \dfrac{{\partial g}}{{\partial \lambda }}\dfrac{{\mathrm{d}} \lambda }{{\mathrm{d}} t} + \lambda \int_0^\infty {[g(N + 1,\lambda + z,t) - g(N,\lambda ,t)]} \,{\mathrm{d}} {G_Z}(z).\end{equation*}

Based on the above analysis, we see that if the process is Markovian, the results obtained by using the Dynkin formula can also be obtained by using our method. Furthermore, our method is very naïve and simple. However, our method is more general because, as the following example shows, it can also be applied in non-Markovian cases for which the Dynkin formula is not applicable. In fact, another non-Markovian example will be presented in Section 4 as well.

We consider any regular counting process such that the probability of multiple events in an interval of length $\Delta $ is of order $\mathrm{o}(\Delta )$ as $\Delta \to 0$ . Then we have the following theorem.

Theorem 1. For a counting process such that $\mathbb{P}\{ N(t + \Delta ) - N(t) \geq 2\} = \mathrm{o}(\Delta )$ , we have

(5) \begin{equation}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N^m}(t)] = \sum_{j = 0}^{m - 1} \binom m j \mathbb{E}[{N^{\kern0.5ptj}}(t)\lambda (t)]\quad (m \geq 1).\end{equation}

3.1. Applications to a simple Hawkes process

Consider a simple Hawkes process $N(t),t \geq 0$ with an intensity function given in Section 2. In the following steps we demonstrate our method for $\mathbb{E}[{\lambda ^n}(t)]$ , $t \geq 0$ , and derive all general equations (i.e. (6), (8), and (9)). We then consider special cases such as $n = m = 2$ .

Step 1. Set our target as $\mathbb{E}[{\lambda ^n}(t)]$ for any positive integer n $(n \geq 1)$ .

Step 2. Find two probabilities such as

\begin{align*}\mathbb{P}\{N(t + \Delta ) - N(t) = 0 \mid \mathcal{F}_t\} & =1-\lambda (t)\Delta + \mathrm{o}(\Delta ),\\*\mathbb{P}\{N(t + \Delta ) - N(t) = 1 \mid \mathcal{F}_t\} & =\lambda (t)\Delta + \mathrm{o}(\Delta ),\end{align*}

where the filtration ${\mathcal{F}_t} = \sigma (N(s),0 \leq s \lt t)$ represents the history of the simple Hawkes process up to time t, and $\Delta \gt 0$ . We also know that $\mathbb{P}\{N(t + \Delta ) - N(t) \geq 2 \mid \mathcal{F}_t\}=\mathrm{o}(\Delta )$ due to the assumption.

Step 3. Calculate the intensity function $\lambda (t + \Delta )$ given $\mathcal{F}_t$ and two events

\begin{align*}\{ N(t + \Delta ) - N(t) = 0\},\quad\{ N(t + \Delta ) - N(t) = 1\}.\end{align*}

1. (i) When $\{ N(t + \Delta ) - N(t) = 0\} $ occurs,

   \[ {\lambda _0}(t + \Delta ) = v + \sum_{{T_i} < t } {\alpha \,{{\mathrm{e}}^{ - \beta (t - {T_i} + \Delta )}}} = (1 - \beta \Delta )\lambda (t) + \beta v\Delta + \mathrm{o}(\Delta ){.} \]

2. (ii) When $\{ N(t + \Delta ) - N(t) = 1\} $ occurs,

   \begin{align*}{\lambda _1}(t + \Delta ) &= v + \sum_{{T_i} \lt t} {\alpha \,{{\mathrm{e}}^{ - \beta (t - {T_i} + \Delta )}}} + \alpha \,{{\mathrm{e}}^{ - \beta {\Delta _1}}} \\*&= {\lambda _0}(t + \Delta ) + \alpha (1 - \beta {\Delta _1}) + \mathrm{o}({\Delta _1}),\quad 0 \lt {\Delta _1} < \Delta .\end{align*}

Step 4. Calculate $\mathbb{E}[{\lambda ^n}(t + \Delta ) \mid {\mathcal{F}_t}]$ . Based on the values of ${\lambda _0}(t + \Delta )$ and ${\lambda _1}(t + \Delta )$ given in Step 3, we have

\begin{align*}\mathbb{E}[{\lambda ^n}(t + \Delta ) \mid \mathcal{F}_t]&= \lambda _0^n(t + \Delta )[1 - \lambda (t)\Delta ] + \lambda _1^n(t + \Delta )\lambda (t)\Delta + \mathrm{o}(\Delta ) \\*&= \lambda _0^n(t + \Delta ) + \lambda (t)\Delta \sum_{i = 0}^{n - 1} \binom n i \lambda _0^i(t + \Delta ){(\alpha - \beta {\Delta _1})^{n - i}} + \mathrm{o}(\Delta ).\end{align*}

$\mathbb{E}[{\lambda ^n}(t + \Delta ) \mid \mathcal{F}_t]$ is obtained because when $\{ N(t + \Delta ) - N(t) = 0\} $ occurs, $\lambda (t + \Delta ) = {\lambda _0}(t + \Delta )$ . When $\{ N(t + \Delta ) - N(t) = 1\} $ occurs, $\lambda (t + \Delta ) = {\lambda _1}(t + \Delta )$ .

Step 5. Get a differential equation on $\mathbb{E}[{\lambda ^n}(t)]$ . Let $\Delta \to 0$ , and using the result obtained in Step 4 and the equality

\[\mathbb{E}[\mathbb{E}[{\lambda ^n}(t + \Delta ) \mid \mathcal{F}_t]] = \mathbb{E}[{\lambda ^n}(t + \Delta )],\]

we have

(6) \begin{equation}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{\lambda ^n}(t)] = n\beta v\mathbb{E}[{\lambda ^{n - 1}}(t)] - n\beta \mathbb{E}[{\lambda ^n}(t)] + \sum_{i = 0}^{n - 1} \binom n i{\alpha ^{n - i}}\mathbb{E}[{\lambda ^{i + 1}}(t)] \quad (n \geq 1),\end{equation}

Step 6. Solve the established differential equation for getting $\mathbb{E}[{\lambda ^n}(t)]$ . Now we consider the case of $\mathbb{E}[{N^2}(t)]$ . From (5) we have

(7) \begin{equation}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[ {{N^2}(t)} ] = \mathbb{E}[ {\lambda (t)} ] + 2\mathbb{E}[ {N(t)\lambda (t)} ],\end{equation}

so we need two terms, $\mathbb{E}[\lambda (t)]$ and $\mathbb{E}[N(t)\lambda (t)]$ , to obtain $\mathbb{E}[N(t)^2]$ .

Similarly, for $\mathbb{E}[N(t)\lambda (t)]$ , following Steps 1–5, we can obtain differential equations as follows:

(8) \begin{equation}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N(t)\lambda (t)] = \beta v \mathbb{E}[N(t)] + \alpha\mathbb{E}[\lambda (t)] + (\alpha - \beta )\mathbb{E}[N(t)\lambda (t)] + \mathbb{E}[{\lambda ^2}(t)].\end{equation}

We can thus set up and solve a chain of differential equations:

\[\mathbb{E}[\lambda (t)] \to \mathbb{E}[{\lambda ^2}(t)] \to \cdots \to \mathbb{E}[{\lambda ^n}(t)] \quad (n \ge 1).\]

When $n = 1$ , we have

(9) \begin{equation}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda (t)] = (\alpha - \beta )\mathbb{E}[\lambda (t)] + \beta v\end{equation}\vskip-\lastskip\pagebreak

with initial condition $\mathbb{E}[\lambda (0)] = v$ . This differential equation is easily solved as

(10) \begin{equation}\mathbb{E}[\lambda (t)] =\begin{cases}\dfrac{{ - \beta v}}{{\alpha - \beta }} + \dfrac{{\alpha v}}{{\alpha - \beta }}\,{{\mathrm{e}}^{(\alpha - \beta )t}} &\text{when } \alpha \ne \beta ,\\[7pt]v + v\alpha t &\text{when } \alpha = \beta .\end{cases}\end{equation}

Taking $m = 1$ in (5) gives

(11) \begin{align}&\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[ {N(t)} ] = \mathbb{E}[ {\lambda (t)} ] \notag \\*&\Rightarrow\mathbb{E}[N(t)] = \int_0^t \mathbb{E}[ {\lambda (u)} ]\,{\mathrm{d}} u = \begin{cases}\dfrac{{ - \beta vt}}{{\alpha - \beta }} + \dfrac{{\alpha v}}{{{{(\alpha - \beta )}^2}}}[{{\mathrm{e}}^{(\alpha - \beta )t}} - 1] &\text{when } \alpha \ne \beta,\\[9pt]vt + \dfrac{1}{2}v\alpha {t^2} &\text{when } \alpha = \beta.\end{cases}\end{align}

Furthermore, from (6) we have

(12) \begin{equation}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{\lambda ^2}(t)] = (2\beta v + {\alpha ^2})\mathbb{E}[\lambda (t)] + (2\alpha - 2\beta )\mathbb{E}[{\lambda ^2}(t)]\quad (n = 2).\end{equation}

Solving (12), we have, with initial condition $\mathbb{E}[{\lambda ^2}(0)] = {v^2}$ ,

\begin{equation*}\mathbb{E}[{\lambda ^2}(t)] = \begin{cases}\bigg[ {\dfrac{{{\Delta _{11}}}}{{\alpha - \beta }} + {v^2} - \dfrac{{{\Delta _{12}}}}{{\alpha - \beta }}} \bigg]\,{{\mathrm{e}}^{ - 2(\alpha - \beta )t}} &\text{when } \alpha \ne \beta ,\\[9pt]v({\alpha ^2} + 2\beta v)\bigg(\dfrac{1}{2}{t^2} + t\bigg) + {v^2} &\text{when } \alpha = \beta ,\end{cases}\end{equation*}

where

\[ {\Delta _{11}} = ({\alpha ^2} + 2\beta v)v\bigg[ {\dfrac{{\alpha \,{{\mathrm{e}}^{3(\alpha - \beta )t}}}}{{3(\alpha - \beta )}} - \dfrac{{\beta \,{{\mathrm{e}}^{2(\alpha - \beta )t}}}}{{2(\alpha - \beta )}}} \bigg],\quad {\Delta _{12}} = ({\alpha ^2} + 2\beta v)v\bigg[ {\dfrac{\alpha }{{3(\alpha - \beta )}} - \dfrac{\beta }{{2(\alpha - \beta )}}} \bigg]. \]

Similarly, solving (8), we get, with initial condition $\mathbb{E}[N(0)\lambda (0)] = 0,$

(13) \begin{equation} \mathbb{E}[N(t)\lambda (t)] = \begin{cases}\displaystyle {{\mathrm{e}}^{(\alpha - \beta )t}}\int_0^t {[\beta v \mathbb{E}[N(u)] + \mathbb{E}[\lambda (u)] + \mathbb{E}[{\lambda ^2}(u)]]\,{\mathrm{d}} u} &\text{when } \alpha \neq \beta ,\\[7pt]\displaystyle \int_0^t [\beta v \mathbb{E}[N(u)] + \mathbb{E}[\lambda (u)] + \mathbb{E}[{\lambda ^2}(u)]]\,{\mathrm{d}} u &\text{when } \alpha = \beta .\end{cases}\end{equation}

Based on (7), we have, with initial condition $\mathbb{E}[ {{N^2}(0)} ] = 0$ ,

(14) \begin{equation}\mathbb{E}[ {{N^2}(t)} ] = \int_0^t {\mathbb{E}[ {\lambda (u)} ] + 2\mathbb{E}[ {N(u)\lambda (u)} ]}\,{\mathrm{d}} u,\end{equation}

which is a closed-form expression for the second-order moment of a simple Hawkes process. In fact these are not so easy to describe in general, but results for low-order moments are easy to obtain. Results for all first- and second-order moments can be found: from (10) and (11) for first-order moments, and from (8), (12), (13), and (14) for second-order moments; see Example 1 below. All expectations of order m can be found as follows. First calculate all expectations of order less than m. Then find expectation $\mathbb{E}[{\lambda ^m}(t)]$ using the recursion of powers of $\lambda (t)$ from (6). Finally, use the appropriate versions of (15) (to be given in Section 3.2) to obtain successively all expectations of order m:

\[\mathbb{E}[{\lambda ^m}(t)] \to \mathbb{E}[{N}(t){\lambda ^{m - 1}}(t)] \to \cdots \to \mathbb{E}[N^{m - 1}(t)\lambda (t)]\to \mathbb{E}[N^m(t)] \quad (m \ge 1).\]

The whole recursion can be represented as follows.

