where we recall that $\displaystyle j_{n-1}$ is the particle removed at time
$\displaystyle\sigma_{n}$ and $\displaystyle j_{n-1}^{\prime}$ is the index of
the particle duplicated during the branching event at this time (note also
that
$\displaystyle\bar{N}_{\sigma_{n}}=\bar{N}_{\sigma_{n}-}=|\bar{S}_{\sigma_{n}-}|$).
We assume further that $\displaystyle\mathcal{M}$ is constant except at these
jump times.

Then,

$\displaystyle\displaystyle\mu_{t}^{f}-\mu_{0}^{f}=\mathbb{M}_{t}$
$\displaystyle\displaystyle+\mathcal{M}_{t}+\sum_{n=1}^{A_{t}}\frac{1}{\bar{N}_{\rho_{n}}-1}\sum_{i\in\bar{S}_{\rho_{n}-}\setminus\\{i_{n-1}^{\prime}\\}}Q_{T-\rho_{n}}f(X_{\rho_{n}}^{i})$
$\displaystyle\displaystyle-\sum_{n=1}^{B_{t}}\frac{1}{\bar{N}_{\sigma_{n}}+1}\left(\sum_{i\in\bar{S}_{\sigma_{n}-}}Q_{T-\sigma_{n}}f(X_{\sigma_{n}}^{i})+Q_{T-\sigma_{n}}f(X_{\sigma_{n}}^{j^{\prime}_{n-1}})\right).$
(43)

For each $\displaystyle n\geq 1$, we define $\displaystyle\mu_{\tau_{n}-}^{f}$
and $\displaystyle\nu_{\tau_{n}-}^{f}$ by

$\displaystyle\mu_{\tau_{n}-}^{f}=\sum_{i\in
S_{n-1}}f(X^{i}_{\tau_{n}})-{\mathbf{1}}_{\tau_{n}\in\mathcal{K}}f(X^{i^{\prime}_{n-1}}_{\tau_{n}})+\mathbf{{\mathbf{1}}}_{\tau_{n}\in\mathcal{B}}f(X^{j^{\prime}_{n-1}}_{\tau_{n}})\text{
and }\nu_{\tau_{n}-}=\Pi_{\tau_{n-1}}^{A}\Pi_{\tau_{n-1}}^{B}\mu_{\tau_{n}-}$

where $\displaystyle\mathcal{K}$ is the set of killing times and
$\displaystyle\mathcal{B}$ is the set of branching times. Informally,
$\displaystyle\mu_{\tau_{n}-}$ and $\displaystyle\nu_{\tau_{n}-}$ represent
the state of the particle system at time $\displaystyle\tau_{n}$ before the
resampling or selection eventually occurring at time $\displaystyle\tau_{n}$.
Note that, at any time $\displaystyle\tau_{n}$ that is not a resampling or a
selection time, we have $\displaystyle\mu_{\tau_{n}-}=\mu_{\tau_{n}}$ and
$\displaystyle\nu_{\tau_{n}-}=\nu_{\tau_{n}}$. Hence, denoting by
$\displaystyle(\theta_{n})_{n}$ the sequence of events from the families
$\displaystyle\rho$ or $\displaystyle\sigma$, we obtain for all $\displaystyle
t\geq 0$,

$\displaystyle\displaystyle\nu_{t}^{f}-\nu_{0}^{f}$
$\displaystyle\displaystyle=\nu_{t}-\nu_{\tau_{C_{t}}}+\sum_{n=1}^{C_{t}}\left(\nu_{\tau_{n}}-\nu_{\tau_{n}-}\right)+\sum_{n=1}^{C_{t}}\left(\nu_{\tau_{n}-}-\nu_{\tau_{n-1}}\right)$
$\displaystyle\displaystyle=\Pi_{t}^{A}\Pi_{t}^{B}(\mu_{t}-\mu_{\tau_{C_{t}}})+\sum_{n=1}^{A_{t}}\nu_{\rho_{n}}-\nu_{\rho_{n}-}+\sum_{n=1}^{B_{t}}\nu_{\sigma_{n}}-\nu_{\sigma_{n}-}+\sum_{n=1}^{C_{t}}\nu_{\tau_{n}-}-\nu_{\tau_{n-1}}.$

At resampling times $\displaystyle\rho_{n}$, we have
$\displaystyle\Pi_{\rho_{n}}^{B}-\Pi_{\rho_{n}-}^{B}=0$ and hence,

$\displaystyle\displaystyle\nu_{\rho_{n}}^{f}-\nu_{\rho_{n}-}^{f}$
$\displaystyle\displaystyle=\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}\mu_{\rho_{n}}^{f}-\Pi_{\rho_{n-1}}^{A}\Pi_{\rho_{n-1}}^{B}\mu_{\rho_{n}-}^{f}$
$\displaystyle\displaystyle=\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}(\mu_{\rho_{n}}^{f}-\mu_{\rho_{n}-}^{f})+\Pi_{\rho_{n}}^{B}\mu_{\rho_{n}-}^{f}\left(\Pi_{\rho_{n}}^{A}-\Pi_{\rho_{n-1}}^{A}\right)$
$\displaystyle\displaystyle=\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}\,Q_{T-\rho_{n}}f(X_{\rho_{n}}^{i_{n-1}})+\Pi_{\rho_{n}}^{B}\mu_{\rho_{n}-}^{f}\left(\Pi_{\rho_{n}}^{A}-\Pi_{\rho_{n-1}}^{A}\right)$
(44)

For the increment of $\displaystyle\Pi^{A}$, we note that

$\Pi_{\rho_{n}}^{A}-\Pi_{\rho_{n-1}}^{A}=\Pi_{\rho_{n}}^{A}-\left(\frac{\bar{N}_{\rho_{n}}}{\bar{N}_{\rho_{n}}-1}\right)\Pi_{\rho_{n}}^{A}=-\frac{1}{\bar{N}_{\rho_{n}}-1}\Pi_{\rho_{n}}^{A}.$

Substituting this into (44) and using (41), we obtain

$\displaystyle\displaystyle\nu_{\rho_{n}}^{f}-\nu_{\rho_{n}-}^{f}$
$\displaystyle\displaystyle=\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}\left(Q_{T-\rho_{n}}f(X_{\rho_{n}}^{i_{n-1}})-\frac{1}{\bar{N}_{\rho_{n}}-1}\mu_{\rho_{n}-}^{f}\right)$
$\displaystyle\displaystyle=\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}\left(\mathcal{M}_{\rho_{n}}-\mathcal{M}_{\rho_{n}-}\right).$
(45)

Similarly, for $\displaystyle n\in\\{1,\dots,B_{t}\\}$,

$\displaystyle\displaystyle\nu_{\sigma_{n}}^{f}-\nu_{\sigma_{n}-}^{f}=\Pi_{\sigma_{n}}^{A}\Pi_{\sigma_{n}}^{B}(\mu_{\sigma_{n}}^{f}-\mu_{\sigma_{n}-}^{f})+\Pi_{\sigma_{n}}^{A}\mu_{\sigma_{n}-}^{f}\left(\Pi_{\sigma_{n}}^{B}-\Pi_{\sigma_{n}-}^{B}\right).$
(46)

In this case, we have

$\Pi_{\sigma_{n}}^{B}-\Pi_{\sigma_{n}-}^{B}=\Pi_{\sigma_{n}}^{B}-\left(\frac{\bar{N}_{\sigma_{n}}}{\bar{N}_{\sigma_{n}}+1}\right)\Pi_{\sigma_{n}}^{B}=\frac{1}{\bar{N}_{\sigma_{n}}+1}\Pi_{\sigma_{n}}^{B},$
(47)

and

$\mu_{\sigma_{n}}^{f}-\mu_{\sigma_{n}-}^{f}=-Q_{T-\sigma_{n}}f(X_{\sigma_{n}}^{j_{n-1}}),$
(48)

where $\displaystyle j_{n-1}$ is the index of the particle removed from the
system at time $\displaystyle\sigma_{n}$. Again, substituting these equalities
back into (46) and using (42) gives,

