Expectation of Hawkes process with exponential kernel

Question

Let N be a point process adapted to a filtration $\mathcal{F}_{t}$. The left-continuous intensity process is defined as \begin{equation} \begin{split} \lambda(t|\mathcal{F}_{t})&=\lim_{h\downarrow0}\mathbb{E}\Big[\frac{N(t+h)-N(t)}{h}\Big]\\ &=\lim_{h\downarrow0}\frac{1}{h}\mathbb{P}[N(t+h)-N(t)>0|\mathcal{F}_{t}] \end{split} \end{equation} with \begin{equation} \Lambda(0,t)=\int^{t}_{0}\lambda(s)ds \end{equation} For an inhomogeneous Poisson point process, we have \begin{equation} \mathbb{E}[N(t)]=\Lambda(0,t) \end{equation} however, for a simple Hawkes process with exponential kernel the intensity is stochastic and defined as follows \begin{equation} \lambda(t)=\lambda_{0}+\sum_{t_{i}<t}\alpha\cdot e^{-\beta(t-t_{i})} \end{equation} with $\{t_{1},t_{2},...,t_{n}\}$ the times of past events before time $t$ and $n\in\mathbb{N}$.

Assuming stationarity implies $\mathbb{E}[\lambda(t)]$ is constant and gives in this case (derivation is left out, but it is confirmed in literature that stationarity holds for $\alpha<\beta$) \begin{equation} \mathbb{E}[\lambda(t)]=\frac{\lambda_{0}}{1-\frac{\alpha}{\beta}} \end{equation} Now, I would say that given the dynamics above, the following holds \begin{equation} \mathbb{E}[N(t)]=\mathbb{E}[\lambda(t)]\cdot t=\frac{\lambda_{0}}{1-\frac{\alpha}{\beta}}\cdot t \end{equation} however, I cannot find any literature on this. Furthermore, plotting $N(t)$ together with $\frac{\lambda_{0}}{1-\frac{\alpha}{\beta}}\cdot t$ does confirm my suspicion (unfortunately I am not allowed to insert images yet). Additionally, as the interarrival times should be exponentially distributed (confirmed by literature) a QQ-plot confirms that the simulation is appropriate (good fit up to the 5th quantile). Finally, I am left with two questions:

1. Why is $\mathbb{E}[\lambda(t)]$ always constant and can't it be infinitely random or some sort?
2. Where does one start with defining $\mathbb{E}[N(t)]$ in case of a simple Hawkes process with exponential kernel?

Thank you

Answer 1

I have just started reading about Hawkes processes and was trying to calculate the same thing. I will answer the second question since it should be obvious why the quantity in the first question cannot be random, or even infinite given your condition on $\alpha$ and $\beta$. We can use the following theorem:

Theorem for any counting process $N(\cdot)$ with conditional intensity function $\lambda (\cdot)$ such that $\mathbb{P} \left ( N(t + \Delta) - N(t) > 1 \right ) = o(\Delta)$ it holds that $\frac{\mathrm{d}}{\mathrm{d}t} \mathbb{E} \left ( N^m(t) \right) = \sum_{j=0}^{m-1} \binom{m}{j} \mathbb{E} \left ( N^j(t) \lambda(t) \right )$

So immediately $\mathbb{E} \left ( N(t) \right ) = \int_0^t \mathbb{E} \left ( \lambda (s) \right ) \mathrm{d}s$. Evaluating $\mathbb{E} \left ( \lambda (t) \right )$ we have that

\begin{equation} \mathbb{E} \left ( \lambda (t) \right ) = \mathbb{E} \left \{ \lambda_0 + \int_0^t \alpha e^{-\beta(t-s)} N \left ( \mathrm{d}s \right ) \right \} = \lambda_0 \int_0^t \alpha e^{-\beta(t-s)} \mathbb{E} \left ( N \left ( \mathrm{d}s \right ) \right ) \end{equation}

For the remaining expectation writing $\mathcal{H}_t$ for the sigma algebra of events occurring up to but not including time $t$ using iterated expectations we have that

\begin{equation} \mathbb{E} \left ( \lambda(s) \right ) = \mathbb{E} \left \{ \frac{\mathbb{E} \left ( N(\mathrm{d}s) \mid \mathcal{H_t} \right )}{\mathrm{d}s} \right \} = \frac{\mathbb{E} \left ( N (\mathrm{d}s) \right )}{\mathrm{d}s} \end{equation}

So $\mathbb{E} \left ( N(\mathrm{d}s) \right ) = \mathbb{E} \left ( \lambda (s) \right ) \mathrm{d}s$ and plugging back into the previous equation gives

\begin{equation} \mathbb{E} \left ( \lambda (t) \right ) = \lambda_0 + \int_0^t \alpha e^{-\beta(t-s)} \mathbb{E} \left ( \lambda (s) \right ) \mathrm{d}s \end{equation}

Using the initial condition $\mathbb{E} \left ( \lambda (0) \right ) = \lambda_0$ we can solve for

\begin{equation} \mathbb{E} \left ( \lambda(t) \right ) = \frac{\lambda_0}{1-\alpha/\beta} + \frac{\lambda_0}{1-\beta/\alpha} e^{-t(\beta-\alpha)} \end{equation}

Finally solving the initial integral we get

\begin{equation} \mathbb{E} \left ( N(t) \right ) = \frac{\lambda_0 t}{1-\alpha/\beta} + \frac{\alpha\lambda_0}{(\alpha - \beta)^2} \left ( e^{-t(\beta - \alpha)} - 1 \right ) \end{equation}