Order 1: $\mathbb{E}[\lambda (t)] \to \mathbb{E}[N(t)] \to$ order 2.

Order 2: $\mathbb{E}[{\lambda ^2}(t)] \to \mathbb{E}[N(t)\lambda (t)] \to \mathbb{E}[{N^2}(t)] \to$ order 3.

. . . .

Order m − 1: $\cdots \to$ order m.

Order m: $\mathbb{E}[{\lambda ^m}(t)] \to \mathbb{E}[N(t){\lambda ^{m - 1}}(t)] \to \cdots \to \mathbb{E}[{N^{m - 1}}(t)\lambda (t)] \to \mathbb{E}[{N^m}(t)] \quad (m \geq 1)$ .

In the following, a numerical example is given for a simple Hawkes process.

Example 1. For a simple Hawkes process, if $\alpha = 2$ , $\beta = 3$ , $v = 1$ , we have

\[\mathbb{E}[\lambda (t)] = 3 - 2\,{{\mathrm{e}}^{ - t}}{.}\]

From (10) and (11)

\begin{gather*}\mathbb{E}[N(t)] = \int_0^t \mathbb{E}[\lambda (u)]\,{\mathrm{d}} u = - 2 + 3t + 2\,{{\mathrm{e}}^{ - t}};\\ \begin{aligned} & \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{\lambda ^2}(t)] + 2\mathbb{E}[{\lambda ^2}(t)] = 10\mathbb{E}[ {\lambda (t)} ]\quad {\text{from (6) }}\\* & \Rightarrow \mathbb{E}[{\lambda ^2}(t)] = 15 - 20\,{{\mathrm{e}}^{ - t}} + 6\,{{\mathrm{e}}^{ - 2t}};\end{aligned} \\ \begin{aligned} &\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N(t)\lambda (t)] + \mathbb{E}[N(t)\lambda (t)] = 3\mathbb{E}[ {N(t)} ]+2\mathbb{E}[\lambda (t)] + \mathbb{E}[{\lambda ^2}(t)]\quad\text{from (8)}\\*& \Rightarrow \mathbb{E}[N(t)\lambda (t)] = 6 + 9t - 18t\,{{\mathrm{e}}^{ - t}} - 6e{}^{ - 2t};\end{aligned} \\ \begin{aligned} &\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N^2}(t)] = \mathbb{E}[ {\lambda (t)} ] + 2\mathbb{E}[ {N(t)\lambda (t)} ]\quad\text{from (5)}\\* & \Rightarrow \mathbb{E}[{N^2}(t)] = \int_0^t {\mathbb{E}[ {\lambda (u)} ] + 2\mathbb{E}[ {N(u)\lambda (u)} ]\,{\mathrm{d}} u}\\*& \Rightarrow \mathbb{E}[{N^2}(t)] = - 44 + 15t + 9{t^2} + 36t\,{{\mathrm{e}}^{ - t}} + 38\,{{\mathrm{e}}^{ - t}} + 6\,{{\mathrm{e}}^{ - 2t}}.\end{aligned} \end{gather*}

We further obtain, in the usual way,

\begin{align*}\text{Var} [\lambda (t)] & = 6 - 8\,{{\mathrm{e}}^{ - t}} + 2\,{{\mathrm{e}}^{ - 2t}},\\*\text{Var} [N(t)] & = - 48 + 27t + 24t\,{{\mathrm{e}}^{ - t}} + 46\,{{\mathrm{e}}^{ - t}} + 2\,{{\mathrm{e}}^{ - 2t}},\\*{\text{Cov}} [N(t),\lambda (t)] &=\mathbb{E}[N(t)\lambda (t)] - \mathbb{E}[N(t)]\mathbb{E}[\lambda (t)] = 12 - 12t\,{{\mathrm{e}}^{ - t}} - 10\,{{\mathrm{e}}^{ - t}} - 2\,{{\mathrm{e}}^{ - 2t}}.\end{align*}

As shown in Figure 1, we see that correlation coefficient $\rho (t)$ of N(t) and $\lambda (t)$ is a decreasing function of time t and $\mathop {\lim }_{t \downarrow 0} \rho (t) = 1$ , $\mathop {\lim }_{t \uparrow \infty } \rho (t) = 0,$ as we intuitively expect.

Figure 1:

The correlation coefficient $\rho (t)$ of N(t) and $\lambda (t)$ for a simple Hawkes process.

3.2. Applications to a simple self-exciting marked Hawkes process

In this subsection we shall use our method to obtain some results for the simple self-exciting marked Hawkes process defined as follows (see (15)). A simple self-exciting marked Hawkes process ${N_H}(t)$ has intensity

(15) \begin{equation}\lambda (t) = v + \sum_{{T_r} \lt t} {{Z_i}\,{{\mathrm{e}}^{ - \beta (t - {T_r})}}},\quad t \gt 0,\end{equation}

with $\{Z_r\} $ a sequence of i.i.d. random variables with distribution function ${G_Z}(z)$ . An even simpler model is obtained if ${G_Z}(z)$ is a degenerate distribution with all values equal to a constant $\alpha$ , which is a simple Hawkes process we have discussed previously. It is easy to see that the pair $( {{N_H}(t),\lambda (t)} )$ forms a Markov process. In this subsection we simplify the notation by writing $\mathbb{E}[{Z^r}] = {\mu _r}$ , $r \gt 0.$ Note that throughout this paper we always use ${\mu _i}$ as the ith moments for random variables; their distribution functions are known from the context.

3.2.1. A result for $\mathbb{E}[N_{H}^{m}(t){\lambda^{n}}(t)]$

As in Section 3.1, we find that if there is no event occurring in $(t,t + \Delta )$ , with probability ${\rm{1}} - \lambda (t)\Delta + \mathrm{o}(\Delta )$ , then ${N_H}(t + \Delta ) = {N_H}(t)$ and

\[ {\lambda _0}(t + \Delta ) = v + \sum_{{T_i} \lt t + \Delta } {{Z_i}\,{{\mathrm{e}}^{ - \beta (t + \Delta - {T_i})}}} = v + {{\mathrm{e}}^{ - \beta \Delta }}\sum_{{T_i} < t} {{Z_i}\,{{\mathrm{e}}^{ - \beta (t - {T_i})}}} = \lambda (t) - {\nabla _1} + \mathrm{o}(\Delta ). \]

If there is an event occurring in $(t,t + \Delta )$ , with probability $\lambda (t)\Delta + \mathrm{o}(\Delta )$ , then ${N_H}(t + \Delta ) = {N_H}(t) + 1$ and

\begin{equation*}{\lambda _1}(t + \Delta ) = v + \sum_{{T_i} < t + \Delta } {{Z_i}\,{{\mathrm{e}}^{ - \beta (t + \Delta - {T_i})}}}= \lambda (t) - {\nabla _1} + {Z_{{N_H}(t) + 1}}{\nabla _2} + \mathrm{o}(\Delta ),\end{equation*}

where ${\nabla _1}$ was given in Step 4 in Section 2.1, and ${\nabla _2}=1-\beta\Delta_1$ . Thus we can get Theorem 2 given below, which is a differential equation that can be solved recursively.

Theorem 2. For any non-negative integers m and n, suppose the random marks ${Z_i}$ $(i = 1,2, \ldots )$ with distribution function ${G_Z}(z)$ are independent of previous values of Hawkes process ${N_H}(t)$ and its intensity $\lambda (t)$ , and its kernel function is $\gamma (t) = {{\mathrm{e}}^{ - \beta t}}$ . Then, using the notation ${f_{m,n}}(t) = \mathbb{E}[N_H^m(t){\lambda ^n}(t)]$ , we have

\begin{align*} \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}{f_{m,n}}(t) &= n\beta v{f_{m,n - 1}}(t) - n\beta {f_{m,n}}(t) + \sum_{j = 0}^{m - 1} \binom m j {\,f_{j,n + 1}}(t)+ \sum_{j = 0}^m {\sum_{i = 0}^{n - 1} \binom m j \binom n i {\mu _{n - i}}{\,f_{j,i + 1}}(t)},\end{align*}

where $\sum_{i = 0}^w [\! \bullet \!] = 0$ when $w<0$ , and

\[\mu _r = \mathbb{E}[{Z^r}] = \int_0^\infty {u^r}\,{\mathrm{d}} {G_Z}(u) < \infty.\]

We see that ${f_{m,n}}(t)$ satisfies a first-order differential difference equation, which can be solved easily. Furthermore, we know that obtaining ${f_{m,n}}(t)$ needs all values for ${f_{j,i}}(t)$ , $(\,j,i) \in \{ (0,1, \ldots ,m - 1),(0,1, \ldots ,n + 1)\} $ .

In addition, Theorem 2 can also be obtained by using the Dynkin formula approach. We now show that the same general result can be derived from Dynkin’s formula. Because of the Markov property for stochastic process $({N_H}(t),\lambda (t))$ , and using the results in [Reference Davis9] and [Reference Dassios and Zhao8], we consider the function $g(N,\lambda ,t) = N_H^m(t){\lambda ^n}(t)$ . Then we have

\begin{align*}\mathcal{A} g(N,\lambda ,t) &= \beta (v - \lambda )\dfrac{{\partial g}}{{\partial \lambda }} + \lambda \bigg[\int_0^\infty g(N + 1,\lambda + x,t)\,{\mathrm{d}} G(x) - g(N,\lambda ,t)\bigg] \\*&= n\beta (v - \lambda ){N^m}{\lambda ^{n - 1}} + {(N + 1)^m}\lambda \sum_{i = 0}^n \binom n i {\lambda ^i}{\mu _{n - i}} - {N^m}{\lambda ^{n + 1}}\end{align*}

provided that

\[\bigg|{(N + 1)^m}\sum_{i = 0}^n \binom n i {\lambda ^i}{\mu _{n - i}} - {N^m}{\lambda ^n}\bigg| < \infty,\]

where the operator $\mathcal{A}$ is the infinitesimal generator of the Hawkes process with intensity function

\[\lambda (t)= v + \sum_{{T_i} < t} {Z_i}\,{{\mathrm{e}}^{ - \beta (t - {T_i})}} , \quad{\mu _i} = \int_0^\infty {z^i}\,{\mathrm{d}} {G_Z}(z),\quad i = 0,1, \ldots.\]

Based on the Dynkin formula, we have, for $t \lt T$ ,

\[\mathbb{E}[N_H^m(T){\lambda ^n}(T) \mid \mathcal{F}_t] = N_H^m(t){\lambda ^n}(t) + \mathbb{E}\bigg[\int_t^T {\mathcal{A} g({N_H}(u),\lambda (u),u)\,{\mathrm{d}} u} \mid \mathcal{F}_t\bigg].\]

Taking expectations of both sides of the above equation and setting $t = 0$ , we have

\[\mathbb{E}[N_H^m(T){\lambda ^n}(T)] = \int_0^T {\mathbb{E}[\mathcal{A} g({N_H}(u),\lambda (u),u)\,{\mathrm{d}} u} ].\]

Differentiating the above equation, we get

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}{f_{m,n}}(t) & = \mathbb{E}[\mathcal{A}g({N_H}(t),\lambda (t),t)]\\*& = n\beta v{f_{m,n - 1}}(t) - n\beta {f_{m,n}}(t) + \sum_{j = 0}^{m - 1} \dbinom m j {\,f_{j,n + 1}}(t) + \sum_{j = 0}^m {\sum_{i = 0}^{n - 1} \dbinom m j\dbinom n i{\mu _{n - i}}{\,f_{j,i + 1}}(t)},\end{align*}

which is consistent with our result presented in Theorem 2. This again shows that our method gets the same result as the Dynkin formula in the cases where the latter method is valid.

3.2.2. Auto-covariance and cross-covariance

Here we consider the calculations of auto-covariance and cross-covariance functions for the simple self-exciting marked Hawkes process ${N_H}(t)$ and its intensity $\lambda (t)$ , as follows.