$\displaystyle\displaystyle\nu_{\sigma_{n}}^{f}-\nu_{\sigma_{n}-}^{f}$
$\displaystyle\displaystyle=\Pi_{\sigma_{n}}^{A}\Pi_{\sigma_{n}}^{B}\left(-Q_{T-\sigma_{n}}f(X_{\sigma_{n}}^{j_{n-1}})+\frac{1}{\bar{N}_{\sigma_{n}}+1}\mu_{\sigma_{n}-}^{f}\right)$
$\displaystyle\displaystyle=\Pi_{\sigma_{n}}^{A}\Pi_{\sigma_{n}}^{B}(\mathcal{M}_{\sigma_{n}}-\mathcal{M}_{\sigma_{n}-}).$
(49)

Finally, we have
$\displaystyle\nu_{\tau_{n}-}-\nu_{\tau_{n-1}}=\Pi^{A}_{\tau_{n-1}}\Pi^{B}_{\tau_{n-1}}(\mathbb{M}_{\tau_{n}}-\mathbb{M}_{\tau_{n-1}})$
for all $\displaystyle n\geq 1$, hence

$\displaystyle\displaystyle\nu_{t}^{f}-\nu_{0}^{f}$
$\displaystyle\displaystyle=\Pi_{t}^{A}\Pi_{t}^{B}(\mathbb{M}_{t}-\mathbb{M}_{\tau_{C_{t}}})+\sum_{n=1}^{A_{t}}\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}\left(\mathcal{M}_{\rho_{n}}-\mathcal{M}_{\rho_{n}-}\right)+\sum_{n=1}^{B_{t}}\Pi_{\sigma_{n}}^{A}\Pi_{\sigma_{n}}^{B}(\mathcal{M}_{\sigma_{n}}-\mathcal{M}_{\sigma_{n}-})$
$\displaystyle\displaystyle\phantom{\Pi_{t}^{A}\Pi_{t}^{B}(\mu_{t}-\mu_{\tau_{C_{t}}}}\phantom{+\sum_{n=1}^{B_{t}}\Pi^{A}_{\sigma_{n-1}}\Pi^{B}_{\rho_{n-1}}(\mathbb{M}_{\rho_{n}}-\mathbb{M}_{\rho_{n-1}})}+\sum_{n=1}^{C_{t}}\Pi^{A}_{\tau_{n-1}}\Pi^{B}_{\tau_{n-1}}(\mathbb{M}_{\tau_{n}}-\mathbb{M}_{\tau_{n-1}}),$

which entails (35) and thus concludes Step 3.

Step 4. Control of the $\displaystyle L^{2}$ norm of the martingale
increments. In this section, we assume without loss of generality (taking
$\displaystyle e^{-\|b\|_{\infty}T}f/\|f\|_{\infty}$ instead of $\displaystyle
f$) that

$\displaystyle\sup_{t\in[0,T]}\|Q_{t}f\|_{\infty}\leq 1.$

We now show that the $\displaystyle L^{2}$ norm of each of these quantities on
the righthand side of (35) is finite. To this end, let us first consider
$\displaystyle\mathcal{M}_{\rho_{n}}-\mathcal{M}_{\rho_{n}-}$. We have,
according to (41),

$\displaystyle\displaystyle|\mathcal{M}_{\rho_{n}}-\mathcal{M}_{\rho_{n}-}|$
$\displaystyle\displaystyle=\left|Q_{T-\rho_{n}}f(X_{\rho_{n}}^{i_{n-1}})-\frac{1}{\bar{N}_{\rho_{n}}-1}\sum_{i\in\bar{S}_{\rho_{n}-}\setminus\\{i^{\prime}_{n-1}\\}}Q_{T-\rho_{n}}f(X_{\rho_{n}}^{i})\right|\leq
2.$ (50)

Then, setting $\displaystyle R_{T}=\max_{t\in[0,T]}N_{t}$, we obtain

$\displaystyle\displaystyle\mathbb{E}$
$\displaystyle\displaystyle\left[\left(\sum_{n=1}^{A_{T}}\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}\left(\mathcal{M}_{\rho_{n}}-\mathcal{M}_{\rho_{n}-}\right)\right)^{2}\right]$
$\displaystyle\displaystyle\quad\quad=\mathbb{E}\left[\left(\frac{\bar{N}_{\rho_{n}}-1}{\bar{N}_{\rho_{n}}}\,\sum_{n=1}^{A_{T}}\Pi_{\rho_{n}-}^{A}\Pi_{\rho_{n}-}^{B}\left(\mathcal{M}_{\rho_{n}}-\mathcal{M}_{\rho_{n}-}\right)\right)^{2}\right]$
$\displaystyle\displaystyle\quad\quad\leq
4\,\mathbb{E}\left[\sum_{n=1}^{A_{T}}\left(\frac{R_{T}-1}{R_{T}}\right)^{2A_{\rho_{n}}}(\Pi_{\rho_{n}-}^{B})^{2}\right]$
$\displaystyle\displaystyle\quad\quad\leq
4\,\mathbb{E}\left[(\Pi_{T}^{B})^{2}\,\sum_{n=1}^{A_{T}}\left(\frac{R_{T}-1}{R_{T}}\right)^{2n}\right].$
(51)

To control the sum term, since $\displaystyle 1/R_{T}\leq\frac{1}{2}$, we have

$\displaystyle\displaystyle\sum_{n=1}^{A_{T}}\left(1-\frac{1}{R_{T}}\right)^{2n}$
$\displaystyle\displaystyle\leq\sum_{n=1}^{+\infty}\left(1-\frac{1}{R_{T}}\right)^{2n}=\frac{(1-\frac{1}{R_{T}})^{2}}{1-(1-\frac{1}{R_{T}})^{2}}$
$\displaystyle\displaystyle=R_{T}\frac{1-2/R_{T}+R_{T}^{-2}}{2-\frac{1}{R_{T}}}\leq
R_{T}/2\leq(N_{0}+\beta_{T})/2,$

where $\displaystyle\beta_{T}$ denotes the total number of branching events
before time $\displaystyle T$. We now make use of Lemma 23, stated and proved
at the end of this section, which entails that

$\displaystyle\displaystyle\mathbb{E}\left[(\Pi_{T}^{B})^{2}(N_{0}+\beta_{T})/2\right]$
$\displaystyle\displaystyle\leq N_{0}\exp(c\|b\|_{\infty}T),$ (52)

for some (explicit) constant $\displaystyle c>0$. Substituting these estimates
into (51), we obtain

$\displaystyle\displaystyle\mathbb{E}\left[\left(\sum_{n=1}^{A_{T}}\Pi_{\rho_{n}}^{A}\Pi_{\rho_{n}}^{B}\left(\mathcal{M}_{\rho_{n}}-\mathcal{M}_{\rho_{n}-}\right)\right)^{2}\right]\leq
4\,N_{0}\,\exp\left(c\|b\|_{\infty}T\right).$ (53)

Let us now consider $\displaystyle\Delta_{\sigma_{n}}\mathcal{M}$. Using (42),
we have

$\displaystyle\displaystyle|\mathcal{M}_{\sigma_{n}}-\mathcal{M}_{\sigma_{n-}}|$
$\displaystyle\displaystyle\leq 2.$

Then

$\displaystyle\displaystyle\mathbb{E}$
$\displaystyle\displaystyle\left[\left(\sum_{n=1}^{B_{T}}\Pi_{\sigma_{n}}^{A}\Pi_{\sigma_{n}}^{B}(\mathcal{M}_{\sigma_{n}}-\mathcal{M}_{\sigma_{n-}})\right)^{2}\right]$
(54)
$\displaystyle\displaystyle\quad\quad\leq\mathbb{E}\left[\left(\sum_{n=1}^{B_{T}}\Pi_{\sigma_{n-}}^{A}\Pi_{\sigma_{n-}}^{B}\,\frac{N_{\sigma_{n-}}+1}{N_{\sigma_{n-}}}(\mathcal{M}_{\sigma_{n}}-\mathcal{M}_{\sigma_{n-}})\right)^{2}\right]$
$\displaystyle\displaystyle\quad\quad\leq
4\,\mathbb{E}\left[\sum_{n=1}^{B_{T}}\left(\Pi^{B}_{\sigma_{n}}\right)^{2}\right]\leq
4\,\mathbb{E}\left[\beta_{T}\,\left(\Pi^{B}_{T}\right)^{2}\right].$ (55)

Using Lemma 23 stated and proved at the end of this section, we deduce that

$\displaystyle\displaystyle\mathbb{E}\left[\left(\sum_{n=1}^{B_{T}}\Pi_{\sigma_{n}}^{A}\Pi_{\sigma_{n}}^{B}(\mathcal{M}_{\sigma_{n}}-\mathcal{M}_{\sigma_{n-}})\right)^{2}\right]\leq
4\,N_{0}\exp(c\|b\|_{\infty}T),$ (56)

where $\displaystyle c$ can be chosen to be the same as in (53).

