Let us derive the simulation. It will be enough to consider one of the two
process $x^{+}_t$ and $x^{-}_t$, that we can denote by $x(t)$. It has the representation $$ x(t) = \sum_{\{k: \, T_k \leq t\}}
e^{- \alpha [t - T_k ]} \eta_k $$ where $T_k$ are the arrivals of the
Poisson Process. For $\Delta > 0$ we have $$ x(t + \Delta) =
\sum_{\{k: \, T_k \leq t\}} e^{- \alpha [t + \Delta - T_k ]} \eta_k
+ \sum_{\{k: \, t < T_k \leq t + \Delta\}} e^{- \alpha
[t + \Delta - T_k ]} \eta_k = e^{-\alpha \Delta} x(t) + \varepsilon . $$
In the second sum denoted as $\varepsilon$, the
arrivals $T_k$ and the jumps $\eta_k$ are independent of the history
$\{x(s);\, s \leq t \}$, and so is $\varepsilon$. So $x(t)$ is
a Markov process. Moreover, conditional on the number of
arrivals on $(t,\,t+ \Delta)$, say $J$, we know
that the r.vs $T_k - t$ have the distribution of the order statistics
of a sample of size $J$ of the uniform distribution on $(0,\,\Delta)$. So
\begin{equation}
\varepsilon = \sum_{j=1}^J e^{-\alpha U_j}
\eta_j' \tag{1}
\end{equation}
where the $\eta_j'$ are i.i.d. from the
distribution of $\eta_k$ and the $U_j$ are uniform on $(0,\,\Delta)$. The sum is understood as $0$ when $J =0$.
Therefore the simulation of $x(t + \Delta)$ conditional on $\{x(s);\,
s \leq t \}$ can be as follows
Draw the number of jumps $J$ on $(t,\,t+ \Delta)$ from the Poisson distribution with mean
$\lambda \Delta$.
Draw $J$ i.i.d r.vs $U_j$ from the uniform on $(0,\,\Delta)$ and $J$
i.i.d r.vs $\eta_j'$ from
the jump distribution.
Take $x(t + \Delta) = e^{-\alpha \Delta} x(t) + \varepsilon$ where
$\varepsilon$ is obtained by (1).
If $t_i^\star$ is an increasing sequence of observation times which is
independent of the process $x(t)$, then we can simulate the sequence
$x(t_i^\star)$ by simulating $x(t_i^\star)$ conditional on the history
at $t = t_{i-1}^\star$ for $i > 0$, as in the code of the
question. The sequence $x(t_i^\star)$ is a Markov chain, and it is time-homogeneous if $t_i^\star-t_{i-1}^\star$ is constant.
Interestingly, it can be shown that when the jump distribution is
exponential with scale $1/ \gamma$, the stationary distribution of
$x(t)$ is gamma with shape $\lambda / \alpha$ and scale $1 / \gamma$.
This distribution can be used to draw the initial value
$x(t_0^\star)$, and we have an example of Markov chain with gamma margin
as in this question.
A classical reference for shot noise is section 5.6 in D.R Cox and
V. Isham Point Processes, Chapman & Hall (1980).
dt
in code. Your question could be much simplified by using only one process $x(t)$ (as in the simulation) and possibly the notation $x(t) = \sum_{\{i: \,S_i \leq t\}} \eta_i$ where the $S_i$ are the Poisson arrivals. Note that $[x(t)]_t$ is continuous-time Markov, and $[x(t_i)]_i$ is discrete-time Markov both with gamma margin. $\endgroup$x[i] = x[i-1]*exp(-alpha*dt) + jumps
I should have also mentioned, in my case, the jump times are latent, which is where the uniform RV appears (using EM algorithm) $\endgroup$sum(numeric(0)) == 0
. $\endgroup$