Since auto-covariances are symmetric, we need only consider $u \lt t$ when finding $\mathbb{E}[\lambda (t)\lambda (u)]$ and $\mathbb{E}[{N_H}(t){N_H}(u)]$ . Then

\begin{equation*}\mathbb{E}[\lambda (t + \Delta )\lambda (u) \mid \mathcal{F}_t]=\lambda (t)\lambda (u) - {\nabla _1}\lambda (u) + \mathbb{E}[{Z_{{N_H}(t) + 1}}{\nabla _2} - {\nabla _1}]\lambda (u)\lambda (t)\Delta + \mathrm{o}(\Delta ),\end{equation*}

and so

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda (t)\lambda (u)] = \beta v\mathbb{E}[\lambda (u)] + ({\mu _1} - \beta )\mathbb{E}[\lambda (t)\lambda (u)],\]

with solution

\begin{equation*} \mathbb{E}[\lambda (t)\lambda (u)] = \beta v\mathbb{E}[\lambda (u)]\dfrac{{{{\mathrm{e}}^{({\mu _1} - \beta )(t - u)}} - 1}}{{({\mu _1} - \beta )}} + {{\mathrm{e}}^{({\mu _1} - \beta )(t - u)}}\mathbb{E}[{\lambda ^2}(u)],\quad u < t,\end{equation*}

in terms of expectations that we already know how to find.

We define the cross-covariance function of ${N_H}(t)$ and $\lambda (t)$ as

\[{\text{Cov}} [{N_N}(t),\lambda (t)] = \mathbb{E}[{N_H}(t)\lambda (u)] - \mathbb{E}[{N_H}(t)]\mathbb{E}[\lambda (u)].\]

To find $\mathbb{E}[{N_H}(t)\lambda (u)]$ , when $u \gt t$ ,

\begin{align*}\mathbb{E}[{N_H}(t)\lambda (u + \Delta ) \mid \mathcal{F}_u]={N_H}(t)\lambda (u) - {N_H}(t){\nabla _1}(u) + {N_H}(t)\mathbb{E}[Z{\nabla _2} - {\nabla _1}(u)]\lambda (u)\Delta + \mathrm{o}(\Delta ).\end{align*}

Then

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} u}\mathbb{E}[{N_H}(t)\lambda (u)] = \beta v\mathbb{E}[{N_H}(t)] + ({\mu _1} - \beta )\mathbb{E}[{N_H}(t)\lambda (u)],\]

which has solution

\begin{equation*} \mathbb{E}[{N_H}(t)\lambda (u)] = \beta v\mathbb{E}[{N_H}(t)]\dfrac{{{{\mathrm{e}}^{({\mu _1} - \beta )(u - t)}} - 1}}{{({\mu _1} - \beta )}} + {{\mathrm{e}}^{({\mu _1} - \beta )(u - t)}}\mathbb{E}[{N_H}(t)\lambda (t)].\end{equation*}

When $u \lt t$ ,

\[\mathbb{E}[{N_H}(t + \Delta )\lambda (u) \mid \mathcal{F}_t] = {N_H}(t)\lambda (u)[1 - \lambda (t)\Delta ] + [{N_H}(t) + 1]\lambda (u)\lambda (t)\Delta + \mathrm{o}(\Delta ){.}\]

Then

\[\mathbb{E}[{N_H}(t + \Delta )\lambda (u)] = \mathbb{E}[{N_H}(t)\lambda (u)] + \mathbb{E}[\lambda (u)\lambda (t)]\Delta + \mathrm{o}(\Delta ),\]

which means

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N_H}(t)\lambda (u)] = \mathbb{E}[\lambda (u)\lambda (t)] {,}\]

which has solution

\begin{equation*} \mathbb{E}[{N_H}(t)\lambda (u)] = \mathbb{E}[{N_H}(u)\lambda (u)] + \int_u^t \mathbb{E}[\lambda (u)\lambda (\upsilon )]\,{\mathrm{d}} \upsilon, \quad u < t,\end{equation*}

in terms of expectations that we know how to calculate. Finally,

\[\mathbb{E}[{N_H}(t + \Delta ){N_H}(u) \mid \mathcal{F}_t] = {N_H}(t){N_H}(u)[1 - \lambda (t)\Delta ] + ({N_H}(t) + 1){N_H}(u)\lambda (t)\Delta + \mathrm{o}(\Delta )\]

gives us the differential equation

\[\dfrac{{\mathrm{d}}}{{\mathrm{d}} t}\mathbb{E}[{N_H}(t){N_H}(u)] = \mathbb{E}[{N_H}(u)\lambda (t)] {,}\]

with known solution in terms of expectations:

\begin{equation*} \mathbb{E}[{N_H}(t){N_H}(u)] =\mathbb{E}[N_H^2(u)] + \int_u^t {\mathbb{E}[{N_H}(u)\lambda (\upsilon )]\,{\mathrm{d}} \upsilon } , \quad u < t.\end{equation*}

3.2.3. Correlation between ${N_H}(t)$ and $\int_0^t {\lambda (u)\,{\mathrm{d}} u} $

As in Section 3.2.2, the corresponding discussions are presented as follows.

If there is no event occurring in $(t,t + \Delta )$ , with probability $1 - \lambda (t)\Delta + \mathrm{o}(\Delta )$ , then ${N_H}(t + \Delta ) = {N_H}(t)$ and

\begin{align*}\int_0^{t + \Delta } {\lambda (u)\,{\mathrm{d}} u}&= \int_0^t {\lambda (u)\,{\mathrm{d}} u} + \int_t^{t + \Delta } {\Bigg[v + \sum_{{T_i} \lt u} {{Z_i}\,{{\mathrm{e}}^{ - \beta (u - {T_i})}}} \Bigg]\,{\mathrm{d}} u} \\*&= \int_0^t {\lambda (u)\,{\mathrm{d}} u} + \int_t^{t + \Delta } {\Bigg[v + \sum_{{T_i} < t} {{Z_i}\,{{\mathrm{e}}^{ - \beta (u - {T_i})}}} \Bigg]\,{\mathrm{d}} u} \\*&= \int_0^t {\lambda (u)\,{\mathrm{d}} u} + \lambda (t)\Delta + \mathrm{o}(\Delta ).\end{align*}

If there is an event occurring in $(t,t + \Delta )$ , with probability $\lambda (t)\Delta + \mathrm{o}(\Delta )$ , then ${N_H}(t + \Delta ) = {N_H}(t) + 1$ and we obtain

\begin{align*}&\int_0^{t + \Delta } {\lambda (u)\,{\mathrm{d}} u}\\*&\quad = \int_0^t {\lambda (u)\,{\mathrm{d}} u} + \int_t^{{T_{{N_H}(t) + 1}}} {\Bigg[v + \sum_{{T_i} \lt u} {{Z_i}\,{{\mathrm{e}}^{ - \beta (u - {T_i})}}} \Bigg]\,{\mathrm{d}} u} + \int_{{T_{{N_H}(t) + 1}}}^{t + \Delta } {\Bigg[v + \sum_{{T_i} \lt u} {{Z_i}\,{{\mathrm{e}}^{ - \beta (u - {T_i})}}} \Bigg]\,{\mathrm{d}} u} \\&\quad = \int_0^t {\lambda (u)\,{\mathrm{d}} u} + \int_t^{t + \Delta } {\Bigg[v + \sum_{{T_i} < t} {{Z_i}\,{{\mathrm{e}}^{ - \beta (u - {T_i})}}} \Bigg]\,{\mathrm{d}} u} + {Z_{{N_H}(t) + 1}}\int_{{T_{{N_H}(t) + 1}}}^{t + \Delta } {\,{{\mathrm{e}}^{ - \beta (u - {T_{{N_H}(t) + 1}})}}\,{\mathrm{d}} u} \\&\quad = \int_0^t {\lambda (u)\,{\mathrm{d}} u} + \lambda (t)\Delta + {Z_{{N_H}(t) + 1}}{\beta ^{ - 1}}[1 - {{\mathrm{e}}^{ - \beta (t + \Delta - {T_{{N_H}(t) + 1}})}}] + \mathrm{o}(\Delta )\\*&\quad = \int_0^t {\lambda (u)\,{\mathrm{d}} u} + \lambda (t)\Delta + {\nabla _3} + \mathrm{o}(\Delta ),\end{align*}

where ${\nabla _3} = {Z_{{N_H}(t) + 1}}\Delta'$ , $0 \lt \Delta' \lt \Delta$ . Then we have

\begin{align*}&\mathbb{E}\bigg\{ N_H^m(t + \Delta ){\lambda ^n}(t + \Delta ){\bigg[\int_0^{t + \Delta } {\lambda (u)\,{\mathrm{d}} u} \bigg]^l} \mid \mathcal{F}_t\bigg\} \\* &\quad = N_H^m(t){[\lambda (t) - {\Delta _1}]^n}{\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} + \lambda (t)\Delta ]^l}[1 - \lambda (t)\Delta \bigg] + {[{N_H}(t) + 1]^m}\\ &\quad\quad\, \times {[\lambda (t) - {\nabla _1} + {\nabla _2}]^n}{\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} + \lambda (t)\Delta + {\nabla _3}\bigg]^l}\lambda (t)\Delta + \mathrm{o}(\Delta )\\&\quad = \Bigg\{ N_H^m(t){\lambda ^n}(t) + N_H^m(t)\sum_{i = 0}^{n - 1} \dbinom n i{{({-} {\nabla _1})}^{n - i}}{\lambda ^i}(t) - N_H^m(t)\sum_{i = 0}^n \dbinom n i\\&\quad\quad\, \times {{({-} {\nabla _1})}^{n - i}}{\lambda ^{i + 1}}(t) \Delta + \sum_{j = 0}^m \dbinom m j N_H^j(t)\sum_{i = 0}^n \dbinom n i{{({\nabla _2} - {\nabla _1})}^{n - i}}{\lambda ^{i + 1}}(t)\Delta \Bigg\}\\*&\quad\quad\, \times {\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^l}+ lN_H^m(t){\lambda ^{n + 1}}(t){\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^{l - 1}}\Delta + \mathrm{o}(\Delta ).\end{align*}

Thus, as in Theorem 2, we obtain

\begin{align*} & \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ N_H^m(t){\lambda ^n}(t){\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^l}\bigg\} \notag \\*&\quad = n\beta v\mathbb{E}\bigg\{ N_H^m(t){\lambda ^{n - 1}}(t){\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^l}\bigg\} - n\beta \mathbb{E}\bigg\{ N_H^m(t){\lambda ^n}(t){\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^l}\bigg\} \notag \\&\quad\quad\, + \sum_{j = 0}^{m - 1} \dbinom m j\mathbb{E}\bigg\{ N_H^j(t){\lambda ^{n + 1}}(t) {\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^l}\bigg\} + \sum_{j = 0}^m {\sum_{i = 0}^{n - 1} \dbinom m j\dbinom n i {\mu _{n - i}}\mathbb{E}\{ N_H^j(t)} \notag \\* &\quad\quad\, \times {\lambda ^{i + 1}}(t){\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^l} + l\mathbb{E}\bigg\{ N_H^m(t){\lambda ^{n + 1}}(t){\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^{l - 1}}\bigg\}.\end{align*}

As in Section 3.1, we have the following chain recursions for obtaining $\mathbb{E}[N_H^m(t)\int_0^t {\lambda (u)\,{\mathrm{d}} u} ]$ .

Order 0: $\displaystyle\mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] \to$ order 1.

Order 1: $\displaystyle\mathbb{E}\bigg[\lambda (t)\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] \to \mathbb{E}\bigg[{N_H}(t)\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] \to $ order 2.

. . . .

Order m − 1: $\cdots \to$ order m.

Order m: $\displaystyle\mathbb{E}\bigg[{\lambda ^m}(t)\! \int_0^t\!\! {\lambda (u)\,{\mathrm{d}} u} \bigg] \to \mathbb{E}\bigg[{N_H}(t){\lambda ^{m - 1}}(t)\! \int_0^t\!\! {\lambda (u)\,{\mathrm{d}} u} \bigg] \to \cdots \to \mathbb{E}\bigg[N_H^m(t)\! \int_0^t\!\! {\lambda (u)\,{\mathrm{d}} u} \bigg]$ .

Similar recursions can be given for $l = 2$ , $l = 3$ , and so on. For example, for

\[\mathbb{E}\bigg\{ {N_H}(t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\},\]

we have

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] & = \mathbb{E}[\lambda (t)],\\*\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} & = \beta v\mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] + ({\mu _1} - \beta )\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} + \mathbb{E}[{\lambda ^2}(t)],\\*\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ {N_H}(t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} & = \mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} + \mathbb{E}[{N_H}(t)\lambda (t)].\end{align*}

Remark 7. The first equation above always holds; the reason for providing it here is that it will be used below.