It remains to control the increments of $\displaystyle\mathbb{M}$. For all
$\displaystyle n\geq 1$, let $\displaystyle E_{n}\subset\mathbb{N}$ be the set
of indices of particles defined by

$\displaystyle E_{n}=\\{i\in\mathbb{N}\text{ s.t. the $\displaystyle i^{th}$
particle is branching, killed, born, removed or duplicated at time
$\displaystyle\tau_{n}$}\\}.$

Observe that $\displaystyle|E_{n}|\leq 4$ for all $\displaystyle n\geq 1$. We
also define, for all $\displaystyle i\in\mathbb{N}$, the sequence of times
$\displaystyle(\tau_{n}^{i})_{n\geq 0}$ by $\displaystyle\tau^{i}_{0}=0$ and

$\displaystyle\displaystyle\tau_{n+1}^{i}=\min\\{\tau_{k}>\tau^{i}_{n},\text{
with }k\geq 0\text{ and }i\in E_{k}\\}.$

We write

$\displaystyle\displaystyle\mathbb{M}_{T}=\sum_{i\in\mathbb{N}}\sum_{n\geq
1}\mathbb{M}^{i}_{\tau_{n}^{i}\wedge T}-\mathbb{M}^{i}_{\tau_{n-1}^{i}\wedge
T}.$

Since all the martingales increments are orthogonal, we have, for all
$\displaystyle i\in\mathbb{N}$,

$\Delta:=\mathbb{E}\left[\left(\Pi_{T}^{A}\Pi_{T}^{B}(\mathbb{M}_{T}-\mathbb{M}_{\tau_{C_{T}}})+\sum_{n=1}^{C_{T}}\Pi^{A}_{\tau_{n-1}}\Pi^{B}_{\tau_{n-1}}(\mathbb{M}_{\tau_{n}}-\mathbb{M}_{\tau_{n-1}})\right)^{2}\right]\\\
\begin{aligned} &=\mathbb{E}\left[\left(\sum_{i\in\mathbb{N}}\sum_{n\geq
1}\Pi_{\tau_{n-1}^{i}}^{A}\Pi_{\tau_{n-1}^{i}}^{B}(\mathbb{M}^{i}_{\tau_{n}^{i}\wedge
T}-\mathbb{M}^{i}_{\tau_{n-1}^{i}\wedge T})\right)^{2}\right]\\\
&=\mathbb{E}\left[\sum_{i\in\mathbb{N}}\sum_{n\geq
1}\left(\Pi_{\tau_{n-1}^{i}}^{A}\Pi_{\tau_{n-1}^{i}}^{B}\right)^{2}\left(\mathbb{M}^{i}_{\tau_{n}^{i}\wedge
T}-\mathbb{M}^{i}_{\tau_{n-1}^{i}\wedge T}\right)^{2}\right].\end{aligned}$

But the increments $\displaystyle\mathbb{M}^{i}_{\tau_{n+1}^{i}\wedge
T}-\mathbb{M}^{i}_{\tau_{n}^{i}\wedge T}$ are almost surely bounded by
$\displaystyle 3$, so that

$\displaystyle\displaystyle\Delta$ $\displaystyle\displaystyle\leq
9\,\mathbb{E}\left[\sum_{i\in\mathbb{N}}\sum_{n\geq
1}\left(\Pi_{\tau_{n-1}^{i}}^{A}\Pi_{\tau_{n-1}^{i}}^{B}\right)^{2}{\mathbf{1}}_{\tau_{n-1}^{i}<T}\right]$
$\displaystyle\displaystyle\leq 9\,\mathbb{E}\left[\sum_{n\geq 1}\sum_{i\in
E_{n}}\left(\Pi_{\tau_{n-1}}^{A}\Pi_{\tau_{n-1}}^{B}\right)^{2}{\mathbf{1}}_{\tau_{n-1}<T}\right]$
$\displaystyle\displaystyle\leq 36\,\mathbb{E}\left[\sum_{n\geq
1}\left(\Pi_{\tau_{n-1}}^{A}\Pi_{\tau_{n-1}}^{B}\right)^{2}{\mathbf{1}}_{\tau_{n-1}<T}\right]$

Hence, denoting by $\displaystyle\beta_{T}$ (resp. $\displaystyle\kappa_{T}$)
the number of branching (resp. killing without resampling) events occuring
before time $\displaystyle T\geq 0$ (so that $\displaystyle
C_{T}=\beta_{T}+\kappa_{T}+A_{T}$), we have

$\displaystyle\displaystyle\Delta\leq
36\,\mathbb{E}\left[\left(\Pi_{T}^{B}\right)^{2}(1+\beta_{T}+\kappa_{T}+\sum_{n=1}^{A_{T}}\left(\Pi^{A}_{\rho_{n-1}}\right)^{2})\right]$
(57)

We already computed that, almost surely,

$\displaystyle\sum_{n=1}^{A_{T}}\left(\Pi^{A}_{\rho_{n-1}}\right)^{2}\leq\sum_{k=0}^{+\infty}\left(\frac{R_{T}-1}{R_{T}}\right)^{2k}\leq
R_{T}/2\leq(N_{0}+\beta_{T})/2.$

Now, observe that the number of particles in the system as time $\displaystyle
T$ is upper bounded by $\displaystyle N_{0}+\beta_{T}-\kappa_{T}$ and lower
bounded by $\displaystyle 0$, so that $\displaystyle\kappa_{T}\leq
N_{0}+\beta_{T}$, and hence

$\mathbb{E}\left[\left(\Pi_{T}^{B}\right)^{2}(1+\beta_{T}+\kappa_{T}+\sum_{n=1}^{A_{T}}\left(\Pi^{A}_{\rho_{n-1}}\right)^{2})\right]\leq\mathbb{E}\left[\left(\Pi_{T}^{B}\right)^{2}(\beta_{T}+3N_{0})\right]\leq
4N_{0}\exp(c\|b\|_{\infty}T),$

where we used, as above, Lemma 23. We thus obtained, using the last inequality
and (57), that

$\displaystyle\displaystyle\Delta\leq
144\,N_{0}\,\exp\left(c\|b\|_{\infty}T\right).$

Using (35), (53), (56) and the last inequality, we deduce that

$\displaystyle\displaystyle\mathbb{E}\left[\left(\nu_{T}^{f}-\nu_{0}^{f}\right)^{2}\right]\leq
152N_{0}\exp\left(c\|b\|_{\infty}T\right).$

Without assuming that $\displaystyle\sup_{t\in[0,T]}\|Q_{t}f\|_{\infty}\leq
1$, we thus obtain

$\displaystyle\displaystyle\mathbb{E}\left[\left(\nu_{T}^{f}-\nu_{0}^{f}\right)^{2}\right]\leq
152N_{0}\exp\left(c\|b\|_{\infty}T\right)\left(\sup_{t\in[0,T]}\|Q_{t}f\|_{\infty}\right)^{2}.$
(58)

Step 5. Conclusion.

