Importance sampling for a robust and efficient multilevel Monte Carlo estimator for stochastic reaction networks

Abstract

The multilevel Monte Carlo (MLMC) method for continuous-time Markov chains, first introduced by Anderson and Higham (SIAM Multiscal Model Simul 10(1):146–179, 2012), is a highly efficient simulation technique that can be used to estimate various statistical quantities for stochastic reaction networks, in particular for stochastic biological systems. Unfortunately, the robustness and performance of the multilevel method can be affected by the high kurtosis, a phenomenon observed at the deep levels of MLMC, which leads to inaccurate estimates of the sample variance. In this work, we address cases where the high-kurtosis phenomenon is due to catastrophic coupling (characteristic of pure jump processes where coupled consecutive paths are identical in most of the simulations, while differences only appear in a tiny proportion) and introduce a pathwise-dependent importance sampling (IS) technique that improves the robustness and efficiency of the multilevel method. Our theoretical results, along with the conducted numerical experiments, demonstrate that our proposed method significantly reduces the kurtosis of the deep levels of MLMC, and also improves the strong convergence rate from $\beta =1$ for the standard case (without IS), to $\beta =1+\delta$, where $0<\delta <1$ is a user-selected parameter in our IS algorithm. Due to the complexity theorem of MLMC, and given a pre-selected tolerance, $\text {TOL}$, this results in an improvement of the complexity from ${\mathcal {O}}\left( \text {TOL}^{-2} \log (\text {TOL})^2\right)$ in the standard case to ${\mathcal {O}}\left( \text {TOL}^{-2}\right)$, which is the optimal complexity of the MLMC estimator. We achieve all these improvements with a negligible additional cost since our IS algorithm is only applied a few times across each simulated path.

Appendices

Appendix A: Proofs of Lemma 4.1 and Theorems 4.1 and 4.2

Proof of Lemma 4.1

We denote by $K=\sum _{n \in S} k_n$, $L_{\ell }\left( j\right)$ the likelihood evaluated at $K=j$. Then, for $p\ge 1$ and $0\le \delta <1$, and using relation (4.4), we write

$$\begin{aligned}&\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left|\Delta g_{\ell }\right|^p(T) L_{\ell }^p ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] = \sum _{\underset{ i \in {\mathbb {N}}\setminus \{0\}}{\left|\Delta g_{\ell }(T)\right|=K =i}} i ^p L_{\ell }^p \left( i\right) {\overline{\pi }}_{\ell } \left( \left|\Delta g_{\ell }(T)\right|=i, K= i ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right) \nonumber \\&\qquad + \sum _{\underset{i \ne j, \, (i,j) \in {\mathbb {N}}^2}{\left|\Delta g_{\ell }(T)\right|=i,\, K =j}} i ^p L_{\ell }^p \left( j\right) {\overline{\pi }}_{\ell } \left( \left|\Delta g_{\ell }(T)\right|=i, K= j ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right) \nonumber \\&\quad = \sum _{\underset{ i \in {\mathbb {N}}\setminus \{0\}}{\left|\Delta g_{\ell }\right|=K =i}} \underset{A_{i}}{\underbrace{ i ^p e^{p(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} \Delta t_{\ell }^{i p \delta } {\overline{\pi }}_{\ell } \left( \left|\Delta g_{\ell }(T)\right|=i, K= i ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right) }}\nonumber \\&\qquad + \sum _{\underset{\underset{i \ne j, \, (i,j) \in {\mathbb {N}}^2}{K =j}}{\left|\Delta g_{\ell }(T)\right|=i}} \underset{B_{ij}}{\underbrace{i^p e^{p(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} \Delta t_{\ell }^{j p \delta } {\overline{\pi }}_{\ell } \left( \left|\Delta g_{\ell }(T)\right|=i, K= j ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right) }} =\sum _{\underset{ i \in {\mathbb {N}}\setminus \{0\}}{\left|\Delta g_{\ell }\right|=K =i}} A_i + \sum _{\underset{i \ne j, \, (i,j) \in {\mathbb {N}}^2}{\left|\Delta g_{\ell }\right|=i,\, K =j}} B_{ij}. \end{aligned}$$

(A.1)