In fact, we can use the Dynkin formula to give the equation

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} = \beta v\mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] + ({\mu _1} - \beta )\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} + \mathbb{E}[{\lambda ^2}(t)].\]

Let

\[g({N_H}(t),\lambda (t),t) = \lambda (t)\int_0^t {\lambda (u)\,{\mathrm{d}} u} .\]

Then we have

\begin{align*}\mathcal{A}g({N_H}(t),\lambda (t),t) &= \dfrac{{\partial g}}{{\partial t}} + \dfrac{{\partial g}}{{\partial \lambda }}\dfrac{{{\mathrm{d}} {\lambda (t)} }}{{\mathrm{d}} t} + \lambda (t)\bigg\{ [\lambda (t) + Z]\int_0^t {\lambda (u)\,{\mathrm{d}} u} - \lambda (t)\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg\} \\*&= {\lambda ^2}(t) + \dfrac{{{\mathrm{d}} {\lambda (t)} }}{{\mathrm{d}} t}\int_0^t {\lambda (u)\,{\mathrm{d}} u} + \lambda (t)Z\int_0^t {\lambda (u)\,{\mathrm{d}} u} \\*&= {\lambda ^2}(t) + [\beta v - \beta \lambda (t)]\int_0^t {\lambda (u)\,{\mathrm{d}} u} + \lambda (t)Z\int_0^t {\lambda (u)\,{\mathrm{d}} u} ,\end{align*}

and hence

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} & = \mathbb{E}[\mathcal{A}g({N_H}(t),\lambda (t),t)]\\*&= \beta v\mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] + ({\mu _1} - \beta )\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} + \mathbb{E}[{\lambda ^2}(t)],\end{align*}

which is the equation we obtained by using our elementary approach.

Remark 8. When we see the procedure for using the Dynkin formula above, we know it can sometimes cause confusion for the derivation process for $({{\partial g}}/{{\partial \lambda }})$ , because there is a $\lambda (t)$ in the integral. This tells us that our method is more simple and direct.

Example 2. For the simple self-exciting Hawkes processes in Example 1, we have

\begin{gather*} \mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] = \int_0^t {\mathbb{E}[\lambda (u)]\,{\mathrm{d}} u} = - 2 + 3t + 2\,{{\mathrm{e}}^{ - t}}{;}\\ \begin{aligned} &\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} + \mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} = 3\mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] + \mathbb{E}[{\lambda ^2}(t)]\\* &\quad\, = 9 + 9t - 14\,{{\mathrm{e}}^{ - t}} + 6\,{{\mathrm{e}}^{ - 2t}}\\* & \Rightarrow \mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} = {{\mathrm{e}}^{ - t}}\int_0^t {\,{{\mathrm{e}}^u}(9 + 9u - 14\,{{\mathrm{e}}^{ - u}} + 6\,{{\mathrm{e}}^{ - 2u}})\,{\mathrm{d}} u}\\* &\quad\, = 9t - 14t\,{{\mathrm{e}}^{ - t}} + 6\,{{\mathrm{e}}^{ - t}} - 6\,{{\mathrm{e}}^{ - 2t}}{;}\end{aligned} \\ \begin{aligned} &\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ {N_H}(t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} = \mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} + \mathbb{E}[{N_H}(t)\lambda (t)]\\* &\quad\, = 6 + 18t - 32t\,{{\mathrm{e}}^{ - t}} + 6\,{{\mathrm{e}}^{ - t}} - 12\,{{\mathrm{e}}^{ - 2t}}\\*& \Rightarrow \mathbb{E}\bigg\{ {N_H}(t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} = - 32 + 6t + 9{t^2} + 32t\,{{\mathrm{e}}^{ - t}} + 26\,{{\mathrm{e}}^{ - t}} + 6\,{{\mathrm{e}}^{ - 2t}},\end{aligned} \\ \begin{aligned} & \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}\bigg\{ {\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^2}\bigg\} = 2\mathbb{E}\bigg\{ \lambda (t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} = 18t - 28t\,{{\mathrm{e}}^{ - t}} + 12\,{{\mathrm{e}}^{ - t}} - 12\,{{\mathrm{e}}^{ - 2t}}\\*& \Rightarrow \mathbb{E}\bigg\{ {\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]^2}\bigg\} = - 22 + 9{t^2} + 28t\,{{\mathrm{e}}^{ - t}} + 16\,{{\mathrm{e}}^{ - t}} + 6\,{{\mathrm{e}}^{ - 2t}}.\end{aligned} \end{gather*}

Then we have

\begin{align*}\text{Var} [{N_H}(t)] & = 2\,{{\mathrm{e}}^{ - 2t}} + 46\,{{\mathrm{e}}^{ - t}} + 24t\,{{\mathrm{e}}^{ - t}} + 27t - 48,\\*\text{Var} \bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] & = 2\,{{\mathrm{e}}^{ - 2t}} + 24\,{{\mathrm{e}}^{ - t}} + 16t\,{{\mathrm{e}}^{ - t}} + 12t - 26,\displaybreak\\{\text{Cov}} \bigg[{N_H}(t),\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg] & = \mathbb{E}\bigg\{ {N_H}(t)\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\bigg\} - \mathbb{E}[{N_H}(t)]\mathbb{E}\bigg[\int_0^t {\lambda (u)\,{\mathrm{d}} u} \bigg]\\* & = 2\,{{\mathrm{e}}^{ - 2t}} + 34\,{{\mathrm{e}}^{ - t}} + 20t\,{{\mathrm{e}}^{ - t}} + 18t - 36.\end{align*}

Thus we have the correlation coefficient

\[\tilde \rho (t) = \dfrac{{2\,{{\mathrm{e}}^{ - 2t}} + 34\,{{\mathrm{e}}^{ - t}} + 20t\,{{\mathrm{e}}^{ - t}} + 18t - 36}}{{\sqrt {2\,{{\mathrm{e}}^{ - 2t}} + 46\,{{\mathrm{e}}^{ - t}} + 24t\,{{\mathrm{e}}^{ - t}} + 27t - 48} \sqrt {2\,{{\mathrm{e}}^{ - 2t}} + 24\,{{\mathrm{e}}^{ - t}} + 16t\,{{\mathrm{e}}^{ - t}} + 12t - 26} }}.\]

As shown in Figure 2, we see that correlation coefficient $\tilde \rho (t)$ of ${N_H}(t)$ and $\int_0^t {\lambda (u)\,{\mathrm{d}} u} $ is an increasing function of time t and $\mathop {\lim }_{t \downarrow 0} \tilde \rho (t) \approx 0.8662$ , $\mathop {\lim }_{t \uparrow \infty } \tilde \rho (t) = 1$ , which is interesting to compare with Figure 1.

Figure 2:

The curve of the correlation coefficient $\tilde \rho (t)$ .

3.2.4. Generating functions

This subsection contains results for probability generating functions for the distribution of ${N_H}(t)$ , Laplace transforms for the distribution of $\lambda (t)$ , and the joint distribution of the two. As usual, we are concerned with obtaining the appropriate partial differential equation. We start with the joint distribution.

Theorem 3. For the Hawkes process with intensity function $\lambda (t) = v + \sum_{{T_r} \lt t} {{Z_r}\,{{\mathrm{e}}^{ - \beta (t - {T_r})}}} $ , we have

\begin{equation*} \dfrac{\partial }{{\partial t}}G(\theta,s,t) = [1 - \beta s - \theta \phi (s)]\dfrac{\partial }{{\partial s}}G(\theta ,s,t) - \beta vsG(\theta ,s,t),\end{equation*}

where $G(\theta ,s,t)\,:\!= \mathbb{E}[{\theta ^{{N_H}(t)}}\,{{\mathrm{e}}^{ - s\lambda (t)}}]$ is the p.g.f./Laplace transform of the joint distribution of $({N_H}(t),\lambda (t))$ , and

\[\phi(s)=\mathbb{E}[{{\mathrm{e}}^{ - sZ}}] = \int_0^\infty {\,{{\mathrm{e}}^{ - sz}}\,{\mathrm{d}} {G_Z}(z)}\]

is the Laplace transform of the density function of the marked random variables. The initial condition is $G(\theta ,s,0) = {{\mathrm{e}}^{ - sv}}$ .

Proof. We have

\begin{align*}G(\theta ,s,t + \Delta \mid \mathcal{F}_t) & = \mathbb{E}[{\theta ^{{N_H}(t + \Delta )}}\,{{\mathrm{e}}^{ - s\lambda (t + \Delta )}} \mid \mathcal{F}_t] \\& = {\theta ^{{N_H}(t)}}\,{{\mathrm{e}}^{ - s[\lambda (t) - \Delta (\beta \lambda (t) - \beta v)]}}[1 - \lambda (t)\Delta ]\\&\quad\, + \theta \mathbb{E}[ {{\theta ^{{N_H}(t)}}\,{{\mathrm{e}}^{ - s[\lambda (t) - \Delta (\beta \lambda (t) - \beta v) + {Z_{{N_H}(t) + 1}}(1 - \beta \Delta ')]}}\lambda (t)} ]\Delta + \mathrm{o}(\Delta )\displaybreak\\& = {\theta ^{{N_H}(t)}}\,{{\mathrm{e}}^{ - s\lambda (t)}}\{ 1 + s[\beta \lambda (t) - \beta v]\Delta \} [1 - \lambda (t)\Delta ]\\&\quad\, + \theta \mathbb{E}[ {{\theta ^{{N_H}(t)}}\,{{\mathrm{e}}^{ - s[\lambda (t) - \Delta (\beta \lambda (t) - \beta v)]}}\lambda (t)\,{{\mathrm{e}}^{ - s{Z_{{N_H}(t) + 1}}(1 - \beta \Delta ')}}} ]\Delta\\&\quad\, + \mathrm{o}(\Delta ),\quad 0 \lt \Delta ' \lt \Delta .\end{align*}

Then we have

\begin{align*}\dfrac{\partial }{{\partial t}}G(\theta ,s,t) &= [\beta s - 1 + \theta \phi (s)]\mathbb{E}[{\theta ^{{N_H}(t)}}\,{{\mathrm{e}}^{ - s\lambda (t)}}\lambda (t)] - \beta vs\mathbb{E}[{\theta ^{{N_H}(t)}}\,{{\mathrm{e}}^{ - s\lambda (t)}}]\\*&= [1 - \beta s - \theta \phi (s)]\dfrac{\partial }{{\partial s}}G(\theta ,s,t) - \beta vsG(\theta ,s,t),\end{align*}

and the proof is completed.

Corollary 1. Under the same conditions as above, the Laplace transform of the distribution of $\lambda (t)$ , $g(s,t)\,:\!= \mathbb{E}[ {{{\mathrm{e}}^{ - s\lambda (t)}}} ] = G(1,s,t)$ , and satisfies the partial differential equation

\begin{equation*} \dfrac{\partial }{{\partial t}}g(s,t) = [1 - \beta s - \phi (s)]\dfrac{\partial }{{\partial s}}g(s,t) - \beta vsg(s,t),\end{equation*}

with boundary/initial conditions $g(0,t) = 1$ , $g(s,0) = {{\mathrm{e}}^{ - sv}}$ .

The proof follows readily from the theorem by taking $g(s,t) = G(1,s,t)$ .

Corollary 2. Under the same conditions as above, the p.g.f. of the distribution of $N_H(t)$ is given by $G(\theta ,t) = G(\theta ,0,t)$ and satisfies the differential equation

\begin{equation*} \dfrac{\partial }{{\partial t}}G(\theta ,t) = (\theta - 1)\mathbb{E}[ {{\theta ^{{N_H}(t)}}\lambda (t)} ],\end{equation*}

subject to conditions $G(1,t) = 1$ , $G(\theta ,0) = 1$ .

The joint probability generating function for $({N_H}(t),\lambda (t))$ can be given by solving a partial differential equation, which provides a way to find the moments and count distribution for Hawkes processes.