Recall the occupation measure $\displaystyle m_{t}$ defined in (5) for all
$\displaystyle t\geq 0$. With this notation, the conclusion of Step 4 reads

$\displaystyle\mathbb{E}\left[\left(m_{0}Q_{T}f-\Pi_{T}^{A}\Pi_{T}^{B}m_{T}f\right)^{2}\right]\leq
152N_{0}\exp\left(c\|b\|_{\infty}T\right)\left(\sup_{t\in[0,T]}\|Q_{t}f\|_{\infty}\right)^{2}.$

We deduce that

$\displaystyle\displaystyle\left\|m_{0}Q_{T}\mathbf{1}_{E}\,\frac{m_{T}f}{m_{T}\mathbf{1}_{E}}\mathbf{1}_{m_{T}\neq
0}-m_{0}Q_{T}f\right\|_{2}$ $\displaystyle\displaystyle\hskip
56.9055pt\leq\left\|(m_{0}Q_{T}\mathbf{1}_{E}-\Pi_{T}^{A}\Pi_{T}^{B}m_{T}\mathbf{1}_{E})\,\frac{m_{T}f}{m_{T}\mathbf{1}_{E}}\mathbf{1}_{m_{T}\neq
0}\right\|_{2}+\left\|\Pi_{T}^{A}\Pi_{T}^{B}m_{T}f-m_{0}Q_{T}f\right\|_{2}$
$\displaystyle\displaystyle\hskip 56.9055pt\leq
2\,\sqrt{152N_{0}}\exp\left(c\|b\|_{\infty}T/2\right)\|f\|_{\infty}\sup_{t\in[0,T]}\|Q_{t}{\mathbf{1}}_{E}\|_{\infty}.$

We conclude that

$\displaystyle\displaystyle\left\|\frac{m_{0}Q_{T}f}{m_{0}Q_{T}\mathbf{1}_{E}}-\frac{m_{T}f}{m_{T}(\mathbf{1}_{E})}\mathbf{1}_{m_{T}\neq
0}\right\|_{2}$
$\displaystyle\displaystyle\leq\frac{2\,\sqrt{152}\exp\left(c\|b\|_{\infty}T/2\right)\|f\|_{\infty}\sup_{t\in[0,T]}\|Q_{t}{\mathbf{1}}_{E}\|_{\infty}}{m_{0}Q_{T}\mathbf{1}_{E}/N_{0}}\,\frac{1}{\sqrt{N_{0}}}.$

Observing that
$\displaystyle\sup_{t\in[0,T]}\|Q_{t}{\mathbf{1}}_{E}\|_{\infty}\leq
e^{\|b\|_{\infty}T}$, this concludes the proof of Theorem 6. ∎

###### Proof of Lemma 23.

We use the same notations as in the proof of Theorem 6.

Fix $\displaystyle C>c_{\ell}\|b\|_{\infty}$ and $\displaystyle\delta>0$ such
that $\displaystyle C-\delta C^{2}2^{\ell}\geq c_{\ell}\|b\|_{\infty}$. We
have

$\displaystyle\displaystyle\mathbb{E}\left(\left(\Pi^{B}_{\tau_{1}\wedge\delta}\right)^{\ell}e^{-C(\tau_{1}\wedge\delta)}\right)$
$\displaystyle\displaystyle=\mathbb{E}\left(\left(\frac{N_{0}+1}{N_{0}}\right)^{\ell}\mathbf{1}_{\sigma_{1}=\tau_{1}\wedge\delta}e^{-C(\tau_{1}\wedge\delta)}\right)+\mathbb{E}\left(\mathbf{1}_{\sigma_{1}>\tau_{1}\wedge\delta}e^{-C(\tau_{1}\wedge\delta)}\right)$
$\displaystyle\displaystyle=\left(\frac{N_{0}+1}{N_{0}}\right)^{\ell}\mathbb{E}\left(e^{-C(\tau_{1}\wedge\delta)}\right)+\left(1-\left(\frac{N_{0}+1}{N_{0}}\right)^{\ell}\right)\,\mathbb{E}\left(\mathbf{1}_{\sigma_{1}>\tau_{1}\wedge\delta}e^{-C(\tau_{1}\wedge\delta)}\right).$

On the one hand, we have

$\displaystyle\displaystyle\mathbb{E}\left(e^{-C(\tau_{1}\wedge\delta)}\right)\leq
1-e^{-C\delta}C\mathbb{E}(\tau_{1}\wedge\delta)\leq 1-(C-\delta
C^{2})\mathbb{E}(\tau_{1}\wedge\delta),$

and, on the other hand,

$\displaystyle\displaystyle\mathbb{E}\left(\mathbf{1}_{\sigma_{1}>\tau_{1}\wedge\delta}e^{-C(\tau_{1}\wedge\delta)}\right)\geq\mathbb{E}\left(e^{-(\tau_{1}\wedge\delta)(N_{0}\|b\|_{\infty}+C)}\right)\geq
1-(N_{0}\|b\|_{\infty}+C)\mathbb{E}\left(\tau_{1}\wedge\delta\right).$

Setting, for the sake of readability,
$\displaystyle\alpha=\mathbb{E}(\tau_{1}\wedge\delta)$ and
$\displaystyle\beta=\left(\frac{N_{0}+1}{N_{0}}\right)^{\ell}$, we deduce that

$\displaystyle\displaystyle\mathbb{E}\left(\left(\Pi^{B}_{\tau_{1}\wedge\delta}\right)^{\ell}e^{-C(\tau_{1}\wedge\delta)}\right)$
$\displaystyle\displaystyle\leq 1-\alpha\left(C-\delta
C^{2}\beta+N_{0}\|b\|_{\infty}(1-\beta)\right)$
$\displaystyle\displaystyle\leq 1-\alpha\left(C-\delta
C^{2}2^{\ell}-\|b\|_{\infty}c_{\ell}\right)\leq 1,$ (59)

where we used the fact that $\displaystyle\beta\leq 2^{\ell}$ and
$\displaystyle\beta-1\leq c_{\ell}/N_{0}$.

We define the sequence of stopping times $\displaystyle(\theta_{n})_{n\geq 0}$
by $\displaystyle\theta_{0}=0$ and

$\displaystyle\displaystyle\theta_{n+1}:=\inf\\{((n+1)\delta)\wedge\tau_{k},\
k\geq 1,\tau_{k}>\theta_{n}\\}.$

Then

$\mathbb{E}\left(\left(\Pi_{T\wedge\theta_{n+1}}^{B}\right)^{\ell}e^{-C(T\wedge\theta_{n+1})}\mid\mathcal{F}_{T\wedge\theta_{n}}\right)=\left(\Pi_{T\wedge\theta_{n}}^{B}\right)^{\ell}\,e^{-C(T\wedge\theta_{n})}\\\
\times\mathbb{E}_{\mathbb{X}_{T\wedge\theta_{n}}}\left(\left(\Pi_{(T-u)\wedge\theta_{1}}^{B}\right)^{\ell}e^{-C((T-u)\wedge\theta_{1})}\right)_{|u=T\wedge\theta_{n}}.$

Since $\displaystyle T-u\leq\delta$ almost surely in the right hand term, we
deduce from (59) that

$\displaystyle\displaystyle\mathbb{E}\left(\left(\Pi_{T\wedge\theta_{n+1}}^{B}\right)^{\ell}e^{-C(T\wedge\theta_{n+1})}\mid\mathcal{F}_{T\wedge\theta_{n}}\right)$
$\displaystyle\displaystyle\leq\left(\Pi_{T\wedge\theta_{n}}^{B}\right)^{\ell}\,e^{-C(T\wedge\theta_{n})}.$

In particular, using Fatou’s Lemma, we deduce, letting $\displaystyle
n\to+\infty$, that

$\displaystyle\displaystyle\mathbb{E}\left(\left(\Pi_{T}^{B}\right)^{\ell}\,e^{-CT}\right)\leq
1.$

Since this is true for any $\displaystyle C>c_{\ell}\|b\|_{\infty}$, this
concludes the proof of the first part of the lemma.

