Proving Hawkes intensity has all the moments. Let $N_t$ be a point process whose intensity follows Hawkes:
\begin{align*}
  \lambda_t = \mu_t + \phi * dN_t,
\end{align*}
where $*$ is a convolution opeartor, $\phi * dN_t = \int_0^t \phi(t-s)dN_s$, and $\mu_t$ is some bounded process which has all the moments,
The question is

Question. For all $k\ge0$, $\mathbb{E} [\lambda_t^k] < \infty$?

My intuition says "yes", but I have no idea how to prove it mathematically because it includes some tricky recursive relation:
\begin{align*}
  \mathbb{E} [\lambda_t^k] &\le C\mathbb{E}[\mu_t^k] + C\mathbb{E}[(\phi * dN_t)^k]  \\
& C\mathbb{E}[\mu_t^k] + C\int_0^t \int_{t_1}^t \cdots \int_{t_{k-1}}^t \phi(t-t_1)\cdots \phi(t-t_k) \mathbb{E} [dN_{t_1} \cdots dN_{t_k} ]
\end{align*}
Any help would be appreciated.
Thanks,
 A: This is for the self-reference.
I will use the induction argument to show this.
For $k=1$, we have $\mathbb{E} \lambda_t = \mu_t + \phi * \mathbb{E} \lambda_t$, and this is in fact the Volterra integral equation. So we have its solution as $\mathbb{E} \lambda_t = \mu_t + \psi * \mu_t$, where $\psi$ is resolvent kernel which satisfies $\psi = \phi + \phi * \psi$. Hence, obviously $\mathbb{E} \lambda_t < C$.
For $k>1$, the inequality $(x+y)^k \le (1+\epsilon)^{k-1} x^k + (1+\epsilon^{-1})^{k-1}y^k$ for any $x,y,\epsilon > 0$ gives
\begin{align*}
\mathbb{E}\lambda_t^k \le (1+\epsilon)^{k-1}\mu_t^k + (1+\epsilon^{-1})^{k-1} \mathbb{E} [ (\phi * dN_t)^k].
\end{align*}
Then by martingale representation of $N_t$, we have $dN_t = \lambda_t\, dt + dM_t$, and
\begin{align*}
  \mathbb{E} [ (\phi * dN_t)^k] \le (1+\delta)^{k-1}\mathbb{E} [ (\phi * \lambda_t)^k] + (1+\delta^{-1})^{k-1} \mathbb{E} [ (\phi * dM_t)^k]
\end{align*}
To bound the first term, we apply Hölder's inequality $(\int \phi \lambda)^k \le (\int \phi \lambda^k)(\int \phi)^{k-1}$:
\begin{align*}
\mathbb{E} [ (\phi * \lambda_t)^k] &\le \mathbb{E} [\phi * \lambda^k_t] \bigg( \int_0^t \phi(t-s)\,ds\bigg)^{k-1} \\
&\le \| \phi \|_1^k \mathbb{E} \lambda_t^k.
\end{align*}
For the second term, by Lemma 2.1.5 in Jacod and Protter's book (2012), we have
\begin{align*}
\mathbb{E} [ (\phi * dM_t)^k] &\le C \mathbb{E} [ \phi^k * \lambda_t] + C\mathbb{E}\big[ (\phi^2 * \lambda_t)^{\frac{k}{2}} \big] \\
&\le C \|\phi\|_k^k \mathbb{E} \lambda_t + C\mathbb{E}[\phi*\lambda^{\frac{k}{2}}_t] \bigg( \ \int_0^t \phi(t-s) ds \bigg)^{\frac{k}{2} - 1} \\
&\le C \|\phi\|_1^k C + C \|\phi\|_1 \mathbb{E}\lambda_t^{\frac{k}{2}} \|\phi\|_1^{\frac{k}{2}-1}
\end{align*}
and this is bound because all the previous value is bounded.
By combining all of these, we have
\begin{align*}
\mathbb{E} \lambda_t^k &\le (1+\epsilon)^{k-1}\mu_t^k + (1+\epsilon^{-1})^{k-1} \mathbb{E} [ (\phi * dN_t)^k] \\
&\le C_{\epsilon} + (1+\epsilon^{-1})^{k-1}(1+\delta)^{k-1}\|\phi\|_1^k \mathbb{E} \lambda_t^k + C_{\epsilon,\delta}.
\end{align*}
Thus, by choosing $\epsilon,\delta$ sufficiently small or large that makes
$$
(1+\epsilon^{-1})^{k-1}(1+\delta)^{k-1}\|\phi\|_1^k < 1,
$$
we have the result.