4.1. Mutually exciting processes

We now consider a fairly general model comprising K mutually exciting marked point processes with exponential exciting kernels. Suppose the intensity of ${N_l}(t)$ is given by

\begin{equation*} {\lambda _l}(t) = {v _l} + \sum_{j = 1}^K {\lambda _{lj}}(t),\quad l = 1, \ldots ,K,\end{equation*}

where

\begin{equation*} {\lambda _{lj}}(t) = \sum_{{T_{j,r}} < t} {{Z_{j,r}}{\alpha _{lj}}\,{{\mathrm{e}}^{ - {\beta _{lj}}(t - {T_{j,r}})}}},\end{equation*}

and ${\{ {{T_{j,r}}} \}_{r = 1,2, \ldots }}$ is a sequence of occurrence times of type j events and ${\{ {{Z_{j,r}}} \}_{r = 1,2, \ldots }}$ the marks associated with them. Note that there is an extra factor ${\alpha _{lj}}$ that contributes to the effect of a type j event, and its mark, on the future intensity of type l events: this is because the same event and the same mark may have different effects on various other types of event. Then

\begin{align*}&\mathbb{E}[ N_k^m(t + \Delta )\lambda _l^n(t + \Delta ) \mid \mathcal{F}_t] = N_k^m(t)\Bigg\{ {\lambda _l}(t) - \Delta \sum_{j = 1}^K {{\lambda _{lj}}(t){\beta _{lj}} } \Bigg\}^n\Bigg\{ {1 - \Delta \sum_{j = 1}^K {{\lambda _j}(t)} } \Bigg\}\\*&\quad\, + \mathbb{E}\Bigg[ {N_k^m(t)\sum_{i = 1,i \ne k}^K {\Delta {\lambda _i}(t)} {\Bigg\{ {{\lambda _l}(t) - \Delta \sum_{j = 1}^K {{\lambda _{lj}}(t){\beta _{lj}} + {Z_{i,{N_i}(t) + 1}}{\alpha _{li}}(1 - \Delta '{\beta _{li}})} } \Bigg\}^n}} \Bigg]\\*&\quad\, + \mathbb{E}\Bigg[ {{( {{N_k}(t) + 1} )}^m}\Delta {\lambda _k}(t){\Bigg\{ {{\lambda _l}(t) - \Delta \sum_{j = 1}^K {{\lambda _{lj}}(t){\beta _{lj}} + {Z_{k,{N_k}(t) + 1}}{\alpha _{lk}}(1 - \Delta '{\beta _{lk}})} }\Bigg\}^n} \Bigg] + \mathrm{o}(\Delta ),\end{align*}

where $0 \lt \Delta ' \lt \Delta$ . Taking expectations and appropriate limits as $\Delta \to 0$ ,

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[ {N_k^m(t)\lambda _l^n(t)} ] &= - n\mathbb{E}\Bigg[ {N_k^m(t)\lambda _l^{n - 1}(t)\sum_{j = 1}^K {{\lambda _{lj}}(t){\beta _{lj}}} } \Bigg] - \mathbb{E}\Bigg[ {\sum_{j = 1}^K {N_k^m(t)\lambda _l^n(t){\lambda _j}(t)} } \Bigg]\\*&\quad\, + \mathbb{E}\Bigg[ {N_k^m(t)\sum_{j = 1,j \ne k}^K {{\lambda _j}(t)} \sum_{i = 0}^n\binom n i \lambda _l^i(t)\mu _{n - i}^{(q)}\alpha _{lj}^{n - i}} \Bigg]\\*&\quad\, + \mathbb{E}\Bigg[ {{{( {{N_k}(t) + 1} )}^m}{\lambda _k}(t)\sum_{i = 0}^n \binom n i\lambda _l^i(t)\mu _{n - i}^{(k)}\alpha _{lk}^{n - i}} \Bigg],\end{align*}

where $\mu _{n - i}^{(\,j)} = \mathbb{E}[ {Z_{j,r}^{n - i}} ]$ is the $(n - i)$ th moment of ${Z_{j,r}}$ for all $r = 1,2, \ldots .$

This can be written as

(16) \begin{align}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[ {N_k^m(t)\lambda _l^n(t)} ]&= - n\sum_{j = 1}^K {{\beta _{lj}}\mathbb{E}[ {N_k^m(t){\lambda _{lj}}(t)\lambda _l^{n - 1}(t)} ]} + \sum_{j = 0}^{m - 1} \binom m j\mathbb{E}[ {N_{k}^{\kern0.5ptj}(t){\lambda _k}(t)\lambda _l^n(t)} ] \notag\\*&\quad\, + \sum_{i = 0}^{n - 1} \binom n i \sum_{j = 1,j \ne k}^K {\mu _{n - i}^{(\,j)}\alpha _{lj}^{n - i}\mathbb{E}[ {N_k^m(t)\lambda _l^i(t){\lambda _j}(t)} ]} \notag\\*&\quad\, + \sum_{j = 0}^m {\sum_{i = 0}^{n - 1} \binom m j\binom n i \mu _{n - i}^{(k)}\alpha _{lk}^{n - i}\mathbb{E}[ {N_{k}^{\kern0.5ptj}(t)\lambda _l^i(t){\lambda _k}(t)} ]} .\end{align}

4.2. Cox processes

Cox [Reference Cox4] introduced a doubly stochastic model of point processes, now usually called Cox processes. More recent treatments of the properties of general classes of Cox processes are given, for example, by [Reference Brémaud and Massoulie2] and [Reference Møller20]. For this paper, we consider a simple version used by [Reference Dassios and Jang7] in the context of catastrophe insurance. It has intensity

(17) \begin{equation}\lambda (t) = v + \sum_{{T_r} < t} {{Y_r}\,{{\mathrm{e}}^{ - \beta (t - {T_r})}}} ,\quad t \gt 0,\end{equation}

where ${\{ {{Y_r}} \}_{r = 1,2, \ldots }}$ is a sequence of i.i.d. random variables with distribution function ${G_Y}(y)$ , $y \gt 0$ . This looks much like (1) for the Hawkes process when $v(t) = v$ . But there is one very big difference: the events $\{ {{T_r}} \}$ on the right of (17) arise from a time-homogeneous Poisson process with rate $\eta $ , and the $\{ {{Y_r}} \}$ are marks associated with events of the Poisson process; in contrast, the events on the right of (1) belong to the same process that is generated by the intensity $\lambda (t)$ on the left. It is this that drives the contagious property of the Hawkes process, with the occurrence of past events increasing the intensity that governs future events.

Note that the above Cox model involves two point processes, the Cox process ${N_C}(t)$ and the Poisson process, ${N_P}(t)$ , that drives it. These can be considered as forming a bivariate mutually exciting pair. In this case the general form of (2) can be written as the special case

(18) \begin{equation}{\lambda _C}(t) = v + \sum_{{T_{P,r}} < t} {Y_i}{\mathrm{e}}^{ - \beta (t - T_{P;r})}, \quad{\lambda _P}(t) = \eta .\end{equation}

In general, each intensity of a bivariate mutually exciting Hawkes process consists of a baseline intensity and two exciting terms: a self-exciting term and a cross-exciting term. In this case we see that the Cox intensity contains a term that is excited by the Poisson process, but no self-exciting term; the Poisson intensity has no exciting terms at all. So they are not really mutually exciting, because the Cox process is not excited at all!

Similarly, we know that the triple $( {{N_C}(t),{N_P}(t),{\lambda _C}(t)} )$ forms a Markov process.

For the Cox process we have two point processes, ${N_C}(t)$ and the Poisson process, ${N_P}(t)$ , that drives it. Therefore, for small time increment $\Delta$ , we need four probabilities:

\begin{align*}& \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 0\} \cap \{ {N_P}(t + \Delta ) - {N_P}(t) = 0\} \mid \mathcal{F}_t\}\\*&\quad = [1 - \lambda (t)\Delta ][1 - \eta \Delta ] + \mathrm{o}(\Delta ) = 1 - (\eta + \lambda (t))\Delta + \mathrm{o}(\Delta ),\\& \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 0\} \cap \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \mathcal{F}_t \} = \eta \Delta + \mathrm{o}(\Delta ),\\& \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} \cap \{ {N_P}(t + \Delta ) - {N_P}(t) = 0\} \mid \mathcal{F}_t \} = \lambda (t)\Delta + \mathrm{o}(\Delta ),\\*& \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} \cap \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \mathcal{F}_t \} = \mathrm{o}(\Delta ).\end{align*}

Note that the reasons for the above equality are as follows. We can consider two cases: (i) $\{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} $ occurs no later than the occurrence of $\{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} $ ; (ii) $\{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} $ occurs after the occurrence of $\{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} $ .

For case (i), we have

\begin{align*}& \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} \cap \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \mathcal{F}_t \} \\*& \quad =\mathbb{P}\{ \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} ,\mathcal{F}_t \} \\& \quad\quad\, \times \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} \mid \mathcal{F}_t \} \\& \quad= \mathbb{P}\{ \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \mathcal{F}_t \} \mathbb{P}\{\{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} \mid \mathcal{F}_t\} \\& \quad\quad\, (\text{{as} $ N_C (t)$ does not affect $ N_P(t)$, and $ \mathbb{P}\{ {N_C}(t + \Delta ) - {N_C}(t) \ge 2\} = \mathrm{o}(\Delta )$ if}\\& \quad\quad\, \text{$\{{N_P}(t + \Delta ) - {N_P}(t) = 1\}$ affects $ N_C(t))$}\\*& \quad= \eta \Delta \lambda (t)\Delta + \mathrm{o}(\Delta ) = \mathrm{o}(\Delta ).\end{align*}

For case (ii), we have

\begin{align*}& \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} \cap \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \mathcal{F}_t\} \\* &\quad = \mathbb{P}\{ \{ {N_C}(t + \Delta ) - {N_C}(t) = 1\} \mid \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} ,\mathcal{F}_t \} \\* & \quad\quad\, \times \mathbb{P}\{ \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \mathcal{F}_t\} \\&\quad = \lambda (t + \Delta )\lambda (t)\Delta \mathbb{P}\{ \{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} \mid \mathcal{F}_t\}+ \mathrm{o}(\Delta )\\*&\quad = \lambda (t + \Delta )\lambda (t)\Delta \eta \Delta + \mathrm{o}(\Delta ) = \mathrm{o}(\Delta ),\end{align*}

where

\begin{align*}\lambda (t + \Delta )&= v + \sum_{{T_{P,i}} \lt t + \Delta } {{Y_i}\,{{\mathrm{e}}^{ - \beta (t - {T_{P,i}} + \Delta )}}} \\*&= v + {{\mathrm{e}}^{ - \beta \Delta }}[\lambda (t) - v] + {Y^*}\,{{\mathrm{e}}^{ - \beta {\Delta _1}}} \quad (0 \lt {\Delta _1} < \Delta )\\*&= (1 - \beta \Delta )\lambda (t) + \beta v\Delta + {Y^*}\,{{\mathrm{e}}^{ - \beta {\Delta _1}}} + \mathrm{o}(\Delta ).\end{align*}

For the Cox process we have

\begin{equation*}\lambda (t + \Delta ) = \begin{cases}\lambda (t) - \Delta [\beta \lambda (t) - \beta v] + \mathrm{o}(\Delta ) &\text{given } \{ {N_P}(t + \Delta ) - {N_P}(t) = 0\} ,\\\lambda (t) - \Delta [\beta \lambda (t) - \beta v] + {Y_{{N_P}(t) + 1}} + \mathrm{o}(\Delta ) &\text{given }\{ {N_P}(t + \Delta ) - {N_P}(t) = 1\} .\end{cases}\end{equation*}

Thus, apart from the constant term, the intensity decays exponentially with an added jump, Y, when a new Poisson event occurs at a time that precedes the end of the time increment by $\Delta '$ $(\Delta ' \lt \Delta )$ .