Using Assumption 4.1(a), we have

$$\begin{aligned} 0 \le \frac{\sum _{\underset{ i \in {\mathbb {N}}\setminus \{0,1\}}{\left|\Delta g_{\ell }(T)\right|=K =i}} A_i}{A_1} \le \underset{\underset{\Delta t_{\ell } \rightarrow 0}{\longrightarrow 0 }}{\underbrace{\sum _{i \in {\mathbb {N}}\setminus \{0,1\}} i ^{p} \Delta t_{\ell }^{(i-1) p \delta }}}. \end{aligned}$$

(A.2)

Now, let us examine the second sum in the right-hand side of (A.1). First, observe that $B_{0j}=0, \, \forall j \ge 1$ and $B_{i0}=0, \, \forall i \ge 1$. Although the first observation is clear, we need to explain the second observation, which is mainly due to the fact that ${\overline{\pi }}\left( \left|\Delta g_{\ell }(T)\right|=i, K= 0 ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right) =0$, $\forall i \ge 1$. For the purpose of simplification, let us consider $g_{\ell }={\overline{X}}_{\ell }$; then considering the first interval in the coarse level, and using the coupling equation (2.6), we have: (i) At $t=0$: ${\overline{X}}_{\ell }(0)={\overline{X}}_{\ell -1}(0)$ and $\Delta a^1_{\ell -1,0}=0$. (ii) At $t=\Delta t_{\ell }$: ${\overline{X}}_{\ell }(\Delta t_{\ell })={\overline{X}}_{\ell -1}(\Delta t_{\ell })$ and $\Delta a^2_{\ell -1,0}=a({\overline{X}}_{\ell }(\Delta t_{\ell }))-a({\overline{X}}_{\ell -1}(0))$. (iii) At $t=t_1=2\Delta t_{\ell }$: if $\Delta a^2_{\ell -1,0}=0$, then we simulate this step under the old measure and consequently we will have ${\overline{X}}_{\ell }(t_1)={\overline{X}}_{\ell -1}(t_1)$ otherwise if $\Delta a^2_{\ell -1,0} \ne 0$, then we simulate this step under the IS measure, but since $j=0$, then we will have ${\overline{X}}_{\ell }(t_1)={\overline{X}}_{\ell -1}(t_1)$. Therefore, in both scenarios, we will have the same situation at the start, $t_0=0$. Therefore, we conclude that ${\overline{\pi }}\left( \left|\Delta g_{\ell }(T)\right|=i, K= 0 ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right) =0$, $\forall \, i \ge 1$ and $B_{i0}=0$, $\forall \, i \ge 1$.

Then, using Assumptions 4.1(b) and 4.1(c), we obtain

$$\begin{aligned}&0 \le \frac{\sum _{\underset{i \ne j, \, (i,j) \in {\mathbb {N}}^2}{\left|\Delta g_{\ell } (T)\right|=i,\, K =j}} B_{ij}}{A_1} \nonumber \\&\quad \le \frac{ \sum _{i \ne j, \, 1 \le i,j} i^p \Delta t_{\ell }^{j p\delta } {\overline{\pi }}_{\ell } \left\{ \left|\Delta g_{\ell } (T)\right|= i ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right\} }{\left( \Delta t_{\ell }^{p \delta } \right) e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*}\right) \left( 1+{o}\left( 1\right) \right) }\nonumber \\&\quad \le \frac{ \sum _{i \ne j, \, 1 \le i,j} \eta _{i,\ell } i^p \Delta t_{\ell }^{j p\delta } \Delta t_{\ell }^{i (1-\delta )} }{\left( \Delta t_{\ell }^{p \delta } \right) e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*}\right) \left( 1+{o}\left( 1\right) \right) }\nonumber \\&\quad = \underset{\underset{\Delta t_{\ell } \rightarrow 0}{\longrightarrow 0}}{\underbrace{ \left( 1+{o}\left( 1\right) \right) ^{-1} \left( e^{ \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \Delta a_{\ell ,n^*}^{-1} \sum _{i \ne j, \, 1 \le i,j} \eta _{i,\ell } i^p \Delta t_{\ell }^{p\delta ( j-1)} \Delta t_{\ell }^{(1-\delta ) ( i-1)} \right) }}. \end{aligned}$$

(A.3)

Therefore, using (A.1), (A.2) and (A.3), we conclude Lemma 4.1, that is

$$\begin{aligned}&\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left|\Delta g_{\ell }\right|^p(T) L_{\ell }^p ; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \\&\quad = \Delta t_{\ell }^{(p -1)\delta +1} \left( \Delta a_{\ell ,n^*}\right) e^{p(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} \\&\qquad e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( 1+h_{p,\ell }\right) , \end{aligned}$$

such that $h_{p,\ell } \underset{\Delta t_{\ell } \rightarrow 0}{\longrightarrow 0}$. $\square$