For the second part, we observe that $\displaystyle\beta_{T}$ is
stochastically dominated by a continuous time counting process
$\displaystyle(Z_{t})_{t\geq 0}$ on $\displaystyle\mathbb{N}$ starting from
$\displaystyle N_{0}$ and with increment rate $\displaystyle\|b\|_{\infty}n$
from state $\displaystyle n$ to state $\displaystyle n+1$. In particular,
denoting by $\displaystyle L$ the infinitesimal generator of this process and
$\displaystyle V(n)=n^{2}$ for all $\displaystyle n\geq 1$, we obtain

$\displaystyle
LV(n)=\|b\|_{\infty}n((n+1)^{2}-n^{2})=\|b\|_{\infty}n(2n+1)\leq
3\|b\|_{\infty}V(n),$

and deduce that

$\displaystyle\mathbb{E}(\beta_{T}^{2})\leq\mathbb{E}(Z_{T}^{2})\leq
e^{3\|b\|_{\infty}T}Z_{0}^{2}=e^{3\|b\|_{\infty}T}N_{0}^{2}.$

Hence, we obtain

$\displaystyle\displaystyle\mathbb{E}\left(\beta_{T}\left(\Pi_{T}^{B}\right)^{2}\right)\leq\sqrt{\mathbb{E}(\beta_{T}^{2})}\sqrt{\mathbb{E}\left(\left(\Pi_{T}^{B}\right)^{4}\right)}\leq
N_{0}\exp((3+c_{4})\|b\|_{\infty}T/2).$

This concludes the proof of the second part of the Lemma. ∎

######  Remark 24.

Our proof adapts to the case where $\displaystyle\tau_{\infty}<+\infty$ with
positive probability (that is when Assumption 2 does not holds true), using
the following strategy. In this situation, one shows using the same approach
that, for all $\displaystyle n\geq 0$,

$\displaystyle m_{0}Q_{T}f=\mathbb{E}\left(\Pi^{A}_{\tau_{n}\wedge
T}\,\Pi^{B}_{\tau_{n}\wedge T}m_{\tau_{n}\wedge T}Q_{T-\tau_{n}\wedge
T}f\right).$

Since the branching rate is uniformly bounded and since
$\displaystyle\Pi^{A}_{\tau_{n}\wedge T}$ is bounded by $\displaystyle 1$, one
checks that $\displaystyle\Pi^{A}_{\tau_{n}\wedge T}\,\Pi^{B}_{\tau_{n}\wedge
T}m_{\tau_{n}\wedge T}Q_{T-\tau_{n}\wedge T}f$ can be uniformly bounded (in
$\displaystyle n$) by an integrable random variable. Since in addition
$\displaystyle\Pi^{A}_{\tau_{n}\wedge T}\,\Pi^{B}_{\tau_{n}\wedge
T}m_{\tau_{n}\wedge T}Q_{T-\tau_{n}\wedge T}f\to 0$ on the event
$\displaystyle\tau_{\infty}\leq T$ (since this events necessarily corresponds
to the accumulation of resampling events) and
$\displaystyle\Pi^{A}_{\tau_{n}\wedge T}\,\Pi^{B}_{\tau_{n}\wedge
T}m_{\tau_{n}\wedge T}Q_{T-\tau_{n}\wedge
T}f\to\Pi^{A}_{T}\,\Pi^{B}_{T}m_{T}f$ on the complementary event, one deduces
from the dominated convergence theorem that

$\displaystyle
m_{0}Q_{T}f=\mathbb{E}\left(\Pi^{A}_{T}\,\Pi^{B}_{T}m_{T}f\mathbf{1}_{T<\tau_{\infty}}\right).$

The rest of the proof (see Steps 4 and 5) then proceeds almost identically.

######  Remark 25.

We comment on some direct and natural continuations of the proof, which are
easily derived from the fact that the proof is based on the control of
martingale increments.

First, if the martingale $\displaystyle\nu_{t}^{f}-\nu_{0}^{f}$ is càdlàg,
then Doob’s maximal inequality and (58) entail that

$\displaystyle\mathbb{E}\left[\sup_{t\in[0,T]}\left(\nu_{T}^{f}-\nu_{0}^{f}\right)^{2}\right]\leq
c_{T}\,\|f\|_{\infty}^{2}\sup_{t\in[0,T]}\|Q_{t}\mathbf{1}_{E}\|_{\infty}^{2}N_{0}$

for some (explicit) constant $\displaystyle c_{T}$. Then the conclusion reads

$\displaystyle\displaystyle\left\|\sup_{t\in{[0,T]}}\left|\frac{m_{0}Q_{t}f}{m_{0}Q_{t}\mathbf{1}_{E}}-\frac{m_{t}f}{m_{t}(\mathbf{1}_{E})}\mathbf{1}_{m_{t}\neq
0}\right|\right\|_{2}$
$\displaystyle\displaystyle\leq\frac{c_{T}\,\|f\|_{\infty}}{m_{0}Q_{T}\mathbf{1}_{E}/N_{0}}\,\frac{1}{\sqrt{N_{0}}},$

Second, if the killing rate is bounded, then the statement of Lemma 23 holds
true with $\displaystyle\beta_{T}$ replaced by the total number of
killing/branching events $\displaystyle C_{T}$, and with different constants.
Then, for any $\displaystyle p\geq 2$, using the same approach as in Step 4 of
the proof, but using the Burkholder-Davis-Gundy inequality instead, one
deduces similarly that

$\displaystyle\mathbb{E}\left[\left|\nu_{T}^{f}-\nu_{0}^{f}\right|^{p}\right]\leq
c_{T}N_{0}^{p/2},$

and hence

$\displaystyle\displaystyle\left\|\frac{m_{0}Q_{t}f}{m_{0}Q_{t}\mathbf{1}_{E}}-\frac{m_{t}f}{m_{t}(\mathbf{1}_{E})}\right\|_{p}$
$\displaystyle\displaystyle\leq\frac{c^{\prime}_{T}\,\|f\|_{\infty}}{m_{0}Q_{T}\mathbf{1}_{E}/N_{0}}\,\frac{1}{\sqrt{N_{0}}},$

for some (explicit) constants $\displaystyle c_{T}$ and $\displaystyle
c^{\prime}_{T}$. Note that the assumption that the killing rate is uniformly
bounded is technical: after using Burkholder-Davis-Gundy inequality, one needs
to control the term
$\displaystyle\mathbb{E}\left(C_{T}^{p/2}\left(\frac{N+1}{N}\right)^{pC_{T}}\right)$,
however, depending on the particular process at hand, there may be other ways
to obtain upper bounds on this quantity which do not require the killing rate
to be bounded.

##  Appendix A Formal construction of the particle system

In this section, we provide a formal construction of the particle system
described above, and show that it defines a Markov process with respect to its
natural filtration.

We consider the measurable space

$\displaystyle\displaystyle
F=\bigcup_{s\in\mathcal{P}_{f}(\mathbb{N})}\\{s\\}\otimes\Omega^{s}\otimes\mathbb{R}_{+}^{s}\otimes\mathbb{R}_{+}^{s},$

where, as usual, denumerable spaces are endowed with the discrete
$\displaystyle\sigma$-fields, $\displaystyle\mathbb{R}_{+}$ is endowed with
its Borel $\displaystyle\sigma$-field, and product spaces are endowed with the
product $\displaystyle\sigma$-field. We denote generic elements of
$\displaystyle F$ by

$\displaystyle\displaystyle\theta=(s,(\omega_{i})_{i\in s},(e^{b}_{i})_{i\in
s},(e^{\kappa}_{i})_{i\in s}),$

where $\displaystyle s\in\mathcal{P}_{f}(\mathbb{N})$ is used to enumerate the
particle system, where, for all $\displaystyle i\in s$, $\displaystyle
X(\omega_{i}):t\in[0,+\infty)\to X_{t}(\omega_{i})\in E\cup\partial$ is the
trajectory of the $\displaystyle i^{th}$ particle, and where $\displaystyle
e^{b}_{i}\in\mathbb{R}_{+}$ and $\displaystyle
e^{\kappa}_{i}\in\mathbb{R}_{+}$ are auxiliary components used to define the
branching and soft-killing times.

