which implies that $R_{11} = 0 \le a$. Moreover

$$ \mathbf{1}\{|\frac{\gamma(\vartheta_N, \mathbf{x})}{\gamma(\vartheta, \mathbf{x})} - 1| > \epsilon\} \le \mathbf{1}\{|\frac{\gamma(\vartheta_N, \mathbf{x})}{\gamma(\vartheta, \mathbf{x})} - 1 - \frac{h^T \nabla_1 \gamma(\vartheta, \mathbf{x})}{\sqrt{N} \gamma(\vartheta, \mathbf{x})}| > \frac{\epsilon}{2}\} + \mathbf{1}\{|\frac{h^T \nabla_1 \gamma(\vartheta, \mathbf{x})}{\sqrt{N} \gamma(\vartheta, \mathbf{x})}| > \frac{\epsilon}{2}\}. $$

Using this and the fact that for any $z$, $\mathbf{1}_{\{|z| > \frac{\epsilon}{2}\}} \le 4\epsilon^{-2}z^2\mathbf{1}_{\{|z| > \frac{\epsilon}{2}\}}$, we get

$$ \int_0^{tN} \mathbf{1}\left\{|\frac{\gamma(\vartheta_N, \mathbf{x}_{s-})}{\gamma(\vartheta, \mathbf{x}_{s-})} - 1\right| > \epsilon\right\} dN_s^\gamma \\ \le 4\epsilon^{-2} \int_0^{tN} \left(\frac{\gamma(\vartheta_N, \mathbf{X}_{s-})}{\gamma(\vartheta, \mathbf{X}_{s-})} - 1 - \frac{h^T \nabla_1 \gamma(\vartheta, \mathbf{X}_{s-})}{\sqrt{N} \gamma(\vartheta, \mathbf{X}_{s-})}\right)^2 dN_s^\gamma \\ + 4\epsilon^{-2} \int_0^{tN} \left(\frac{h^T \nabla_1 \gamma(\vartheta, \mathbf{X}_{s-})}{\sqrt{N} \gamma(\vartheta, \mathbf{X}_{s-})}\right)^2 \mathbf{1}\left\{|\frac{h^T \nabla_1 \gamma(\vartheta, \mathbf{X}_{s-})}{\sqrt{N} \gamma(\vartheta, \mathbf{X}_{s-})}| > \frac{\epsilon}{2}\right\} dN_s^\gamma. $$

The first term is exactly $4\epsilon^{-2} \int (f-g)^2 dN_s^\gamma$, as already studied in (39), and the second term is $4\epsilon^{-2}A_2$ (up to $\epsilon/2$ instead of $\epsilon$), see (37). Both terms tend to 0 in $\mathbb{P}_{\mathbf{x}_0}^\vartheta$-probability whatever the value of $\epsilon > 0$. This proves that the right-hand side term in (42) tends to 0 for any $b < 1$, which yields the convergence of $R_{11}$ in $\mathbb{P}_{\mathbf{x}_0}^\vartheta$-probability to 0.

For $R_{12}$ given by (41), we choose $\epsilon < \frac{1}{2}$ and since $|\varphi(x)| \le 2|x|^3$ for $|x| \le \frac{1}{2}$, we have

$$ R_{12} \le \int_0^{tN} 2 \left| \frac{\gamma(\vartheta_N, \mathbf{X}_{s-})}{\gamma(\vartheta, \mathbf{X}_{s-})} - 1 \right|^3 \mathbf{1}\left\{ \left| \frac{\gamma(\vartheta_N, \mathbf{X}_{s-})}{\gamma(\vartheta, \mathbf{X}_{s-})} - 1 \right| \le \epsilon \right\} dN_s^\gamma \\ \le 2\epsilon \int_0^{tN} \left( \frac{\gamma(\vartheta_N, \mathbf{X}_{s-})}{\gamma(\vartheta, \mathbf{X}_{s-})} - 1 \right)^2 dN_s^\gamma. $$

This last term is $2\epsilon(R_2 + \int g^2 dN_s^\gamma)$ if we use as previously the notation $R_2 = \int (f^2 - g^2)dN_s^\gamma$ for $R_2$ given by (38). We already know that $U := 2(R_2 + \int g^2 dN_s^\gamma)$ is a $O_{\mathbb{P}_{\mathbf{x}_0}^\vartheta}(1)$ as $N \to \infty$. This means that for any $\eta > 0$, there exists $M$ such that for $N$ sufficiently large $\mathbb{P}_{\mathbf{x}_0}^\vartheta(U > M) < \eta$. This implies that for any $a > 0$ and $\eta > 0$, we can choose $\epsilon$ small enough ($\epsilon < a/M$) so that $\mathbb{P}_{\mathbf{x}_0}^\vartheta(\epsilon U > a) < P_{\mathbf{x}_0}(U > M) < \eta$. So for any $a > 0$ and $\eta > 0$, we can choose $\epsilon$ so that $\mathbb{P}_{\mathbf{x}_0}^\vartheta(R_{12} > a) < \eta$ for $N$ large enough. The same inequality holds true for $R_{11}$ since we have proven that $R_{11}$ tends to 0 in probability for any $\epsilon > 0$. So $R_1 = R_{11} + R_{12}$ tends to 0 in $\mathbb{P}_{\mathbf{x}_0}^\vartheta$-probability, which concludes the proof.
## B.2 Proof of Lemma 2

The proof follows the same scheme as the proof of Lemma 1. Consider

$$ \mathrm{Rem}_{N, \vartheta, h}^{k_{\gamma}}(t) = \Lambda_{k_{\gamma}}^{\vartheta_N/\vartheta}(tN) - h^T M_{k_{\gamma}}^{N, \vartheta}(t) + \frac{1}{2} h^T \langle M_{k_{\gamma}}^{N, \vartheta} \rangle(t)h, $$

where $M_{k_{\gamma}}^{N, \vartheta}(t)$ is given by (10). The fact that $M_{k_{\gamma}}^{N, \vartheta}$ is a $\mathcal{F}_{tN}$-martingale is a consequence of Lemma 4. We deduce that

$$ [M_{k_{\gamma}}^{N, \vartheta}](t) = \frac{1}{N} \int_{0}^{tN} \frac{\nabla_{1} k_{\gamma} (\nabla_{1} k_{\gamma})^{T}}{k_{\gamma}^{2}} (\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s}) dN_{s}^{\gamma}, $$

$$ \langle M_{k_{\gamma}}^{N, \vartheta} \rangle(t) = \frac{1}{N} \int_{0}^{tN} \int_{E} \mathbf{1}_{k_{\gamma}>0} \frac{\nabla_{1} k_{\gamma} (\nabla_{1} k_{\gamma})^{T}}{k_{\gamma}} (\vartheta, \mathbf{X}_{s}, \mathbf{y}) \nu_{\gamma}(\mathbf{X}_{s}, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_{s}) ds. $$

