I think I am missing something on an exercise about Hawkes process. Here is the context :
Let $h:\mathbb{R}_{+}\to\mathbb{R}_{+}$ such that $\int_{0}^{\infty}h(t)dt<\infty$ and let $\mu>0$. We consider $N$ and $\bar{N}$ be the associated Hawkes processes having intensity
$$ \lambda(t)=\mu+\sum_{n:T_n<t}h(t-T_n)\quad\text{and}\quad \bar{\lambda}(t)=\frac{\mu}{2}+\sum_{n:\bar{T}_n<t}h(t-\bar{T}_n) $$
The question is to prove that $N_t\geq\bar{N}_t$ a.s.
My question is the following : why the underlying point process $T_i$ is not the same for these two Hawkes processes ?
As it is written, I cannot see how to properly compare theses processes since the function $h$ has not the same point process given $\lambda(t)$ and $\bar{\lambda}(t)$.
Thank you a lot
Here is an attempt where I assume that the the $T_n$ and $\bar{T}_n$ are the same, otherwise I cannot see how to proceed.
Assume that $N_t(\omega)<\bar{N}_t(\omega)$ for some $\omega\in E$ such that $\mathbb{P}(E)>0$.
By construction of the Hawkes process, this means that at fixed $t$, there exists a set of measure positive such that $\lambda(t)(\omega)<\bar{\lambda}(t)(\omega)$ which is not possible since by construction $\lambda(t)>\bar{\lambda}(t)$ (since I assume they have the same jump as I said before starting the proof).
If someone knows how to prove this without this assumption, I would like to have a taste of this please !
Thank you
The intuition behind the Hawkes process is that immigrants arrive according to a rate-$\lambda$ Poisson process (this is for $N$). An immigrant arriving at time $u$ (say) give birth to descendants according to a non-homogeneous Poisson process of rate $h(t-u)$ for $t>u$. $N_t$ counts all arrivals or births that occurred befor time $t$. Now imagine constructing $\tilde N$ by coupling it with $N$ as follows. Each immigrant is colored blue or orange according to the toss of a fair coin (different tosses for each immigrant). The $\tilde N$ process is constructed using only the blue immigrants–its immigration rate will be $\lambda/2$! Because the orange immigrants don't contribute to $\tilde N$, it is clear that, so constructed, $\tilde N_t\le N_t$ for all $t$, almost surely.