Fix $\displaystyle s\in\mathcal{P}_{f}(\mathbb{N})$ and let us consider the
probability kernel $\displaystyle N^{1}$ from $\displaystyle\\{s\\}\otimes
E^{s}$ to $\displaystyle F$ defined by

$\displaystyle\displaystyle N^{1}(s,(x_{i})_{i\in
s}):=\delta_{s}\otimes\left(\bigotimes_{i\in
s}{\mathbf{P}}_{x_{i}}\right)\otimes\mathcal{E}^{\otimes
s}\otimes\mathcal{E}^{\otimes s},$

where $\displaystyle\mathcal{E}$ denotes the exponential distribution with
parameter $\displaystyle 1$. This kernel is used to model the evolution of the
particles before the occurence of a branching/killing event.

Given an element $\displaystyle\theta\in F$, our aim is now to define the
position of the system at the first time of branching/killing. In order to do
so, we define, for all $\displaystyle\theta\in F$ and all $\displaystyle i\in
s$, the functions

$\displaystyle\displaystyle\tau^{i,b}(\theta)=\inf\\{t\geq 0,\
\int_{0}^{t}b^{i_{0}}((X_{u}(\omega_{j}))_{j\in
s})\,\mathrm{d}u=e_{i}^{b}\\},$
$\displaystyle\displaystyle\tau^{i,\kappa}(\theta)=\inf\\{t\geq 0,\
\int_{0}^{t}\kappa^{i_{0}}((X_{u}(\omega_{j}))_{j\in
s})\,\mathrm{d}u=e_{i}^{\kappa}\\},$
$\displaystyle\displaystyle\tau^{i,\partial}(\theta)=\inf\\{t\geq 0,\
X_{t}(\omega_{i})\in\partial\\}.$

Note that, for all $\displaystyle t\geq 0$,
$\displaystyle(u,\omega)\in[0,+\infty)\times\mathcal{F}\mapsto X_{u}(\omega)$
is measurable with respect
$\displaystyle\mathcal{B}([0,+\infty))\otimes\mathcal{F}$, which entails that
$\displaystyle\omega\mapsto\int_{0}^{t}b^{i}((X_{u}(\omega_{j}))_{j\in
s})\,\mathrm{d}u$ is $\displaystyle\mathcal{F}^{\otimes s}$-measurable. In
addition, since $\displaystyle
t\mapsto\int_{0}^{t}b^{i}((X_{u}(\omega_{j}))_{j\in s})\,\mathrm{d}u$,
$\displaystyle t\mapsto\int_{0}^{t}\kappa^{i}((X_{u}(\omega_{j}))_{j\in
s})\,\mathrm{d}u$ and $\displaystyle
t\mapsto{\mathbf{1}}_{X_{t}(\omega_{i})\in\partial}$ are right-continuous, the
infima in the above definitions can be taken over $\displaystyle\mathbb{Q}$
and are thus measurable functions of $\displaystyle(\omega_{j})_{j\in s}$,
$\displaystyle(e^{b}_{j})_{j\in s}$ and $\displaystyle(e^{\kappa}_{j})_{j\in
s}$ and thus of $\displaystyle\theta$. This is also the case for the functions
$\displaystyle\tau$ and $\displaystyle i_{0}$ defined as

$\displaystyle\displaystyle\tau(\theta):=\min_{i\in
s}\tau^{i,b}(\theta)\wedge\tau^{i,\kappa}(\theta)\wedge\tau^{i,\partial}(\theta),$
$\displaystyle\displaystyle i_{0}(\theta):=\min\\{i\in S,\
\tau^{i,b}(\theta)\wedge\tau^{i,\kappa}(\theta)\wedge\tau^{i,\partial}(\theta)=\tau(\theta)\\}.$

For all $\displaystyle\theta\in F$, we define the final position of
$\displaystyle\theta$, denoted by $\displaystyle
Y(\theta)\in(E\cup\partial)^{s}$, by

$\displaystyle Y(\theta):=(X_{\tau(\omega)}(\omega_{i}))_{i\in s},$

which is also a measurable function of $\displaystyle\theta$.

Then we construct a kernel which, to any given $\displaystyle\theta\in F$ and
its final position $\displaystyle Y(\theta)$, associates its position after
the branching/killing and selection/resampling event. In order to do so, we
define the kernel $\displaystyle N^{2}$ from $\displaystyle F$ to
$\displaystyle\cup_{s\in\mathcal{P}_{f}(\mathbb{N})}\\{s\\}\times(E\cup{\partial})^{s}$
by

$\displaystyle\displaystyle N^{2}(\theta)$
$\displaystyle\displaystyle={\mathbf{1}}_{\tau(\theta)=\tau^{i_{0},b}}p^{b,i_{0}(\theta)}(Y(\theta))\frac{1}{|s|+1}\sum_{i\in
s_{+}}\delta_{s_{+}^{-i}}\otimes\delta_{Y(\theta)^{+i_{0},-i}}$
$\displaystyle\displaystyle\qquad+{\mathbf{1}}_{\tau(\theta)=\tau^{i_{0},b}}\left(1-p^{b,i_{0}(\theta)}(Y(\theta))\right)\delta_{s_{+}}\otimes\delta_{Y(\theta)^{+i_{0}}}$
$\displaystyle\displaystyle\qquad+{\mathbf{1}}_{\tau(\theta)=\tau^{i_{0},\kappa}\wedge\tau^{i_{0},{\partial}}}\,p^{\kappa,i_{0}(\theta)}(Y(\theta))\frac{1}{|s|-1}\sum_{i\in
s^{-i_{0}}}\delta_{s_{+}^{-i_{0}}}\otimes\delta_{Y(\theta)^{+i,-i_{0}}}$
$\displaystyle\displaystyle\qquad+{\mathbf{1}}_{\tau(\theta)=\tau^{i_{0},\kappa}\wedge\tau^{i_{0},{\partial}}}\,\left(1-p^{\kappa,i_{0}(\theta)}(Y(\theta))\right)\delta_{s^{-i_{0}}}\otimes\delta_{Y(\theta)^{-i_{0}}}$

where $\displaystyle s_{+}=\\{1+\max s\\}$, $\displaystyle
s_{+}^{-i}=\\{1+\max s\\}\cup s\setminus\\{i\\}$, $\displaystyle
Y(\theta)^{i_{0},-i}\in(E\cup{\partial})^{s_{+}^{-i}}$ is such that

$\displaystyle Y(\theta)^{i_{0},-i}_{j}=\begin{cases}Y(\theta)_{j}&\text{ for
all }j\in s\setminus\\{i\\},\\\ Y(\theta)_{i_{0}}&\text{ for }j=1+\max
s,\end{cases}$

$\displaystyle Y(\theta)^{i_{0}}\in(E\cup{\partial})^{s_{+}}$ is such that

$\displaystyle Y(\theta)^{i_{0}}_{j}=\begin{cases}Y(\theta)_{j}&\text{ for all
}j\in s,\\\ Y(\theta)_{i_{0}}&\text{ for }j=1+\max s,\end{cases}$

According to Ionescu Tulcea’s extension Theorem, for any initial distribution
$\displaystyle\mu_{0}$ on $\displaystyle
F_{0}:=\cup_{s\in\mathcal{P}_{f}(\mathbb{N})}\\{s\\}\times(E\cup{\partial})^{s}$,
there exists a probability measure $\displaystyle\mathbb{P}_{\mu_{0}}$ on
$\displaystyle\bar{\Omega}=F_{0}\times F^{\mathbb{N}}$, such that, for all
$\displaystyle n\geq 2$,

$\mathbb{P}_{\mu_{0}}(A_{0}\in\mathrm{d}a_{0},\Theta_{1}\in\mathrm{d}\theta_{1},\,\ldots,\,A_{n-1}\in\mathrm{d}a_{n-1},\,\Theta_{n}\in\mathrm{d}\theta_{n})\\\
=\mu_{0}\circ N^{1}\circ N^{2}\circ N^{1}\circ\cdots\circ N^{2}\circ
N^{1}(\mathrm{d}a_{0},\mathrm{d}\theta_{1},\cdots,\,\mathrm{d}a_{n-1},\mathrm{d}\theta_{n}).$