Let ${\nabla _4} = {Y_{{N_P}(t) + 1}}(1 - \beta \Delta ')$ , so $\mathop {\lim }_{\Delta \downarrow 0} ({\nabla _4} - {\nabla _1}) = {Y_{{N_P}(t) + 1}}$ . Then we have

\begin{align*}&\mathbb{E}[N_C^m(t + \Delta ){\lambda ^n}(t + \Delta ) \mid \mathcal{F}_t]\\*&\quad =N_C^m(t){( {\lambda (t) - {\nabla _1}} )^n}( {1 - (\lambda (t) + \eta )\Delta } ) + N_C^m(t){( {\lambda (t) - {\nabla _1} + {\nabla _4}} )^n}\eta \Delta \\*&\quad\quad\, + {( {{N_C}(t) + 1} )^m}{( {\lambda (t) - {\nabla _1}} )^n}\lambda (t)\Delta + \mathrm{o}(\Delta ),\end{align*}

and

(19) \begin{align}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N_C^m(t){\lambda ^n}(t)]&= \mathop {\lim }_{\Delta \downarrow 0} \dfrac{{\mathbb{E}[N_C^m(t + \Delta ){\lambda ^n}(t + \Delta )] - \mathbb{E}[ {N_C^m(t){\lambda ^n}(t)} ]}}{\Delta }\notag\\*&= n\beta v\mathbb{E}[N_C^m(t){\lambda ^{n - 1}}(t)] - n\beta \mathbb{E}[N_C^m(t){\lambda ^n}(t)]\notag\\&\quad\, +\eta \sum_{i = 0}^{n - 1} \binom n i {\mu _{n - i}}\mathbb{E}[N_C^m(t){\lambda ^i}(t)] \notag\\*&\quad\, + \sum_{j = 0}^{m - 1} \binom m j \mathbb{E}[N_C^j(t){\lambda ^{n + 1}}(t)],\end{align}

where ${\mu _r} = \mathbb{E}[{Y^r}]$ is the rth moment of the mark Y. In fact, because of the Markov property of $({N_C}(t),\lambda (t))$ , we have the same result for the Cox process by using the Dynkin formula for the objective function $\mathbb{E}[N_C^m(t){\lambda ^n}(t)]$ .

Differential (19) can be solved, subject to initial conditions $\mathbb{E}[N_C^i(0)] = 0$ , $\mathbb{E}[{\lambda ^i}(0)] = {v ^i}$ $(i = 1,2, \ldots )$ . We see that this involves similar functions with various powers, so that such equations must be solved recursively. Start with $m = 0$ , $n = 1$ , and we have

\[\dfrac{{\mathrm{d}} \mathbb{E}[\lambda (t)]}{{\mathrm{d}} t} + \beta \mathbb{E}[\lambda (t)] = \beta v + \eta {\mu _1},\]

which has solution

\begin{equation*} \mathbb{E}[\lambda (t)] = v + \dfrac{{\eta {\mu _1}}}{\beta }\, ( {1 - {{\mathrm{e}}^{ - \beta t}}} ).\end{equation*}

Alternatively, this is easily obtained by taking expectations of (17), because it is easy to obtain the expectation of a function of Poisson random variables.

If we let $m = 1$ , $n = 0$ , then (19) becomes

\[\dfrac{{\mathrm{d}} \mathbb{E}[ {{N_C}(t)} ]}{{\mathrm{d}} t} = \mathbb{E}[\lambda (t)]\]

or, rather obviously,

\begin{equation*} \mathbb{E}[ {{N_C}(t)} ] = \int_0^t \mathbb{E}[\lambda (u)]\,{\mathrm{d}} u = \bigg( {\dfrac{{\eta {\mu _1}}}{\beta } + v } \bigg)t - \dfrac{{\eta {\mu _1}}}{{{\beta ^2}}}\, ( {1 - {{\mathrm{e}}^{ - \beta t}}} ).\end{equation*}

The general equation using $m = 0$ is

\[\dfrac{{\mathrm{d}} \mathbb{E}[{\lambda ^n}(t)]}{{\mathrm{d}} t} + n\beta \mathbb{E}[{\lambda ^n}(t)] = n\beta v \mathbb{E}[{\lambda ^{n - 1}}(t)] + \eta \sum_{i = 0}^{n - 1} \binom n i{\mu _{n - i}}\mathbb{E}[{\lambda ^i}(t)],\]

so we have a simple recursion for $\mathbb{E}[{\lambda ^n}(t)]$ in terms of lower powers of n. Recursion is not so simple for $m \gt 0$ , but we give the first two equations that are obviously soluble in terms of powers that we already know. Thus

\begin{align*}\dfrac{{\mathrm{d}} \mathbb{E}[{N_C}(t)\lambda (t)]}{{\mathrm{d}} t} + \beta \mathbb{E}[{N_C}(t)\lambda (t)] & = \beta v \mathbb{E}[{N_C}(t)] + \eta {\mu _1}\mathbb{E}[{N_C}(t)] + \mathbb{E}[{\lambda ^2}(t)],\\*\dfrac{{\mathrm{d}} \mathbb{E}[N_C^2(t)]}{{\mathrm{d}} t} & = \mathbb{E}[\lambda (t)] + 2\mathbb{E}[{N_C}(t)\lambda (t)].\end{align*}

4.3. Dynamic contagion processes

Dassios and Zhao [Reference Dassios and Zhao8] introduced a dynamic contagion process, ${N_D}(t)$ , with intensity function

(20) \begin{equation}{\lambda _D}(t) = v + ({\lambda _0} - v)\,{{\mathrm{e}}^{ - \beta t}} + \sum_{{S_r} \lt t} {{Y_r}\,{{\mathrm{e}}^{ - \beta (t - {S_r})}}} + \sum_{{T_r} < t} {{Z_r}\,{{\mathrm{e}}^{ - \beta (t - {T_r})}}} .\end{equation}

Here ${\{ {{S_r}} \}_{r = 1,2, \ldots }}$ are occurrence times of a homogeneous Poisson process with rate $\eta $ , and ${\{ {{Y_r}} \}_{r = 1,2, \ldots }}$ are their associated marks, assumed to be i.i.d. random variables with distribution function ${G_Y}(y)$ , $y \gt 0$ ; ${\{ {{T_r}} \}_{r = 1,2, \ldots }}$ are occurrence times of a dynamic contagion process with corresponding marks, ${\{ {{Z_r}} \}_{r = 1,2, \ldots }}$ , assumed to be i.i.d. random variables with distribution function ${G_Z}(z)$ , $z \gt 0$ . It is assumed that the random variables ${Y_r}$ , ${Z_r}$ are independent of the history of the process prior to their occurrences. In this case the baseline intensity is not constant.

Note that, apart from changing subscript C into D, and having a time-varying baseline intensity, (20) differs from the first part of (18) simply by adding a self-exciting term to the cross-exciting term that is already there. Together with the expression for ${\lambda _P}(t)$ in (18), we once again have a bivariate mutually exciting process $( {{N_D}(t),{N_P}(t)} )$ and, because of the exponential form of the exciting terms, the triple $( {{N_D}(t),{N_P}(t),{\lambda _D}(t)} )$ forms a Markov process: in fact, because ${N_P}(t)$ is Poisson, $( {{N_D}(t),{\lambda _D}(t)} )$ is a Markov process. In introducing the dynamic contagion process, Dassios and Zhao [Reference Dassios and Zhao8] described it as a generalization of both the Hawkes process (with exponential decay) and the Cox process with shot noise intensity (with exponential decay). We see that it is in fact a special case of a bivariate mutually exciting Hawkes process.

The dynamic contagion process was defined in (20). We restate it here in a slightly different form. The intensity is

\[\lambda (t) = v (t) + \sum_{{S_r} \lt t} {{Y_i}\,{{\mathrm{e}}^{ - \beta (t - {S_r})}}} + \sum_{{T_r} < t} {{Z_r}\,{{\mathrm{e}}^{ - \beta (t - {T_r})}}} ,\]

where the baseline intensity is the time-varying function $v (t) = v + ({\lambda _0} - v)\,{{\mathrm{e}}^{ - \beta t}}$ .

In this model we have two point processes: the dynamic contagion process ${N_D}(t)$ , which has intensity $\lambda (t)$ , and the Poisson process ${N_P}(t)$ , which has constant intensity $\eta $ .

Similarly, we seek to find a differential equation for $\mathbb{E}[ {N_D^m(t)N_P^k(t){\lambda ^n}(t)} ]$ . The result is as follows:

(21) \begin{align}&\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N_D^m(t)N_P^k(t){\lambda ^n}(t)] \notag \\*&\quad = n[v'(t) + \beta v(t)]\mathbb{E}[N_D^m(t)N_P^k(t){\lambda ^{n - 1}}(t)] - n\beta \mathbb{E}[N_D^m(t)N_P^k(t){\lambda ^n}(t)] \notag \\&\quad\quad\, + \eta \sum_{r = 0}^k \dbinom k r \sum_{i = 0}^{n - 1} \dbinom n i\mu _{n - i}^{(Y)}\mathbb{E}[N_D^m(t)N_p^r(t){\lambda ^i}(t)] + \eta \sum_{r = 0}^{k - 1} \dbinom k r \mathbb{E}[N_D^m(t)N_p^r(t){\lambda ^n}(t)] \notag \\*&\quad\quad\, + \sum_{j = 0}^m \dbinom m j \sum_{i = 0}^{n - 1} \dbinom n i\mu _{n - i}^{(Z)}\mathbb{E}[N_D^j(t)N_P^k(t){\lambda ^{i + 1}}(t)] + \sum_{j = 0}^{m - 1} \dbinom m j\mathbb{E}[N_D^j(t) N_P^k(t){\lambda ^{n + 1}}(t)],\end{align}

where

\[v '(t) = \dfrac{{\mathrm{d}} v (t)}{{\mathrm{d}} t} = - \beta ({\lambda _0} - v )\,{{\mathrm{e}}^{ - \beta t}}.\]

Equation (21) is of course a special case of (16) when $v(t) \equiv v $ is constant. We can consider special cases when some of the parameters m, k, n are zero. We can also consider other cases such as $\mathbb{E}[N_D^m(t)\lambda _1^{{n_1}}(t)\lambda _2^{{n_2}}(t)]$ , where

\[ {\lambda _1}(t) = \sum_{{S_r} \lt t} {{Y_r}\,{{\mathrm{e}}^{ - \beta (t - {S_r})}}} , \quad {\lambda _2}(t) = \sum_{{T_r} < t} {{Z_r}\,{{\mathrm{e}}^{ - \beta (t - {T_r})}}} . \]

With the given baseline intensity the process is Markovian, so (21) can also be derived by Dynkin’s formula. However, our proof is valid for more general baseline intensity functions.

4.4. Inhomogeneous Poisson processes

Here we simply take $\lambda (t)$ to be a given deterministic function: there are no exciting kernels of the kind we have assumed so far. Using our method, we can get the differential equation

\begin{equation*} \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[g(N(t))] = \{ \mathbb{E}[g(N(t) + 1)] - \mathbb{E}[g(N(t))]\} \lambda (t),\end{equation*}

which can be solved analytically for some functions $g \colon {\Re ^ + } \times {\emph{\textbf{N}}} \to \Re $ under some mild conditions.

Example 3. For $g(x) = {\theta ^x}$ , $\theta \gt 0$ , we have

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{\theta ^{N(t)}}] = \{ \mathbb{E}[{\theta ^{N(t) + 1}}] - \mathbb{E}[{\theta ^{N(t)}}]\} = (\theta - 1)\mathbb{E}[{\theta ^{N(t)}}]\lambda (t),\]

so that

\begin{equation*} \mathbb{E}[{\theta ^{N(t)}}] = \exp \bigg[ {\int_0^t {(\theta - 1)\lambda (u)\,{\mathrm{d}} u} } \bigg],\end{equation*}

which is of course the very well known expression for the p.g.f. of the inhomogeneous Poisson process.

Example 4. For $g(x) = {x^n}$ , $n = 1,2, \ldots ,$ we have

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N^m}(t)] = \{ \mathbb{E}[{(N(t){\rm{ + }}1)^m}] - \mathbb{E}[{N^m}(t)]\} \lambda (t) = \lambda (t)\sum_{j = 0}^{m - 1}\binom m j\mathbb{E}[{N^{\kern0.5ptj}}(t) ],\]

which can be integrated to give

(22) \begin{equation}\mathbb{E}[{N^m}(t)] = \sum_{j = 1}^{m - 1} \binom m j\int_0^t {\lambda (u)\mathbb{E}[{N^{\kern0.5ptj}}(u)]\,{\mathrm{d}} u }.\end{equation}

In fact, for an inhomogeneous Poisson process N(t) whose intensity function is $\lambda (t)$ , and

\[\mathbb{P}\{N(t) = k\}= \dfrac{[\Lambda (t)]^k}{k!}\,{{\mathrm{e}}^{ - \Lambda (t)}},\quad k = 0,1, \ldots ,\]

where $\Lambda (t) = \int_0^t {\lambda (u)\,{\mathrm{d}} u} ,$ we have

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N^m}(t)] &= \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\Bigg\{ \sum_{k = 0}^\infty {k^m}\dfrac{[\Lambda (t)]^k}{k!}\,{{\mathrm{e}}^{ - \Lambda (t)}}\Bigg\} \\*&= \lambda (t)\sum_{k = 0}^\infty {{k^m}\dfrac{{{{[\Lambda (t)]}^{k - 1}}}}{{(k - 1)!}}\,{{\mathrm{e}}^{ - \Lambda (t)}}} - \lambda (t)\sum_{k = 0}^\infty {{k^m}\dfrac{{{{[\Lambda (t)]}^k}}}{{k!}}\,{{\mathrm{e}}^{ - \Lambda (t)}}} \\*&= \sum_{j = 0}^{m - 1} \binom m j \mathbb{E}[{N^{\kern0.5ptj}}(t)\lambda (t)],\end{align*}

which is the same as (22) with initial condition $\mathbb{E}[{N^m}(0)] = 0$ .