Starting from (27) and using the above representations we can decompose $\mathrm{Rem}_{N, \vartheta, h}^{k_{\gamma}}(t)$ as in the proof of Lemma 1, that is $\mathrm{Rem}_{N, \vartheta, h}^{k_{\gamma}}(t) = S_{t}^{N} + A_{t}^{N} + R_{t}^{N}$, where

$$ S_{t}^{N} = \int_{0}^{tN} \left( \frac{k_{\gamma}(\vartheta_{N}, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} - 1 - \frac{h^T \nabla_{1} k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{\sqrt{N} k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} \right) dN_{s}^{\gamma} \\ + \frac{1}{\sqrt{N}} \int_{0}^{tN} \int_{E} h^T \nabla_{1} k_{\gamma}(\vartheta, \mathbf{X}_{s}, \mathbf{y}) \nu_{\gamma}(\mathbf{X}_{s}, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_{s}) ds, $$

$$ A_{t}^{N} = \frac{1}{2N} \int_{0}^{tN} \int_{E} \mathbf{1}_{k_{\gamma}>0} \frac{(h^T \nabla_{1} k_{\gamma})^2}{k_{\gamma}} (\vartheta, \mathbf{X}_{s}, \mathbf{y}) \nu_{\gamma}(\mathbf{X}_{s}, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_{s}) ds \\ - \frac{1}{2N} \int_{0}^{tN} \frac{(h^T \nabla_{1} k_{\gamma})^2}{k_{\gamma}^2} (\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s}) dN_{s}^{\gamma}, $$

$$ R_{t}^{N} = \int_{0}^{tN} \left( \log \left( \frac{k_{\gamma}(\vartheta_{N}, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} \right) - \frac{k_{\gamma}(\vartheta_{N}, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} + 1 \right. \\ \left. + \frac{1}{2N} \frac{(h^T \nabla_{1} k_{\gamma})^2}{k_{\gamma}^2} (\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s}) \right) dN_{s}^{\gamma}. $$

We show that each of these three terms tends to 0 in $\mathbb{P}_{\mathbf{x}_{0}}^{\vartheta}$-probability as in the proof of Lemma 1.
## B.2 Proof of Lemma 2

The proof follows the same scheme as the proof of Lemma 1. Consider

$$ \mathrm{Rem}_{N, \vartheta, h}^{k_{\gamma}}(t) = \Lambda_{k_{\gamma}}^{\vartheta_N/\vartheta}(tN) - h^T M_{k_{\gamma}}^{N, \vartheta}(t) + \frac{1}{2} h^T \langle M_{k_{\gamma}}^{N, \vartheta} \rangle(t)h, $$

where $M_{k_{\gamma}}^{N, \vartheta}(t)$ is given by (10). The fact that $M_{k_{\gamma}}^{N, \vartheta}$ is a $\mathcal{F}_{tN}$-martingale is a consequence of Lemma 4. We deduce that

$$ [M_{k_{\gamma}}^{N, \vartheta}](t) = \frac{1}{N} \int_{0}^{tN} \frac{\nabla_{1} k_{\gamma} (\nabla_{1} k_{\gamma})^{T}}{k_{\gamma}^{2}} (\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s}) dN_{s}^{\gamma}, $$

$$ \langle M_{k_{\gamma}}^{N, \vartheta} \rangle(t) = \frac{1}{N} \int_{0}^{tN} \int_{E} \mathbf{1}_{k_{\gamma}>0} \frac{\nabla_{1} k_{\gamma} (\nabla_{1} k_{\gamma})^{T}}{k_{\gamma}} (\vartheta, \mathbf{X}_{s}, \mathbf{y}) \nu_{\gamma}(\mathbf{X}_{s}, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_{s}) ds. $$

Starting from (27) and using the above representations we can decompose $\mathrm{Rem}_{N, \vartheta, h}^{k_{\gamma}}(t)$ as in the proof of Lemma 1, that is $\mathrm{Rem}_{N, \vartheta, h}^{k_{\gamma}}(t) = S_{t}^{N} + A_{t}^{N} + R_{t}^{N}$, where

$$ S_{t}^{N} = \int_{0}^{tN} \left( \frac{k_{\gamma}(\vartheta_{N}, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} - 1 - \frac{h^T \nabla_{1} k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{\sqrt{N} k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} \right) dN_{s}^{\gamma} \\ + \frac{1}{\sqrt{N}} \int_{0}^{tN} \int_{E} h^T \nabla_{1} k_{\gamma}(\vartheta, \mathbf{X}_{s}, \mathbf{y}) \nu_{\gamma}(\mathbf{X}_{s}, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_{s}) ds, $$

$$ A_{t}^{N} = \frac{1}{2N} \int_{0}^{tN} \int_{E} \mathbf{1}_{k_{\gamma}>0} \frac{(h^T \nabla_{1} k_{\gamma})^2}{k_{\gamma}} (\vartheta, \mathbf{X}_{s}, \mathbf{y}) \nu_{\gamma}(\mathbf{X}_{s}, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_{s}) ds \\ - \frac{1}{2N} \int_{0}^{tN} \frac{(h^T \nabla_{1} k_{\gamma})^2}{k_{\gamma}^2} (\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s}) dN_{s}^{\gamma}, $$

$$ R_{t}^{N} = \int_{0}^{tN} \left( \log \left( \frac{k_{\gamma}(\vartheta_{N}, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} \right) - \frac{k_{\gamma}(\vartheta_{N}, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})}{k_{\gamma}(\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s})} + 1 \right. \\ \left. + \frac{1}{2N} \frac{(h^T \nabla_{1} k_{\gamma})^2}{k_{\gamma}^2} (\vartheta, \mathbf{X}_{s^{-}}, \mathbf{X}_{s}) \right) dN_{s}^{\gamma}. $$

We show that each of these three terms tends to 0 in $\mathbb{P}_{\mathbf{x}_{0}}^{\vartheta}$-probability as in the proof of Lemma 1.
**First step:** We have

$$[S_t^N] = \int_0^{tN} \left( \frac{k_\gamma(\vartheta_N, \mathbf{X}_{s-}, \mathbf{X}_s)}{k_\gamma(\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s)} - 1 - \frac{h^T \nabla_1 k_\gamma(\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s)}{\sqrt{N} k_\gamma(\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s)} \right)^2 dN_s^\gamma$$

so that by Lemma 4 in Appendix C.2,

$$\langle S^N \rangle_t = \int_0^{tN} \int_E \left( \frac{k_\gamma(\vartheta_N, \mathbf{X}_s, \mathbf{y})}{k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y})} - 1 - \frac{h^T \nabla_1 k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y})}{\sqrt{N} k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y})} \right)^2 \\ \times k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y}) \nu_\gamma(\mathbf{X}_s, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_s) ds.$$

Using the same inequalities as in (34) we obtain that

$$\langle S^N \rangle_t \le \frac{\|h\|^2}{N} \int_0^{tN} f_\vartheta^{k_\gamma} \left( \mathbf{X}_s, \frac{\|h\|}{\sqrt{N}} \right) \gamma(\vartheta, \mathbf{X}_s) ds$$

where $f_\vartheta^{k_\gamma}$ is given by (7). We can then deduce that $\mathbb{E}_{\mathbf{X}_0}^\vartheta (\langle S^N \rangle_t)$ tends to 0 by use of Conditions **B**(k_γ, γ), **(H)** and the dominated convergence theorem, as in the proof of Lemma 1.