Then, we can define, for any
$\displaystyle\bar{\omega}=(a_{0},\theta_{1},\ldots)$, the sequence of times
$\displaystyle(\tau_{n})_{n\in\mathbb{N}}$ as

$\displaystyle\tau_{0}(\bar{\omega})=0\text{ and
}\tau_{n}(\bar{\omega})=\tau(\theta_{1})+\cdots+\tau(\theta_{n}),\ \forall
n\geq 0$

and, for all $\displaystyle t\geq 0$,

$\displaystyle\bar{S}_{t}(\bar{\omega})=\sum_{n\geq
0}{\mathbf{1}}_{\tau_{n-1}\leq t<\tau_{n}}s(\theta_{n}).$

and, for all $\displaystyle i\in\bar{S}_{t}(\bar{\omega})$,

$\displaystyle
X^{i}_{t}(\bar{\omega})=X_{t-\tau_{n-1}(\bar{\omega})}(\omega_{i}(\theta_{n})).$

The process $\displaystyle(\bar{S}_{t},(X^{i}_{t})_{i\in\bar{S}_{t}})_{t\geq
0}$, defined on the probability space $\displaystyle(\bar{\Omega},\mathbb{P})$
and taking its values in $\displaystyle F_{0}$, is then the size constrained
branching process, which concludes the construction of the process.

######  Remark 26.

Observe that the particle system is well defined without Assumptions 1 and 2.
However, if Assumption 1 holds true, then
$\displaystyle(\bar{S}_{t},(X^{i}_{t})_{i\in\bar{S}_{t}})_{t\geq 0}$ takes its
values in
$\displaystyle\bigcup_{s\in\mathcal{P}_{f}(\mathbb{N})}\\{s\\}\otimes E^{s}$
almost surely. Also, if Assumption 2 holds true, then
$\displaystyle\bar{S}_{t}$ is non-empty almost surely, for all $\displaystyle
t\in[0,+\infty)$.

The above construction makes use of the fact that $\displaystyle X$ is a
measurable adapted process. Using the fact that it is also assumed
progressively measurable, we are in position to prove the following.

######  Proposition 27.

The process $\displaystyle(\bar{S}_{t},(X^{i}_{t})_{i\in\bar{S}_{t}})_{t\geq
0}$ is a Markov process with respect to its natural filtration.

###### Proof.

Fix $\displaystyle t\geq s\geq 0$, and $\displaystyle A$ a measurable subset
of $\displaystyle F_{0}$. For all $\displaystyle k\geq 1$, all $\displaystyle
s_{1}<s_{2}<\cdots<s_{k}=s$ and all measurable subsets $\displaystyle
A_{1},\ldots,A_{k}$ of $\displaystyle F_{0}$, we have, setting
$\displaystyle\mathbb{X}_{t}:=(\bar{S}_{t},(X^{i}_{t})_{i\in\bar{S}_{t}})$,

$\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in A_{k},\mathbb{X}_{t}\in A\right)\\\
=\sum_{n_{1}\leq\cdots\leq n_{k}=m\leq
n}\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}}),\,\mathbb{X}_{t}\in
A,\,t\in[\tau_{n},\tau_{n+1})\right).$ (60)

Then, conditioning with respect to
$\displaystyle\mathcal{G}_{m}:=\sigma(\Theta_{1},\ldots,\Theta_{m},A_{m})$, we
have

$\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}}),\,\mathbb{X}_{s+t}\in
A,\,s+t\in[\tau_{n},\tau_{n+1})\mid\mathcal{G}_{m}\right)\\\
={\mathbf{1}}_{\mathbb{X}_{s_{1}}\in A_{1},\ldots,\mathbb{X}_{s_{\ell}}\in
A_{\ell},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}}),\tau_{m}\leq s}\times
f(A_{m},\tau_{m}),$ (61)

where

$\displaystyle\displaystyle
f(a,u):=\mathbb{P}_{a}\left(\mathbb{X}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{X}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\tau_{1}),\mathbb{X}_{t-u}\in
A,\,t-u\in[\tau_{n-m},\tau_{n-m+1})\right).$

There are now two possible situations, either $\displaystyle m=n$ or
$\displaystyle m<n$, which we consider separately.

Case $\displaystyle m=n$. In the former case, we observe that, by definition
of $\displaystyle N^{1}$,

$\displaystyle\displaystyle f(a,u)$
$\displaystyle\displaystyle=\mathbb{P}_{a}\left(\mathbb{X}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{X}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\tau_{1}),\mathbb{X}_{t-u}\in
A,\,t-u\in[0,\tau_{1})\right)$
$\displaystyle\displaystyle=\texttt{P}_{a}\left(\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\texttt{t}),\mathbb{Y}_{t-u}\in
A,\,t-u\in[0,\texttt{t})\right),$

where $\displaystyle\texttt{P}_{a}$ is the product measure
$\displaystyle\Omega^{\prime}=\mathbb{P}^{\otimes s}\otimes\mathcal{E}(1)^{s}$
on the probability space
$\displaystyle\Omega^{s}(\mathbb{R}_{+})^{s}\otimes(\mathbb{R}_{+})^{s}$
endowed with the filtration defined by
$\displaystyle\mathcal{H}_{t}:=\mathcal{F}_{t}^{\otimes
s}\otimes\\{\emptyset,\mathbb{R}_{+}\\}^{s}\otimes\\{\emptyset,\mathbb{R}_{+}\\}^{s}$,
with the variables $\displaystyle\mathbb{Y}$ and t being defined, for all
$\displaystyle\theta^{\prime}=((\omega_{i})_{i\in s},(e^{b}_{i})_{i\in
s},(e^{\kappa}_{i})_{i\in s})\in\Omega^{\prime}$, by

$\displaystyle\displaystyle\mathbb{Y}_{t}=(s,(X_{t}(\omega_{i})_{i\in s}),\
\forall t\geq 0,$

and $\displaystyle\texttt{t}=\min_{i\in
s}\texttt{t}^{b,i}\wedge\texttt{t}^{\kappa,i}\wedge t^{{\partial},i}$, with

$\displaystyle\displaystyle\texttt{t}^{b,i}=\inf\\{t\geq 0,\
\int_{0}^{t}b^{i_{0}}((X_{u}(\omega_{j}))_{j\in
s})\,\mathrm{d}u=e_{i}^{b}\\},$
$\displaystyle\displaystyle\texttt{t}^{\kappa,i}=\inf\\{t\geq 0,\
\int_{0}^{t}\kappa^{i_{0}}((X_{u}(\omega_{j}))_{j\in
s})\,\mathrm{d}u=e_{i}^{\kappa}\\},$
$\displaystyle\displaystyle\texttt{t}^{\partial,i}=\inf\\{t\geq 0,\
X_{t}(\omega_{i})\in\partial\\}.$

We emphasise that the second component of $\displaystyle\mathbb{Y}$ is
distributed as a vector indexed by $\displaystyle s$ of independent copies of
$\displaystyle X$ starting from $\displaystyle x_{i}$, for all $\displaystyle
i\in s$, and that it is a progressively measurable Markov process with respect
to the filtration $\displaystyle\mathcal{H}$. In addition, the random
variables $\displaystyle e^{b}_{i}$ and $\displaystyle e^{\kappa}_{i}$,
$\displaystyle i\in s$, are independent from each others and from
$\displaystyle\mathcal{H}_{t}$, for all $\displaystyle t\geq 0$.