4.5. A multi-marked Hawkes process

We now consider a new type of Hawkes process whose intensity function is given by

\begin{equation*} \lambda (t) = v + \sum_{j = 1}^K {\sum_{{T_r} < t} {{Z_{j,r}}\,{{\mathrm{e}}^{ - {\beta _j}(t - {T_r})}}} } ,\end{equation*}

where ${T_r}$ is the occurrence time of the rth event of the point process ${N_H}(t)$ ; for each $j = 1$ to K, $\{Z_{j,r}\sim G_{Z_j}(z) \}_{r = 1,2, \ldots }$ is a sequence of i.i.d. marks; these series are also independent of each other and independent of previous values of the process N(t). The model looks very similar to the standard mutually exciting process. However, in this case there is only one point process and each event on the right of this equation is operated on by a kernel that consists of K different exponential functions, each multiplied by a different mark sequence, which combine additively to produce the output intensity. Note that the rates ${\beta _j}$ are in general different: if they were all the same then this would become a simpler self-exciting process, with marks that are just the sums of the individual marks $\sum_{j = 1}^K {{Z_{j,r}}} $ , with a distribution that is the convolution of the various ${G_{{Z_j}}}(z)$ distributions. Our method can still work on this more general Hawkes process. In the following, we only consider the case of $K = 2$ .

Theorem 4. For a Hawkes process with intensity function

\begin{equation*} \lambda (t) = v + \sum_{{T_r} \lt t} {{Z_{1,r}}\,{{\mathrm{e}}^{ - {\beta _1}(t - {T_r})}}} + \sum_{{T_r} < t} {{Z_{2,r}}\,{{\mathrm{e}}^{ - {\beta _2}(t - {T_r})}}} ,\end{equation*}

we have

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N_H^m(t){\lambda ^n}(t)]& = - n{\beta _1}\mathbb{E}[N_H^m(t){\lambda ^{n - 1}}(t){\lambda _1}(t)] - n{\beta _2}\mathbb{E}[N_H^m(t){\lambda ^{n - 1}}(t){\lambda _2}(t)]\\*& \quad\, + \sum_{j = 0}^{m - 1} \binom m j \mathbb{E}[N_H^j(t){\lambda ^{n + 1}}(t)] + \sum_{j = 0}^m \sum_{i = 0}^{n - 1} \binom m j\binom n i {w_{n - i}}\mathbb{E}[N_H^j(t){\lambda ^{i + 1}}(t)],\end{align*}

where for $(k = 1,2, \ldots ;\,r = 1,2, \ldots\!)$

\[ {\lambda _1}(t) = \sum_{{T_r} \lt t} {{Z_{1,r}}\,{{\mathrm{e}}^{ - {\beta _1}(t - {T_r})}}} {,} \quad {\lambda _2}(t) = \sum_{{T_r} < t} {{Z_{2,r}}\,{{\mathrm{e}}^{ - {\beta _2}(t - {T_r})}}} {,} \quad {w_k} = \int_0^\infty {{z^k}\,{\mathrm{d}} {G_{{Z_{1,r}} + {Z_2}_{,r}}}(z)}. \]

We can also treat the intensity components separately:

\begin{align*}& \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N_H^m(t)\lambda _1^{{n_1}}(t)\lambda _2^{{n_2}}(t)]\\*&\quad = - ({n_1}{\beta _1} + {n_2}{\beta _2})\mathbb{E}[N_H^m(t)\lambda _1^{{n_1}}(t)\lambda _2^{{n_2}}(t)] + \sum_{j = 0}^{m - 1} \dbinom m j\mathbb{E}[N_H^j(t)\lambda (t)\lambda _1^{{n_1}}(t)\lambda _2^{{n_2}}(t)] \\&\quad\quad\, + \sum_{j = 0}^m \sum_{i = 0}^{{n_1} - 1} \dbinom m j\dbinom {n_1} {i} \mu _{{n_1} - i}^{(1)}\mathbb{E}[N_H^j(t)\lambda (t)\lambda_1^i(t)\lambda _2^{{n_2}}(t)] \\&\quad\quad\, + \sum_{j = 0}^m\sum_{i = 0}^{{n_2} - 1} \dbinom m j\dbinom {n_2} {i}\mu _{{n_2} - i}^{(2)}\mathbb{E}[N_H^j(t)\lambda (t)\lambda _1^{{n_1}}(t)\lambda _2^i(t)]\\*&\quad\quad\, + \sum_{j = 0}^m \sum_{{j_1} = 0}^{{n_1} - 1} \sum_{{j_2} = 0}^{{n_2} - 1} \dbinom m j\dbinom {n_1} {i_1} \dbinom {n_2} {i_2}\mu _{{n_1} - {i_1}}^{(1)}\mu _{{n_2} - {i_2}}^{(2)}\mathbb{E}[N_H^j(t)\lambda (t)\lambda _1^{{i_1}}(t)\lambda _2^{{i_2}}(t)]{,}\end{align*}

where

\[\mu _k^{(i)} = \int_0^\infty z_i^k \,{\mathrm{d}} G_{Z_i}(z_i)\quad (i = 1,2;\, k = 1,2, \ldots ).\]

Some special cases are considered in the following. When ${n_1} = {n_2} = 0$ , by using Theorem 4 we have

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N_H^m(t)] = \sum_{i = 0}^{m - 1} \binom m i \mathbb{E}[N_H^i(t)\lambda (t)],\]

which is consistent with Theorem 1, provided for general counting processes. In particular,

\[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N_H}(t)] = \mathbb{E}[\lambda (t)] \Rightarrow \mathbb{E}[{N_H}(t)] = \int_0^t \mathbb{E}[\lambda (u)]\,{\mathrm{d}} u.\]

When $m = 0$ , by using Theorem 4 we have

\begin{align*} & \dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda _1^{{n_1}}(t)\lambda _2^{{n_2}}(t)] = - ({n_1}{\beta _1} + {n_2}{\beta _2})\mathbb{E}[\lambda _1^{{n_1}}(t)\lambda _2^{{n_2}}(t)] \notag \\ & \quad \quad\, + \sum_{i = 0}^{{n_1} - 1} \binom {n_1} {i} \mu _{{n_1} - i}^{(1)}\mathbb{E}[\lambda (t)\lambda _1^i(t)\lambda _2^{{n_2}}(t)]\notag \displaybreak\\& \quad \quad\, + \sum_{i = 0}^{{n_2} - 1} \binom {n_2} {i}\mu _{{n_2} - i}^{(2)}\mathbb{E}[\lambda (t)\lambda _1^{{n_1}}(t)\lambda _2^i(t)] \notag \\ & \quad \quad\, + \sum_{{i_1} = 0}^{{n_1} - 1} \sum_{{i_2} = 0}^{{n_2} - 1} \binom {n_1} {i_1}\binom {n_2} {i_2} \mu _{{n_1} - {i_1}}^{(1)}\mu _{{n_2} - {i_2}}^{(2)}\mathbb{E}[\lambda (t)\lambda _1^{{i_1}}(t)\lambda _2^{{i_2}}(t)].\end{align*}

4.6. Non-Markovian Hawkes processes

In this subsection, a non-Markovian Hawkes process, denoted by ${N_{HG}}(t)$ , is considered by using our method, whose intensity kernel function is $h(t) = t\,{{\mathrm{e}}^{ - t}}$ $(t \ge 0)$ (gamma decay function), that is, its intensity function is

\begin{equation*} \lambda (t) = v + \int_0^t {h(t - s)\,{\mathrm{d}} {N_{HG}}(t)} = v + \sum_{{T_i} \lt t} {h(t - {T_i})} = v + \sum_{{T_i} < t} {[(t - {T_i})\,{{\mathrm{e}}^{ - (t - {T_i})}}} ],\end{equation*}

where ${\{ {T_i}\} _{i = 1,2, \ldots }}$ is the series of times at which events occur for the Hawkes process ${N_{HG}}(t)$ .

We shall give the results for the moments of ${N_{HG}}(t)$ by using our method, and then present a closed-form expression for $\mathbb{E}[N_{HG}^2(t)]$ as an example. Because the higher-order moments can also be obtained similarly, here they are omitted.

It is easy to know that $h'(t) = {{\mathrm{e}}^{ - t}} - t\,{{\mathrm{e}}^{ - t}} = {h_2}(t) - h(t)$ , where ${h_2}(t) = {{\mathrm{e}}^{ - t}}$ , for $t \geq 0$ , $h_2(t)=0$ , for $t<0$ .

First, when $\{ {N_{HG}}(t + \Delta ) - {N_{HG}}(t) = 0\} $ occurs, we have

\begin{align*}{\lambda _0}(t + \Delta )&= v + \sum_{{T_i} \lt t + \Delta } {h(t - {T_i} + \Delta )} = v + \sum_{{T_i} \lt t} {h(t - {T_i} + \Delta )} \\*&= v + \sum_{{T_i} \lt t} {h(t - {T_i})} + \Delta \sum_{{T_i} \lt t} {h'(t - {T_i})} + \mathrm{o}(\Delta ) \\&= \lambda (t) + \Delta \Bigg[\sum_{{T_i} \lt t} {{h_2}(t - {T_i})} - \sum_{{T_i} < t} {h(t - {T_i})} \Bigg] + \mathrm{o}(\Delta )\\*&= \lambda (t) + \Delta [{\lambda _2}(t) - \lambda (t) + v] + \mathrm{o}(\Delta ),\end{align*}

where ${\lambda _2}(t)$ is a stochastic process defined by

\begin{equation*} {\lambda _2}(t)\,:\!= \int_0^t {{h_2}(t - s)\,{\mathrm{d}} {N_{HG}}(t)} = \sum_{{T_i} \lt t} {{h_2}(t - {T_i})} = \sum_{{T_i} < t} {\,{{\mathrm{e}}^{ - (t - {T_i})}}} .\end{equation*}

Note that ${\lambda _2}(t)$ depends on the point process ${N_{HG}}(t)$ , whose moments we consider now $({\lambda _2}(0) = 0, {\lambda _2}(t) = 0$ , when $t \lt {T_1})$ .

On the other hand, when $\{ {N_{HG}}(t + \Delta ) - {N_{HG}}(t) = 1\} $ occurs, we have

\begin{align*} {\lambda _1}(t + \Delta )&= v + \sum_{{T_i} \lt t + \Delta } {h(t - {T_i} + \Delta )} \\* & = v + \sum_{{T_i} \lt t} {h(t - {T_i} + \Delta )} + h(t - {T^*} + \Delta ) \\*&= {\lambda _0}(t + \Delta ) + h({\Delta _1}) + \mathrm{o}(\Delta ),\quad 0 \lt {\Delta _1} \lt \Delta ,\ t \lt {T^*} < t + \Delta .\end{align*}

Using our elementary method presented in Section 2, we have the following result.