**Second step:** The proof is identical to the second step of the proof of Lemma 1.

It consists in splitting the integrals in $A_t^N$ according to whether $\left| \frac{h^T \nabla_1 k_\gamma}{k_\gamma} (\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s) \right| > \epsilon \sqrt{N}$ or not. The convergence of $A_t^N$ is then proved similarly, by use of Conditions **A**(k_γ, γ), **(H)** and the dominated convergence theorem.

**Third step:** The proof follows exactly the same lines as the proof of the third step of Lemma 1 where $\gamma(., .)$ is replaced by $k_\gamma(., ., .)$ and Condition **A**(k_γ, γ) is used instead of Condition **A**(γ).

## B.3 Proof of Lemma 3

Remember that $\mathbf{X}_t = \{(\mathbf{Z}_{t-T_i}^{(i)}, \mathbf{m}^{(i)})\}$ for all $t \in [T_i, T_{i+1}]$, and that $L(\vartheta, \mathbf{X}_{[T_i, T_{i+1}.]})$ is a shortcut for $L(\mathbf{Z}_{[0, T_{i+1}-T_i]}^{(i)}, \mathbf{m}^{(i)})$ given by (5). Accordingly, we have

$$\log L(\vartheta, \mathbf{X}_{[T_i, T_{i+1}.]}) = \int_{T_i}^{T_{i+1}} \bar{v}_n(\vartheta, \mathbf{X}_s)^T a_n^{-1}(\mathbf{X}_s) d\mathbf{Z}_{s-T_i}^{(i)} \\ - \frac{1}{2} \int_{T_i}^{T_{i+1}} \bar{v}_n(\vartheta, \mathbf{X}_s)^T a_n^{-1}(\mathbf{X}_s) \bar{v}_n(\vartheta, \mathbf{X}_s) ds,$$
**First step:** We have

$$[S_t^N] = \int_0^{tN} \left( \frac{k_\gamma(\vartheta_N, \mathbf{X}_{s-}, \mathbf{X}_s)}{k_\gamma(\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s)} - 1 - \frac{h^T \nabla_1 k_\gamma(\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s)}{\sqrt{N} k_\gamma(\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s)} \right)^2 dN_s^\gamma$$

so that by Lemma 4 in Appendix C.2,

$$\langle S^N \rangle_t = \int_0^{tN} \int_E \left( \frac{k_\gamma(\vartheta_N, \mathbf{X}_s, \mathbf{y})}{k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y})} - 1 - \frac{h^T \nabla_1 k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y})}{\sqrt{N} k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y})} \right)^2 \times k_\gamma(\vartheta, \mathbf{X}_s, \mathbf{y}) \nu_\gamma(\mathbf{X}_s, d\mathbf{y}) \gamma(\vartheta, \mathbf{X}_s) ds.$$

Using the same inequalities as in (34) we obtain that

$$\langle S^N \rangle_t \le \frac{\|h\|^2}{N} \int_0^{tN} f_{\vartheta}^{k_\gamma} \left( \mathbf{X}_s, \frac{\|h\|}{\sqrt{N}} \right) \gamma(\vartheta, \mathbf{X}_s) ds$$

where $f_{\vartheta}^{k_\gamma}$ is given by (7). We can then deduce that $\mathbb{E}_{\mathbf{X}_0}^{\vartheta}(\langle S^N \rangle_t)$ tends to 0 by use of Conditions **B**(k_γ, γ), **(H)** and the dominated convergence theorem, as in the proof of Lemma 1.

**Second step:** The proof is identical to the second step of the proof of Lemma 1.

It consists in splitting the integrals in $A_t^N$ according to whether $\left| \frac{h^T \nabla_1 k_\gamma}{k_\gamma} (\vartheta, \mathbf{X}_{s-}, \mathbf{X}_s) \right| > \epsilon\sqrt{N}$ or not. The convergence of $A_t^N$ is then proved similarly, by use of Conditions **A**(k_γ, γ), **(H)** and the dominated convergence theorem.

**Third step:** The proof follows exactly the same lines as the proof of the third step of Lemma 1 where $\gamma(., .)$ is replaced by $k_\gamma(., . , .)$ and Condition **A**(k_γ, γ) is used instead of Condition **A**(γ).

## B.3 Proof of Lemma 3

Remember that $\mathbf{X}_t = \{(\mathbf{Z}_{t-T_i}^{(i)}, \mathbf{m}^{(i)})\}$ for all $t \in [T_i, T_{i+1}]$, and that $L(\vartheta, \mathbf{X}_{[T_i, T_{i+1}.]})$ is a shortcut for $L(\mathbf{Z}_{[0, T_{i+1}-T_i]}^{(i)}, \mathbf{m}^{(i)})$ given by (5). Accordingly, we have

$$\log L(\vartheta, \mathbf{X}_{[T_i, T_{i+1}.]}) = \int_{T_i}^{T_{i+1}} \bar{v}_n(\vartheta, \mathbf{X}_s)^T a_n^{-1}(\mathbf{X}_s) d\mathbf{Z}_{s-T_i}^{(i)} - \frac{1}{2} \int_{T_i}^{T_{i+1}} \bar{v}_n(\vartheta, \mathbf{X}_s)^T a_n^{-1}(\mathbf{X}_s) \bar{v}_n(\vartheta, \mathbf{X}_s) ds,$$
where $\bar{v}_n(\vartheta, \mathbf{X}_s)$ stands for $\bar{v}_n(\vartheta, \mathbf{Z}_{s-T_i}^{(i)}, \mathbf{m}^{(i)})$, with $n = n(\mathbf{X}_s)$, and similarly for $a_n^{-1}(\mathbf{X}_s)$. Then we have

$$ 
\begin{aligned}
& \log \frac{L(\vartheta_N, \mathbf{X}_{[T_i, T_{i+1}]} )}{L(\vartheta, \mathbf{X}_{[T_i, T_{i+1}]} )} \\
& = \int_{T_i}^{T_{i+1}} (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (d\mathbf{Z}_{s-T_i}^{(i)} - \bar{v}_n(\vartheta, \mathbf{X}_s) ds) \\
& \quad - \frac{1}{2} \int_{T_i}^{T_{i+1}} (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) ds.
\end{aligned}
$$