Using the Markov property at time $\displaystyle s-u$ for the process
$\displaystyle\mathbb{Y}$, we obtain

$\displaystyle\displaystyle\texttt{P}_{a}$
$\displaystyle\displaystyle\left(\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\texttt{t}),\mathbb{Y}_{t-u}\in
A,\,t-u\in[0,\texttt{t})\right)$
$\displaystyle\displaystyle\qquad=\texttt{E}_{a}\left({\mathbf{1}}_{\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k}}{\mathbf{1}}_{\mathbb{Y}_{t-u}\in
A\setminus{\partial}}\,e^{-\int_{0}^{t-u}h(\mathbb{Y}_{v})\,\mathrm{d}v}\right)$
$\displaystyle\displaystyle\qquad=\texttt{E}_{a}\left({\mathbf{1}}_{\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k}}\texttt{E}_{a}\left({\mathbf{1}}_{\mathbb{Y}_{t-u}\in
A\setminus{\partial}}\,e^{-\int_{0}^{t-u}h(\mathbb{Y}_{v})\,\mathrm{d}v}\mid\mathcal{H}_{s-u}\right)\right)$
$\displaystyle\displaystyle\qquad=\texttt{E}_{a}\left({\mathbf{1}}_{\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k}}e^{-\int_{0}^{s-u}h(\mathbb{Y}_{v})\,\mathrm{d}v}\texttt{E}_{\mathbb{Y}_{s-u}}\left({\mathbf{1}}_{\mathbb{Y}_{t-s}\in
A\setminus{\partial}}\,e^{-\int_{0}^{t-s}h(\mathbb{Y}_{v})\,\mathrm{d}v}\right)\right)$
$\displaystyle\displaystyle\qquad=\texttt{E}_{a}\left({\mathbf{1}}_{\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k}}{\mathbf{1}}_{s_{k}-u\in[0,\texttt{t})}\texttt{E}_{\mathbb{Y}_{s-u}}\left({\mathbf{1}}_{\mathbb{Y}_{t-s}\in
A\setminus{\partial}}\,{\mathbf{1}}_{t-s\in[0,\texttt{t})}\right)\right)$

where $\displaystyle h(s,(x_{i})_{i\in s}):=\sum_{i\in
s}\kappa^{i}(x_{i})+b^{i}(x_{i})$, and where we used the fact that
$\displaystyle\mathbb{Y}$ is progressively measurable to deduce that
$\displaystyle\int_{0}^{s-u}h(\mathbb{Y}_{v})\,\mathrm{d}v$ is
$\displaystyle\mathcal{H}_{s-u}$-measurable. We deduce that

$\displaystyle\displaystyle
f(a,u)=\mathbb{E}_{a}\left({\mathbf{1}}_{\mathbb{X}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{X}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\tau_{1})}\,\mathbb{E}_{\mathbb{X}_{s-u}}\left(\mathbb{X}_{t-s}\in
A,\,t-s\in[0,\tau_{1})\right)\right)$

Taking the expectation in (61), we finally deduce that

$\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}}),\,\mathbb{X}_{s+t}\in
A,\,s+t\in[\tau_{n},\tau_{n+1})\right)\\\
=\mathbb{E}_{\mu_{0}}\left({\mathbf{1}}_{\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{\ell}}\in
A_{\ell},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}}),\tau_{m}\leq s}\times
f(A_{m},\tau_{m})\right),$

and then, by definition of the law of
$\displaystyle(A_{0},\Theta_{1},\cdots,A_{m-1},\Theta_{m},A_{m},\ldots)$, we
deduce that

$\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}}),\,\mathbb{X}_{s+t}\in
A,\,s+t\in[\tau_{n},\tau_{n+1})\right)\\\
=\mathbb{E}_{\mu_{0}}\left({\mathbf{1}}_{\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}})}\mathbb{P}_{\mathbb{X}_{s}}\left(\mathbb{X}_{t-s}\in
A,\,t-s\in[0,\tau_{1})\right)\right).$ (62)

Case $\displaystyle m<n$. We can now proceed with the case $\displaystyle
m<n$. In this case, we write

$\displaystyle\displaystyle
f(a,u)=\mathbb{E}_{a}\left({\mathbf{1}}_{\mathbb{X}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{X}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\tau_{1})}\,g_{u}(A_{1},\tau_{1})\right),$

where

$\displaystyle g_{u}(b,v):=\mathbb{E}_{A_{1}}\left(\mathbb{X}_{t-u-v}\in
A,\,t-u-v\in[\tau_{n-m-1},\tau_{n-m})\right)$

Using the same strategy as above (with the same definition of
$\displaystyle\mathbb{Y}$), we obtain, using also the Markov property at time
$\displaystyle s_{k}-u=s-u$,

$\displaystyle\displaystyle f(a,u)$
$\displaystyle\displaystyle=\texttt{E}_{a}\left({\mathbf{1}}_{\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\texttt{t})}\,g_{u}(\mathbb{Y}_{\texttt{t}},\texttt{t})\right)$
$\displaystyle\displaystyle=\texttt{E}_{a}\left({\mathbf{1}}_{\mathbb{Y}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{Y}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\texttt{t})}\,\texttt{E}_{\mathbb{Y}_{s-u}}(g_{u}(\mathbb{Y}_{\texttt{t}},\texttt{t}))\right)$
$\displaystyle\displaystyle=\mathbb{E}_{a}\left({\mathbf{1}}_{\mathbb{X}_{s_{\ell+1}-u}\in
A_{\ell+1},\ldots,\mathbb{X}_{s_{k}-u}\in
A_{k},\,s_{i}-u\in[0,\tau_{1})}\,\mathbb{E}_{\mathbb{X}_{s-u}}(g_{u}(\mathbb{X}_{\tau_{1}},\tau_{1}))\right)$

Using again the definition of the law of
$\displaystyle(A_{0},\Theta_{1},\cdots,A_{m-1},\Theta_{m},A_{m},\ldots)$, we
deduce that

$\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}}),\,\mathbb{X}_{s+t}\in
A,\,s+t\in[\tau_{n},\tau_{n+1})\right)\\\
=\mathbb{E}_{\mu_{0}}\left({\mathbf{1}}_{\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k},\,s_{i}\in[\tau_{n_{i}-1},\tau_{n_{i}})}\mathbb{P}_{\mathbb{X}_{s}}\left(\mathbb{X}_{t-s}\in
A,\,t-s\in[\tau_{n-m},\tau_{n-m+1})\right)\right).$ (63)

Conclusion. Finally, using (62) and (63) and summing over the indices
$\displaystyle n_{i}$ and $\displaystyle n$ as in (60), we deduce that

$\displaystyle\displaystyle\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in A_{k},\mathbb{X}_{t}\in
A\right)=\mathbb{E}_{\mu_{0}}\left({\mathbf{1}}_{\mathbb{X}_{s_{1}}\in
A_{1},\ldots,\mathbb{X}_{s_{k}}\in
A_{k}}\,\mathbb{P}_{\mathbb{X}_{s}}\left(\mathbb{X}_{t-s}\in A\right)\right).$

Since the property holds true for all finite family of time indices
$\displaystyle s_{1}<\ldots<s_{k}=s$ and all $\displaystyle
A_{1},\ldots,A_{k}$, we deduce that

$\displaystyle\displaystyle\mathbb{P}_{\mu_{0}}\left(\mathbb{X}_{t}\in
A\mid\sigma(\mathbb{X}_{u},\,u\leq
s)\right)=\mathbb{P}_{\mathbb{X}_{s}}\left(\mathbb{X}_{t-s}\in A\right),$

which concludes the proof of the proposition. ∎

######  Remark 28.

Although we won’t use this property, the process
$\displaystyle(\bar{S}_{t},(X^{i}_{t})_{t\in\bar{S}_{t}})_{t\geq 0}$ can be
shown to be a Markov process with respect to larger filtrations. On a related
note, it is natural to ask whether the strong Markov property for
$\displaystyle X$ entails the strong Markov property for the particle system.
This is not clear at all in full generality and we leave it as an open
problem. However, it can be shown, using a similar approach as in [39], that,
under additional regularity properties on the process $\displaystyle X$ and on
the functionals $\displaystyle b,\kappa$, and additional requirements on the
state space $\displaystyle E$, the particle system is indeed a strong Markov
process in a quite general setting.

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