Theorem 5. For a Hawkes process ${N_{HG}}(t)$ with intensity function

\[\lambda (t) = v + \sum_{{T_i} < t} {[(t - {T_i})\,{{\mathrm{e}}^{ - (t - {T_i})}}} ],\]

we have

(23) \begin{align}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N_{HG}^m(t){\lambda ^n}(t)\lambda _2^l(t)] &= n\mathbb{E}[N_{HG}^m(t){\lambda ^{n - 1}}(t)\lambda _2^{l + 1}(t)] - (n + l)\mathbb{E}[N_{HG}^m(t){\lambda ^n}(t)\lambda _2^l(t)]\notag\\*&\quad\, + nv\mathbb{E}[N_{HG}^m(t){\lambda ^{n - 1}}(t)\lambda _2^l(t)] \notag\\&\quad\, + \sum_{i = 0}^{m - 1} \binom m i\mathbb{E}[N_{HG}^i(t){\lambda ^{n + 1}}(t){[{\lambda _2}(t) + 1]^l}]\notag\\*&\quad\, + \sum_{k = 0}^{l - 1} \binom l k \mathbb{E}[N_{HG}^m(t){\lambda ^{n + 1}}(t)\lambda _2^k(t)],\end{align}

where $\sum_0^w {[*]} = 0$ if $w \lt 0$ and ${\lambda ^i}(t) = 0$ if $i \lt 0$ .

Special cases

From (23) we have the following special cases.

1. (i) When $m = 0$ , $n = 1$ , $l = 0$ , we have

   \[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda (t)] = \mathbb{E}[{\lambda _2}(t)] - \mathbb{E}[\lambda (t)] - v.\]

2. (ii) When $m = 0$ , $n = 0$ , $l = 1$ , we have

   \[\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda_2 (t)] = \mathbb{E}[{\lambda }(t)] - \mathbb{E}[\lambda_2 (t)].\]

To summarize cases (i) and (ii), we have a set of two differential equations as follows:

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda (t)] & = \mathbb{E}[{\lambda _2}(t)] - \mathbb{E}[\lambda (t)] - v,\\*\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda_2 (t)] & = \mathbb{E}[{\lambda }(t)] - \mathbb{E}[\lambda_2 (t)].\end{align*}

Numerical results, with $\mathbb{E}[{\lambda _2}(0)] = 0$ , $\mathbb{E}[\lambda (0)] = v$ , and set $v = 2$ , are

\begin{align*}\mathbb{E}[\lambda (t)] & = \dfrac{1}{2}\,{{\mathrm{e}}^{ - 2t}} + t + \dfrac{3}{2},\\*\mathbb{E}[{\lambda _2}(t)] & = - \dfrac{1}{2}\,{{\mathrm{e}}^{ - 2t}} + t + \dfrac{1}{2}.\end{align*}

1. (iii) When $m = 0$ , $n = 2$ , $l = 0$ , we have

   \begin{equation*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{\lambda ^2}(t)] = 2\mathbb{E}[\lambda (t){\lambda _2}(t)] - 2\mathbb{E}[{\lambda ^2}(t)] + 2v\mathbb{E}[\lambda (t)].\end{equation*}

2. (iv) When $m = 0$ , $n = 1$ , $l = 1$ , we have

   \begin{equation*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda (t){\lambda _2}(t)] = \mathbb{E}[{\lambda ^2}(t)] + \mathbb{E}[\lambda _2^2(t)] - 2\mathbb{E}[\lambda (t){\lambda _2}(t)] + v\mathbb{E}[{\lambda _2}(t)].\end{equation*}

3. (v) When $m = 0$ , $n = 0$ , $l = 2$ , we have

   \begin{equation*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda _2^2(t)] = 2\mathbb{E}[\lambda (t){\lambda _2}(t)] - 2\mathbb{E}[\lambda _2^2(t)] + \mathbb{E}[\lambda (t)].\end{equation*}

To summarize cases (iii), (iv), and (v), we have a set of three differential equations as follows:

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{\lambda ^2}(t)] & = 2\mathbb{E}[\lambda (t){\lambda _2}(t)] - 2\mathbb{E}[{\lambda ^2}(t)] + 2v\mathbb{E}[\lambda (t)],\\*\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda (t){\lambda _2}(t)] & = \mathbb{E}[{\lambda ^2}(t)] + \mathbb{E}[\lambda _2^2(t)] - 2\mathbb{E}[\lambda (t){\lambda _2}(t)] + v\mathbb{E}[{\lambda _2}(t)],\\*\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[\lambda _2^2(t)] & = 2\mathbb{E}[\lambda (t){\lambda _2}(t)] - 2\mathbb{E}[\lambda _2^2(t)] + \mathbb{E}[\lambda (t)].\end{align*}

Numerical results, with $\mathbb{E}[\lambda _2^2(0)] = 0$ , $\mathbb{E}[{\lambda ^2}(0)] = {v^2} = 4$ , $\mathbb{E}[\lambda (0){\lambda _2}(0)] = 0$ , are

\begin{align*}\mathbb{E}[{\lambda ^2}(t)] & = \dfrac{9}{8}{t^2} + \dfrac{{51}}{{16}}t + \dfrac{{137}}{{16}} + \dfrac{7}{4}\,{{\mathrm{e}}^{ - 2t}} + \dfrac{3}{4}t\,{{\mathrm{e}}^{ - 2t}} + \dfrac{7}{{64}}\,{{\mathrm{e}}^{ - 4t}},\\*\mathbb{E}[\lambda _2^2(t)] & = \dfrac{9}{8}{t^2} + \dfrac{{27}}{{16}}t + \dfrac{{41}}{{64}} - \dfrac{3}{4}\,{{\mathrm{e}}^{ - 2t}} - \dfrac{3}{4}t\,{{\mathrm{e}}^{ - 2t}} + \dfrac{7}{{64}}\,{{\mathrm{e}}^{ - 4t}},\\*\mathbb{E}[\lambda (t){\lambda _2}(t)] & = \dfrac{9}{8}{t^2} + \dfrac{{37}}{{16}}t + \dfrac{{47}}{{64}} - \dfrac{5}{8}\,{{\mathrm{e}}^{ - 2t}} - \dfrac{7}{{64}}\,{{\mathrm{e}}^{ - 4t}}.\end{align*}

1. (vi) When $m = 1$ , $n = 1$ , $l = 0$ , we have

   \begin{equation*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N_{HG}}(t)\lambda (t)] = E[{N_{HG}}(t){\lambda _2}(t)] - \mathbb{E}[{N_{HG}}(t)\lambda (t)] + v\mathbb{E}[\lambda (t)] + \mathbb{E}[{\lambda ^2}(t)].\end{equation*}

2. (vii) When $m = 1$ , $n = 0$ , $l = 1$ , we have

   \begin{equation*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N_{HG}}(t){\lambda _2}(t)] = \mathbb{E}[{N_{HG}}(t)\lambda (t)] - \mathbb{E}[{N_{HG}}(t){\lambda _2}(t)] + \mathbb{E}[\lambda (t){\lambda _2}(t)] +\mathbb{E}[\lambda (t)].\end{equation*}

To summarize cases (vi) and (vii), we have a set of two differential equations as follows:

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N_{HG}}(t)\lambda (t)] & = E[{N_{HG}}(t){\lambda _2}(t)] - \mathbb{E}[{N_{HG}}(t)\lambda (t)] + v\mathbb{E}[\lambda (t)] + \mathbb{E}[{\lambda ^2}(t)],\\*\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[{N_{HG}}(t){\lambda _2}(t)] & = \mathbb{E}[{N_{HG}}(t)\lambda (t)] - \mathbb{E}[{N_{HG}}(t){\lambda _2}(t)] + \mathbb{E}[\lambda (t){\lambda _2}(t)] +\mathbb{E}[\lambda (t)].\end{align*}

Numerical results, with $\mathbb{E}[{N_{HG}}(0)\lambda (0)] = 0$ , $\mathbb{E}[{N_{HG}}(0){\lambda _2}(0)] = 0$ , are

\begin{align*}\mathbb{E}[{N_{HG}}(t)\lambda (t)] & = \dfrac{3}{8}{t^3} + \dfrac{{17}}{8}{t^2} + \dfrac{{943}}{{128}}t + \dfrac{{715}}{{256}} - \dfrac{7}{{256}}\,{{\mathrm{e}}^{ - 2t}} + \dfrac{5}{4}t\,{{\mathrm{e}}^{ - 2t}} + \dfrac{3}{{16}}{t^2}\,{{\mathrm{e}}^{ - 2t}} - \dfrac{7}{{128}}\,{{\mathrm{e}}^{ - 4t}},\\*\mathbb{E}[{N_{HG}}(t){\lambda _2}(t)] & = \dfrac{3}{8}{t^3} + \dfrac{{17}}{8}{t^2} + \dfrac{{823}}{{128}}t - \dfrac{{345}}{{256}} + \dfrac{{331}}{{256}}\,{{\mathrm{e}}^{ - 2t}} - \dfrac{{13}}{8}t\,{{\mathrm{e}}^{ - 2t}} - \dfrac{3}{{16}}{t^2}\,{{\mathrm{e}}^{ - 2t}} + \dfrac{7}{{128}}\,{{\mathrm{e}}^{ - 4t}}.\end{align*}

Based on Theorem 1, we have

\begin{align*}\dfrac{{\mathrm{d}} }{{\mathrm{d}} t}\mathbb{E}[N_{HG}^2(t)] & = \sum_{j = 0}^1 \dbinom 2 j \mathbb{E}[N_{HG}^j(t)\lambda (t)]\\*& = \mathbb{E}[\lambda (t)] + 2\mathbb{E}[{N_{HG}}(t)\lambda (t)]\\& = \dfrac{1}{2}\,{{\mathrm{e}}^{ - 2t}} + t + \dfrac{3}{2} \\*&\quad\, + 2\bigg[ {\dfrac{3}{8}{t^3} + \dfrac{{17}}{8}{t^2} + \dfrac{{943}}{{128}}t + \dfrac{{715}}{{256}} - \dfrac{7}{{256}}\,{{\mathrm{e}}^{ - 2t}} {+} \dfrac{5}{4}t\,{{\mathrm{e}}^{ - 2t}} {+} \dfrac{3}{{16}}{t^2}\,{{\mathrm{e}}^{ - 2t}} - \dfrac{7}{{128}}\,{{\mathrm{e}}^{ - 4t}}} \bigg]\\*& = \dfrac{3}{4}{t^3} + \dfrac{{17}}{4}{t^2} + \dfrac{{999}}{{56}}t + \dfrac{{883}}{{128}} + \dfrac{{49}}{{128}}\,{{\mathrm{e}}^{ - 2t}} + \dfrac{5}{2}t\,{{\mathrm{e}}^{ - 2t}} + \dfrac{3}{8}{t^2}\,{{\mathrm{e}}^{ - 2t}} - \dfrac{7}{{56}}\,{{\mathrm{e}}^{ - 4t}}{,}\\\mathbb{E}[N_{HG}^2(t)] & = \dfrac{{ - 1}}{{5376}}(1008{t^2} + 7728t + 4893)\,{{\mathrm{e}}^{ - 2t}} \\*&\quad\, + \dfrac{3}{{16}}{t^4} + \dfrac{{17}}{{12}}{t^3} + \dfrac{{999}}{{112}}{t^2} + \dfrac{{883}}{{128}}t + \dfrac{1}{{32}}\,{{\mathrm{e}}^{ - 4t}} + \dfrac{{225}}{{256}}.\end{align*}

On the other hand, we also have

\begin{equation*}\mathbb{E}[{N_{HG}}(t)] = \int_0^t {\mathbb{E}[\lambda (u)]\,{\mathrm{d}} u} = \dfrac{1}{4} - \dfrac{1}{4}\,{{\mathrm{e}}^{ - 2t}} + \dfrac{1}{2}{t^2} + \dfrac{3}{2}t.\end{equation*}

The curves for $\mathbb{E}[N_{HG}^2(t)]$ and $\mathbb{E}[N_{HG}(t)]$ and a sample path of $\lambda (t)$ are presented in Figures 3 and 4, respectively.

Figure 3:

The curves of $\mathbb{E}[N_{HG}^2(t)]$ and $\mathbb{E}[N_{HG}(t)]$ .

Figure 4:

A sample path of $\lambda (t)$ with $v = 2$ ( ${T_i} = 0.5,2,2.5,4$ $(i = 1,\ldots,4)$ .

Remark 10. The key to successful use of our method is to introduce a stochastic process ${\lambda _2}(t)$ for this non-Markovian Hawkes process with the gamma decay kernel function. In fact, the process $({N_{HG}}(t),\lambda (t),{\lambda _2}(t))$ has the Markov property. In general, how to introduce some stochastic processes that ensure the Markov property holds for non-Markovian Hawkes processes is a difficult question, which is similar to the method of supplementary variables. When using our method, one does not need to check whether the Markov property holds: if the differential equations can be established successfully, then our method is applicable.