Consequently, we may rewrite (28) as

$$ 
\begin{aligned}
\Lambda_L^{\vartheta_N/\vartheta}(t) = & \sum_{i \ge 0} \int_0^t \mathbf{1}_{[T_i, T_{i+1}]} (s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) dM_{s-T_i}^{(i)} \\
& - \frac{1}{2} \int_0^t (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s)) ds.
\end{aligned}
\quad (43) $$

where $M_s^{(i)} = 0$ if $s < 0$ and

$$ M_s^{(i)} = \mathbf{Z}_s^{(i)} - \mathbf{Z}_0^{(i)} - \int_0^s \bar{v}_n(\vartheta, \mathbf{Z}_u^{(i)}) du, \quad s \ge 0. \quad (44) $$

We know that $\mathbf{Z}^{(i)}$ is the solution of $M^{|n}(\mathbf{z}^{(i)}, \mathbf{m}^{(i)})$ given by (4), where $(\mathbf{z}^{(i)}, \mathbf{m}^{(i)})$ are such that $\mathbf{X}_{T_i} = \{(\mathbf{z}^{(i)}, \mathbf{m}^{(i)})\}$ and $n = n(\mathbf{X}_{T_i})$. This implies that $(M_s^{(i)})$ is a martingale with respect to $\mathcal{F}_t^{\mathbf{Z}^{(i)}}$, the natural filtration associated to $\mathbf{Z}^{(i)}$. Note that $(M_s^{(i)})$ is independent from $T_i$ and that $\mathcal{F}_t$ is generated by $\{T_{N_t}, \mathbf{X}_{T_{N_t}}, \mathbf{Z}_{[0, t-T_{N_s}]}^{(N_t)}, (T_i, \mathbf{X}_{T_i}, \mathbf{Z}_{[0, T_{i+1}-T_i]}^{(i)})_{i=1, \dots, N_t-1}\}$. This implies that for $T_i \le t < T_{i+1}$, if $s \in [T_i, t]$,

$$ \mathbb{E}(M_{t-T_i}^{(i)} | \mathcal{F}_s) = \mathbb{E}(M_{t-T_i}^{(i)} | T_i, \mathcal{F}_{s-T_i}^{\mathbf{Z}^{(i)}}) = M_{s-T_i}^{(i)}, $$

while if $s < T_i$, $\mathbb{E}(M_{t-T_i}^{(i)} | \mathcal{F}_s) = 0$.

Using these properties, we can easily verified that $M_L^{N, \vartheta}$ given by (11) is
where $\bar{v}_n(\vartheta, \mathbf{X}_s)$ stands for $\bar{v}_n(\vartheta, \mathbf{Z}_{s-T_i}^{(i)}, \mathbf{m}^{(i)})$, with $n = n(\mathbf{X}_s)$, and similarly for $a_n^{-1}(\mathbf{X}_s)$. Then we have

$$ 
\begin{aligned}
\log \frac{L(\vartheta_N, \mathbf{X}_{[T_i, T_{i+1}]} )}{L(\vartheta, \mathbf{X}_{[T_i, T_{i+1}]} )} \\
&= \int_{T_i}^{T_{i+1}} (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (d\mathbf{Z}_{s-T_i}^{(i)} - \bar{v}_n(\vartheta, \mathbf{X}_s) ds) \\
&\quad - \frac{1}{2} \int_{T_i}^{T_{i+1}} (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) ds.
\end{aligned}
$$

Consequently, we may rewrite (28) as

$$ 
\begin{aligned}
\Lambda_L^{\vartheta_N/\vartheta}(t) = & \sum_{i \ge 0} \int_0^t \mathbf{1}_{[T_i, T_{i+1}]} (s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) dM_{s-T_i}^{(i)} \\
& - \frac{1}{2} \int_0^t (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s)) ds.
\end{aligned}
\quad (43) $$

where $M_s^{(i)} = 0$ if $s < 0$ and

$$ M_s^{(i)} = \mathbf{Z}_s^{(i)} - \mathbf{Z}_0^{(i)} - \int_0^s \bar{v}_n(\vartheta, \mathbf{Z}_u^{(i)}) du, \quad s \ge 0. \quad (44) $$

We know that $\mathbf{Z}^{(i)}$ is the solution of $M^{|n}(\mathbf{z}^{(i)}, \mathbf{m}^{(i)})$ given by (4), where $(\mathbf{z}^{(i)}, \mathbf{m}^{(i)})$ are such that $\mathbf{X}_{T_i} = \{(\mathbf{z}^{(i)}, \mathbf{m}^{(i)})\}$ and $n = n(\mathbf{X}_{T_i})$. This implies that $(M_s^{(i)})$ is a martingale with respect to $\mathcal{F}_t^{\mathbf{Z}^{(i)}}$, the natural filtration associated to $\mathbf{Z}^{(i)}$. Note that $(M_s^{(i)})$ is independent from $T_i$ and that $\mathcal{F}_t$ is generated by $\{T_{N_t}, \mathbf{X}_{T_{N_t}}, \mathbf{Z}_{[0, t-T_{N_s}]}^{(N_t)}, (T_i, \mathbf{X}_{T_i}, \mathbf{Z}_{[0, T_{i+1}-T_i]}^{(i)})_{i=1, \dots, N_t-1}\}$. This implies that for $T_i \le t < T_{i+1}$, if $s \in [T_i, t]$,

$$ \mathbb{E}(M_{t-T_i}^{(i)} | \mathcal{F}_s) = \mathbb{E}(M_{t-T_i}^{(i)} | T_i, \mathcal{F}_{s-T_i}^{\mathbf{Z}^{(i)}}) = M_{s-T_i}^{(i)}, $$

while if $s < T_i$, $\mathbb{E}(M_{t-T_i}^{(i)} | \mathcal{F}_s) = 0$.

Using these properties, we can easily verified that $M_L^{N, \vartheta}$ given by (11) is
a $\mathcal{F}_{tN}$-martingale. Similarly, since $\langle M^{(i)} \rangle_s = \int_0^s a_n(\mathbf{Z}_u^{(i)}) du$, we have

$$ \begin{aligned} \langle M_L^{N, \vartheta} \rangle (t) &= \frac{1}{N} \sum_{i \ge 0} \int_0^{tN} \mathbf{1}_{[T_i; T_{i+1}]} (s) (\nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) \\ & \qquad \times a_n(\mathbf{X}_s) a_n^{-1}(\mathbf{X}_s) (\mathbf{X}_s) \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) ds \\ &= \frac{1}{N} \int_0^{tN} (\nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) ds. \end{aligned} $$

From this representation, (11) and (43), we have

$$ \mathrm{Rem}_{N, \vartheta, h}^{L}(t) = \Lambda_{L}^{\vartheta N / \vartheta}(tN) - h^{T} M_{L}^{N, \vartheta}(t) + \frac{1}{2} h^{T} \langle M_{L}^{N, \vartheta} \rangle(t) h = S_{t}^{N} - \frac{1}{2} R_{t}^{N}, $$

where

$$ S_t^N = \sum_{i \ge 0} \int_0^{tN} \mathbf{1}_{[T_i; T_{i+1}]} (s) \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right)^T a_n^{-1}(\mathbf{X}_s) dM_{s-T_i}^{(i)}, $$

and

$$ R_t^N = \int_0^{tN} \left( (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s)) - \frac{1}{N} h^T (\nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) h \right) ds. $$

Let us prove that both $S_t^N$ and $R_t^N$ tend to 0 in $\mathbb{P}_{\mathbf{x}_0}^\vartheta$-probability. For $S_t^N$, using the Markov inequality and the fact that $\mathbb{E}_{\mathbf{x}_0}^\vartheta((S_t^N)^2) = \mathbb{E}_{\mathbf{x}_0}^\vartheta(\langle S_t^N \rangle)$, this boils down to proving that $\mathbb{E}_{\mathbf{x}_0}^\vartheta(\langle S_t^N \rangle)$ tends to 0. We have, since $a_n = \bar{\sigma}_n \bar{\sigma}_n^T$,

$$ \begin{aligned} \langle S_t^N \rangle &= \int_0^{tN} \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right)^T a_n^{-1}(\mathbf{X}_s) \\ & \qquad \times \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right) ds \\ &= \int_0^{tN} \left\| \bar{\sigma}_n(\mathbf{X}_s)^{-1} \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right) \right\|^2 ds. \end{aligned} $$
a $\mathcal{F}_{tN}$-martingale. Similarly, since $\langle M^{(i)} \rangle_s = \int_0^s a_n(\mathbf{Z}_u^{(i)}) du$, we have

$$ \begin{aligned} \langle M_L^{N, \vartheta} \rangle (t) &= \frac{1}{N} \sum_{i \ge 0} \int_0^{tN} \mathbf{1}_{[T_i; T_{i+1}]} (s) (\nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) \\ & \qquad \times a_n(\mathbf{X}_s) a_n^{-1}(\mathbf{X}_s) (\mathbf{X}_s) \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) ds \\ &= \frac{1}{N} \int_0^{tN} (\nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) ds. \end{aligned} $$

From this representation, (11) and (43), we have

$$ \mathrm{Rem}_{N, \vartheta, h}^{L}(t) = \Lambda_{L}^{\vartheta N / \vartheta}(tN) - h^{T} M_{L}^{N, \vartheta}(t) + \frac{1}{2} h^{T} \langle M_{L}^{N, \vartheta} \rangle(t) h = S_{t}^{N} - \frac{1}{2} R_{t}^{N}, $$

where

$$ S_t^N = \sum_{i \ge 0} \int_0^{tN} \mathbf{1}_{[T_i; T_{i+1}]} (s) \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right)^T a_n^{-1}(\mathbf{X}_s) dM_{s-T_i}^{(i)}, $$

and

$$ R_t^N = \int_0^{tN} \left( (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s)) - \frac{1}{N} h^T (\nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s))^T a_n^{-1}(\mathbf{X}_s) \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) h \right) ds. $$

Let us prove that both $S_t^N$ and $R_t^N$ tend to 0 in $\mathbb{P}_{\mathbf{x}_0}^\vartheta$-probability. For $S_t^N$, using the Markov inequality and the fact that $\mathbb{E}_{\mathbf{x}_0}^\vartheta((S_t^N)^2) = \mathbb{E}_{\mathbf{x}_0}^\vartheta(\langle S_t^N \rangle)$, this boils down to proving that $\mathbb{E}_{\mathbf{x}_0}^\vartheta(\langle S_t^N \rangle)$ tends to 0. We have, since $a_n = \bar{\sigma}_n \bar{\sigma}_n^T$,

$$ \begin{aligned} \langle S_t^N \rangle &= \int_0^{tN} \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right)^T a_n^{-1}(\mathbf{X}_s) \\ & \qquad \times \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right) ds \\ &= \int_0^{tN} \left\| \bar{\sigma}_n(\mathbf{X}_s)^{-1} \left( \bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s) - \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) \frac{h}{\sqrt{N}} \right) \right\|^2 ds. \end{aligned} $$
Using the same argument as for (34), we get

$$ \langle S_t^N \rangle \le \frac{\|h\|^2}{N} \int_0^{tN} \|\bar{\sigma}_n(\mathbf{X}_s)^{-1}\|^2 f_{\vartheta}^{\text{move}}\left(\mathbf{X}_s, \frac{\|h\|}{\sqrt{N}}\right) ds. $$

where $f_{\vartheta}^{\text{move}}$ is given by (8). We then deduce that $\mathbb{E}_{\mathbf{x}_0}^{\vartheta}(\langle S^N \rangle_t)$ tends to 0 thanks to the dominated convergence theorem, using Conditions **B$_{move}$** and **(H)** as in the first step of the proof of Lemma 1.

Concerning $R_t^N$, note that

$$ R_t^N = \int_0^{tN} \|\bar{\sigma}_n(\mathbf{X}_s)^{-1} (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))\|^2 \\ - \frac{1}{N} \|\bar{\sigma}_n(\mathbf{X}_s)^{-1} \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) h\|^2 ds, $$

that reads $\int (\|f\|^2 - \|g\|^2)ds$ with obvious notation, so that using the same argument as in the third step of the proof of Lemma 1 (term $R_2$), it suffices to show that $\int \|f-g\|^2ds = o(1)$ and $\int \|g\|^2ds = O(1)$ in $\mathbb{P}_{\mathbf{x}_0}^{\vartheta}$-probability. But the former is exactly $\langle S_t^N \rangle$ studied above, that has been proven to converge to 0 in probability. And the expectation of the latter is bounded by use of Condition **A$_{move}$**, proving that it is $O(1)$ in $\mathbb{P}_{\mathbf{x}_0}^{\vartheta}$-probability. This concludes the proof.

# C Ancillary results

## C.1 Conditions for geometric ergodicity

The following proposition provides some conditions ensuring the hypothesis (H), namely non-explosion and geometric ergodicity of the BDMM process. It is based on the study conducted in [14] for birth-death-move processes (without mutations). The arguments in presence of mutations are similar and sketched below. The conditions include the technical assumption that the process is Feller. This property is discussed in [14] and proved for several examples of transition kernels that include all examples of this paper, provided the underlying spatial state $\Lambda$ is compact. The other conditions deal with the intensity functions of the jumps. First, the total intensity function $\alpha$ is assumed to be bounded, to avoid explosion of the process. Second, the death intensity function $\delta$ must compensate in a proper way the birth intensity
Using the same argument as for (34), we get

$$ \langle S_t^N \rangle \le \frac{\|h\|^2}{N} \int_0^{tN} \|\bar{\sigma}_n(\mathbf{X}_s)^{-1}\|^2 f_{\vartheta}^{\text{move}}\left(\mathbf{X}_s, \frac{\|h\|}{\sqrt{N}}\right) ds. $$

where $f_{\vartheta}^{\text{move}}$ is given by (8). We then deduce that $\mathbb{E}_{\mathbf{x}_0}^{\vartheta}(\langle S^N \rangle_t)$ tends to 0 thanks to the dominated convergence theorem, using Conditions **B$_{move}$** and **(H)** as in the first step of the proof of Lemma 1.

Concerning $R_t^N$, note that

$$ R_t^N = \int_0^{tN} \|\bar{\sigma}_n(\mathbf{X}_s)^{-1} (\bar{v}_n(\vartheta_N, \mathbf{X}_s) - \bar{v}_n(\vartheta, \mathbf{X}_s))\|^2 \\ - \frac{1}{N} \|\bar{\sigma}_n(\mathbf{X}_s)^{-1} \nabla_1 \bar{v}_n(\vartheta, \mathbf{X}_s) h\|^2 ds, $$

that reads $\int (\|f\|^2 - \|g\|^2)ds$ with obvious notation, so that using the same argument as in the third step of the proof of Lemma 1 (term $R_2$), it suffices to show that $\int \|f-g\|^2ds = o(1)$ and $\int \|g\|^2ds = O(1)$ in $\mathbb{P}_{\mathbf{x}_0}^{\vartheta}$-probability. But the former is exactly $\langle S_t^N \rangle$ studied above, that has been proven to converge to 0 in probability. And the expectation of the latter is bounded by use of Condition **A$_{move}$**, proving that it is $O(1)$ in $\mathbb{P}_{\mathbf{x}_0}^{\vartheta}$-probability. This concludes the proof.

# C Ancillary results

## C.1 Conditions for geometric ergodicity

The following proposition provides some conditions ensuring the hypothesis (H), namely non-explosion and geometric ergodicity of the BDMM process. It is based on the study conducted in [14] for birth-death-move processes (without mutations). The arguments in presence of mutations are similar and sketched below. The conditions include the technical assumption that the process is Feller. This property is discussed in [14] and proved for several examples of transition kernels that include all examples of this paper, provided the underlying spatial state $\Lambda$ is compact. The other conditions deal with the intensity functions of the jumps. First, the total intensity function $\alpha$ is assumed to be bounded, to avoid explosion of the process. Second, the death intensity function $\delta$ must compensate in a proper way the birth intensity
function $\beta$. Notably, the ergodic properties do not depend on the inter-jump move process, neither on the mutation dynamics. We introduce the following notation:

$$ \beta_n = \sup_{\mathbf{x} \in E_n} \beta(\mathbf{x}), \quad \delta_n = \inf_{\mathbf{x} \in E_n} \delta(\mathbf{x}) \quad \text{and} \quad \alpha_n = \beta_n + \delta_n. \quad (45) $$

In fact the conditions of geometric ergodicity below are exactly the same as the conditions of ergodicity for a simple birth-death process $(\eta_t)_{t \ge 0}$ on $\mathbb{N}$ with birth rate $(\beta_n)$ and death rate $(\delta_n)$, as established in [11]. This is because to address the ergodic properties of a BDMM $(\mathbf{X}_t)_{t \ge 0}$, we construct a coupling between $(\mathbf{X}_t)_{t \ge 0}$ and $(\eta_t)_{t \ge 0}$, in such a way that $\eta_t = 0$ implies $\mathbf{X}_t = \emptyset$. Consequently $\emptyset$ becomes a positive recurrent state for $(\mathbf{X}_t)_{t \ge 0}$ whenever 0 is a positive recurrent state for $(\eta_t)_{t \ge 0}$, which implies the geometric ergodicity of $(\mathbf{X}_t)_{t \ge 0}$.

**Proposition 1.** Let $(\mathbf{X}_t)_{t \ge 0}$ be a Feller BDMM process with a bounded intensity function $\alpha$. Suppose that $\delta_n > 0$ for all $n \ge 1$ and one of the following condition holds:

$$ (i) \text{ there exists } n_0 \ge 1 \text{ such that } \beta_n = 0 \text{ for any } n \ge n_0, \quad (46) $$

$$ (ii) \beta_n > 0 \text{ for all } n \ge 1, \quad \sum_{n=2}^{\infty} \frac{\beta_1 \dots \beta_{n-1}}{\delta_1 \dots \delta_n} < \infty \quad \text{and} \quad \sum_{n=1}^{\infty} \frac{\delta_1 \dots \delta_n}{\beta_1 \dots \beta_n} = \infty, \quad (47) $$

where $\beta_n$ and $\delta_n$ are defined by (45). Assume moreover that

$$ \sum_{n=2}^{\infty} \sqrt{\frac{\beta_1 \dots \beta_{n-1}}{\delta_1 \dots \delta_n}} < \infty \quad \text{and} \quad \exists N \ge 0, \text{ s.t. } \forall n \ge N, \beta_n \le \delta_{n+1}. $$

Then (H) is satisfied.

As mentioned above, following [25] and [14], the proof of this proposition is based on a coupling between $(\mathbf{X}_t)_{t \ge 0}$ and $(\eta_t)_{t \ge 0}$, where the latter is a simple birth-death process with birth and death rates given by (45). We detail below how we construct the coupled process $\tilde{C}_t = (\mathbf{X}_t, \eta_t)$. This is a straightforward generalisation of the construction in [14], where we account for the presence of mutations. Specifically, $\tilde{C}$ is a jump move process on
function $\beta$. Notably, the ergodic properties do not depend on the inter-jump move process, neither on the mutation dynamics. We introduce the following notation:

$$ \beta_n = \sup_{\mathbf{x} \in E_n} \beta(\mathbf{x}), \quad \delta_n = \inf_{\mathbf{x} \in E_n} \delta(\mathbf{x}) \quad \text{and} \quad \alpha_n = \beta_n + \delta_n. \quad (45) $$

In fact the conditions of geometric ergodicity below are exactly the same as the conditions of ergodicity for a simple birth-death process $(\eta_t)_{t \ge 0}$ on $\mathbb{N}$ with birth rate $(\beta_n)$ and death rate $(\delta_n)$, as established in [11]. This is because to address the ergodic properties of a BDMM $(\mathbf{X}_t)_{t \ge 0}$, we construct a coupling between $(\mathbf{X}_t)_{t \ge 0}$ and $(\eta_t)_{t \ge 0}$, in such a way that $\eta_t = 0$ implies $\mathbf{X}_t = \emptyset$. Consequently $\emptyset$ becomes a positive recurrent state for $(\mathbf{X}_t)_{t \ge 0}$ whenever 0 is a positive recurrent state for $(\eta_t)_{t \ge 0}$, which implies the geometric ergodicity of $(\mathbf{X}_t)_{t \ge 0}$.

**Proposition 1.** Let $(\mathbf{X}_t)_{t \ge 0}$ be a Feller BDMM process with a bounded intensity function $\alpha$. Suppose that $\delta_n > 0$ for all $n \ge 1$ and one of the following condition holds:

$$ (i) \text{ there exists } n_0 \ge 1 \text{ such that } \beta_n = 0 \text{ for any } n \ge n_0, \quad (46) $$

$$ (ii) \beta_n > 0 \text{ for all } n \ge 1, \quad \sum_{n=2}^{\infty} \frac{\beta_1 \dots \beta_{n-1}}{\delta_1 \dots \delta_n} < \infty \quad \text{and} \quad \sum_{n=1}^{\infty} \frac{\delta_1 \dots \delta_n}{\beta_1 \dots \beta_n} = \infty, \quad (47) $$

where $\beta_n$ and $\delta_n$ are defined by (45). Assume moreover that

$$ \sum_{n=2}^{\infty} \sqrt{\frac{\beta_1 \dots \beta_{n-1}}{\delta_1 \dots \delta_n}} < \infty \quad \text{and} \quad \exists N \ge 0, \text{ s.t. } \forall n \ge N, \beta_n \le \delta_{n+1}. $$

Then (H) is satisfied.

As mentioned above, following [25] and [14], the proof of this proposition is based on a coupling between $(\mathbf{X}_t)_{t \ge 0}$ and $(\eta_t)_{t \ge 0}$, where the latter is a simple birth-death process with birth and death rates given by (45). We detail below how we construct the coupled process $\tilde{C}_t = (\mathbf{X}_t, \eta_t)$. This is a straightforward generalisation of the construction in [14], where we account for the presence of mutations. Specifically, $\tilde{C}$ is a jump move process on
$E \times \mathbb{N}$ with intensity function $\tilde{\alpha}$ and transition kernel $\tilde{K}$ defined as follows.
The intensity function $\tilde{\alpha} : E \times \mathbb{N} \to \mathbb{R}_+$ is given by

$$ \tilde{\alpha}(\mathbf{x}, n) = \begin{cases} \alpha(\mathbf{x}) + \alpha_n & \text{if } \mathbf{x} \in E_m, m \neq n, \\ \beta_n + \delta(\mathbf{x}) + \tau(\mathbf{x}) & \text{if } \mathbf{x} \in E_n. \end{cases} $$

Letting $K$ be the kernel given by (21), the transition kernel $\tilde{K}$ takes the form,
for any $A \subset E$:

1. If $\mathbf{x} \in E_m, m \neq n$:

$$ \tilde{K}((\mathbf{x}, n); A \times \{n\}) = \frac{\alpha(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)} K(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); \{\mathbf{x}\} \times \{n+1\}) = \frac{\beta_n}{\tilde{\alpha}(\mathbf{x}, n)}; $$

$$ \tilde{K}((\mathbf{x}, n); \{\mathbf{x}\} \times \{n-1\}) = \frac{\delta_n}{\tilde{\alpha}(\mathbf{x}, n)}. $$

2. If $\mathbf{x} \in E_n$:

$$ \tilde{K}((\mathbf{x}, n); A \times \{n+1\}) = \frac{\beta(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)} K_{\beta}(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); \{\mathbf{x}\} \times \{n+1\}) = \frac{\beta_n - \beta(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)}; $$

$$ \tilde{K}((\mathbf{x}, n); A \times \{n-1\}) = \frac{\delta_n}{\tilde{\alpha}(\mathbf{x}, n)} K_{\delta}(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); A \times \{n\}) = \frac{\delta(\mathbf{x}) - \delta_n}{\tilde{\alpha}(\mathbf{x}, n)} K_{\delta}(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); A \times \{n\}) = \frac{\tau(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)} K_{\tau}(\mathbf{x}, A). $$

The inter-jump move process of $\tilde{C}$ is in turn a simple independent coupling between the move $\mathbf{Y}$ of $\mathbf{X}$ and a constant move on $\mathbb{N}$ (i.e. $y_t = y_0$ for all $t \ge 0$).

The fact that the above construction is a proper coupling, in the sense that $\mathbf{X}$ and $\eta$ do constitute the marginal distributions of $\tilde{C}$, can be proven
$E \times \mathbb{N}$ with intensity function $\tilde{\alpha}$ and transition kernel $\tilde{K}$ defined as follows.
The intensity function $\tilde{\alpha} : E \times \mathbb{N} \to \mathbb{R}_+$ is given by

$$ \tilde{\alpha}(\mathbf{x}, n) = \begin{cases} \alpha(\mathbf{x}) + \alpha_n & \text{if } \mathbf{x} \in E_m, m \neq n, \\ \beta_n + \delta(\mathbf{x}) + \tau(\mathbf{x}) & \text{if } \mathbf{x} \in E_n. \end{cases} $$

Letting $K$ be the kernel given by (21), the transition kernel $\tilde{K}$ takes the form,
for any $A \subset E$:

1. If $\mathbf{x} \in E_m, m \neq n$:

$$ \tilde{K}((\mathbf{x}, n); A \times \{n\}) = \frac{\alpha(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)} K(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); \{\mathbf{x}\} \times \{n+1\}) = \frac{\beta_n}{\tilde{\alpha}(\mathbf{x}, n)}; $$

$$ \tilde{K}((\mathbf{x}, n); \{\mathbf{x}\} \times \{n-1\}) = \frac{\delta_n}{\tilde{\alpha}(\mathbf{x}, n)}. $$

2. If $\mathbf{x} \in E_n$:

$$ \tilde{K}((\mathbf{x}, n); A \times \{n+1\}) = \frac{\beta(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)} K_{\beta}(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); \{\mathbf{x}\} \times \{n+1\}) = \frac{\beta_n - \beta(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)}; $$

$$ \tilde{K}((\mathbf{x}, n); A \times \{n-1\}) = \frac{\delta_n}{\tilde{\alpha}(\mathbf{x}, n)} K_{\delta}(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); A \times \{n\}) = \frac{\delta(\mathbf{x}) - \delta_n}{\tilde{\alpha}(\mathbf{x}, n)} K_{\delta}(\mathbf{x}, A); $$

$$ \tilde{K}((\mathbf{x}, n); A \times \{n\}) = \frac{\tau(\mathbf{x})}{\tilde{\alpha}(\mathbf{x}, n)} K_{\tau}(\mathbf{x}, A). $$

The inter-jump move process of $\tilde{C}$ is in turn a simple independent coupling between the move $\mathbf{Y}$ of $\mathbf{X}$ and a constant move on $\mathbb{N}$ (i.e. $y_t = y_0$ for all $t \ge 0$).

The fact that the above construction is a proper coupling, in the sense that $\mathbf{X}$ and $\eta$ do constitute the marginal distributions of $\tilde{C}$, can be proven
exactly as in [14] for the case without mutations. On the other hand, the key point is that if at some point, $C_t = (\mathbf{X}_t, \eta_t)$ is such that $n(\mathbf{X}_t) \le \eta_t$, then by the above construction, $n(\mathbf{X}_s) \le \eta_s$ for all $s \ge t$, almost surely. By this property, $\eta_t = 0$ implies $\mathbf{X}_t = \emptyset$ and the geometric ergodic conditions for $\eta$ stated in the proposition, coming from [11], are sufficient for the geometric ergodicity of $\mathbf{X}$. The rigorous proof of these claims can be found in [14].

## C.2 A useful martingale

The following result is useful in the proof of Theorem 2 in Section B. It is proven in Lemma 54 of [20], the BDMM being a particular case of a jump-move process (see the proof of Theorem 1 in Section A). It is also stated under a slightly different setting in Proposition 3.3 (b) of [18].

**Lemma 4.** Let $(\mathbf{X}_t)_{t \ge 0}$ be a BDMM process on E, let g be a measurable bounded function defined on $E \times E$ and introduce for any $t > 0$

$$M_t^* = \int_0^t g(\mathbf{X}_{s^{-}}, \mathbf{X}_s) dN_s^\gamma - \int_0^t \gamma(\mathbf{X}_s) \int_E g(\mathbf{X}_s, \mathbf{y}) k_\gamma(\mathbf{X}_s, \mathbf{y}) \nu_\gamma(\mathbf{X}_s, d\mathbf{y}) ds.$$

If $N_t^\gamma < \infty$ for any $t \ge 0$, then $(M_t^*)_{t \ge 0}$ is a $\mathcal{F}_t$-martingale.

## References

[1] ATHREYA, K. B., AND NEY, P. E. *Branching Processes*, vol. 196. Springer Science & Business Media, 2012.

[2] BALSOLLIER, L., LAVANCIER, F., SALAMERO, J., AND KERVRANN, C. A generative model to simulate spatiotemporal dynamics of biomolecules in cells. *Biological Imaging* 3 (2023), e22.

[3] BANSAYE, V., AND MÉLÉARD, S. *Stochastic models for structured populations*, vol. 16. Springer, 2015.

[4] BOULANGER, J., GUEUDRY, C., MUNCH, D., CINQUIN, B., PAUL-GILLOTEAUX, P., BARDIN, S., GUÉRIN, C., SENGER, F., BLANCHOIN, L., AND SALAMERO, J. Fast high-resolution 3D total internal reflection fluorescence microscopy by incidence angle scanning and azimuthal averaging. *Proc Natl Acad Sci USA* 111, 48 (2014), 17164–17169.
exactly as in [14] for the case without mutations. On the other hand, the key point is that if at some point, $C_t = (\mathbf{X}_t, \eta_t)$ is such that $n(\mathbf{X}_t) \le \eta_t$, then by the above construction, $n(\mathbf{X}_s) \le \eta_s$ for all $s \ge t$, almost surely. By this property, $\eta_t = 0$ implies $\mathbf{X}_t = \emptyset$ and the geometric ergodic conditions for $\eta$ stated in the proposition, coming from [11], are sufficient for the geometric ergodicity of $\mathbf{X}$. The rigorous proof of these claims can be found in [14].

## C.2 A useful martingale

The following result is useful in the proof of Theorem 2 in Section B. It is proven in Lemma 54 of [20], the BDMM being a particular case of a jump-move process (see the proof of Theorem 1 in Section A). It is also stated under a slightly different setting in Proposition 3.3 (b) of [18].

**Lemma 4.** Let $(\mathbf{X}_t)_{t \ge 0}$ be a BDMM process on $E$, let $g$ be a measurable bounded function defined on $E \times E$ and introduce for any $t > 0$

$$M_t^* = \int_0^t g(\mathbf{X}_{s-}, \mathbf{X}_s) dN_s^\gamma - \int_0^t \gamma(\mathbf{X}_s) \int_E g(\mathbf{X}_s, \mathbf{y}) k_\gamma(\mathbf{X}_s, \mathbf{y}) \nu_\gamma(\mathbf{X}_s, d\mathbf{y}) ds.$$

If $N_t^\gamma < \infty$ for any $t \ge 0$, then $(M_t^*)_{t \ge 0}$ is a $\mathcal{F}_t$-martingale.

## References

[1] ATHREYA, K. B., AND NEY, P. E. *Branching Processes*, vol. 196. Springer Science & Business Media, 2012.

[2] BALSOLLIER, L., LAVANCIER, F., SALAMERO, J., AND KERVRANN, C. A generative model to simulate spatiotemporal dynamics of biomolecules in cells. *Biological Imaging* 3 (2023), e22.

[3] BANSAYE, V., AND MÉLÉARD, S. *Stochastic models for structured populations*, vol. 16. Springer, 2015.

[4] BOULANGER, J., GUEUDRY, C., MUNCH, D., CINQUIN, B., PAUL-GILLOTEAUX, P., BARDIN, S., GUÉRIN, C., SENGER, F., BLANCHOIN, L., AND SALAMERO, J. Fast high-resolution 3D total internal reflection fluorescence microscopy by incidence angle scanning and azimuthal averaging. *Proc Natl Acad Sci USA* 111, 48 (2014), 17164–17169.
[5] BRIANE, V., KERVRANN, C., AND VIMOND, M. Statistical analysis of particle trajectories in living cells. *Phys Rev E.* 97, (6-1):062121 (2018).

[6] BRIANE, V., VIMOND, M., AND KERVRANN, C. An overview of diffusion models for intracellular dynamics analysis. *Briefings in bioinformatics* 21, 4 (2020), 1136–1150.

[7] FATTLER, T., AND GROTHAUS, M. Strong feller properties for distorted brownian motion with reflecting boundary condition and an application to continuous n-particle systems with singular interactions. *Journal of Functional Analysis* 246, 2 (2007), 217–241.

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