Proof of Theorem 4.1

Let $0 \le \delta < 1$. In the first step of the proof, we want to show that

$$\begin{aligned} \kappa _{\ell }:=\frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( Y_{\ell }-\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^4\right] }{\left( \text {Var}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^2} \underset{\Delta t_{\ell } \rightarrow 0}{\sim } \frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }{\left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] \right) ^2} \end{aligned}$$

Let us first show that $\text {Var}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \underset{\Delta t_{\ell } \rightarrow 0}{\sim } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right]$. In fact,

$$\begin{aligned} \frac{\text {Var}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] }=\frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] - \left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^2}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] }=1- \frac{ \left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^2}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] }. \end{aligned}$$

Therefore, we need to show that $I_1:=\frac{ \left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^2}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] } \underset{\Delta t_{\ell }\rightarrow 0}{\longrightarrow }0$.

Due to the order one weak error convergence, there exists a constant $d_1>0$ such that $\left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) \le d_1 \Delta t_{\ell }$. Therefore, using Lemma 4.1 and Assumption 4.2, we obtain

$$\begin{aligned}&0 \le I_1\le \frac{d_1^2 \Delta t_{\ell }^2}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] }\\&\quad = \frac{d_1^2 \Delta t_{\ell }^2}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] }\\&\quad =\frac{d_1^2 \Delta t_{\ell }^2}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( \Delta t_{\ell }^{\delta +1} \right) e^{2(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta a_{\ell ,n^*}\right) \left( 1+ h_{2,\ell }\right) \right] }\\&\quad \le \frac{d_1^2 \Delta t_{\ell }^{1-\delta }}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \Delta a_{\ell ,n^*} \right] }\le \frac{d_1^2 \Delta t_{\ell }^{1-\delta }}{C_1} \underset{\Delta t_{\ell }\rightarrow 0}{\longrightarrow }0. \end{aligned}$$

Therefore, we conclude that

$$\begin{aligned} \text {Var}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \underset{\Delta t_{\ell } \rightarrow 0}{\sim } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] . \end{aligned}$$

(A.4)

Now, let us show that $\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( Y_{\ell }-\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^4\right] \underset{\Delta t_{\ell } \rightarrow 0}{\sim } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right]$. In fact,

$$\begin{aligned}&\frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( Y_{\ell }-\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^4\right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }\\&\quad =\frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] -4 \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^3\right] \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] + 6 \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] \left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^2 -3 \left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^4}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }\\&\quad =1-4\frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^3\right] \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }+ 6\frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] \left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^2}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }- 3 \frac{\left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^4}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }\\&\quad =1-4 I_2+ 6 I_3- 3 I_4 \end{aligned}$$

Therefore, we need to show that $I_2, I_3, I_4 \underset{\Delta t_{\ell }\rightarrow 0}{\longrightarrow }0$.

Using Lemma 4.1 and Assumption 4.2, we obtain

$$\begin{aligned}&0 \le I_4\le \frac{d_1^4 \Delta t_{\ell }^4}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }\\&\quad = \frac{d_1^4 \Delta t_{\ell }^4}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] }\\&\quad =\frac{d_1^4 \Delta t_{\ell }^4}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( \Delta t_{\ell }^{3\delta +1} \right) e^{4(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta a_{\ell ,n^*}\right) \left( 1+ h_{4,\ell }\right) \right] }\\&\quad \le \frac{d_1^4 \Delta t_{\ell }^{3(1-\delta )}}{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \Delta a_{\ell ,n^*} \right] }\le \frac{d_1^4 \Delta t_{\ell }^{3(1-\delta )}}{C_1}\underset{\Delta t_{\ell }\rightarrow 0}{\longrightarrow }0. \end{aligned}$$

Similarly for $I_2$, using Lemma 4.1 and Assumptions 4.2 and 4.3, we obtain

$$\begin{aligned} 0 \le I_2&\le \frac{d_1 \Delta t_{\ell } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left|Y_{\ell }\right|^3\right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }\\&= \frac{d_1 \Delta t_{\ell } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^3; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] }\\&=\frac{d_1 \Delta t_{\ell } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( \Delta t_{\ell }^{2 \delta +1} \right) e^{3(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta a_{\ell ,n^*}\right) \left( 1+ h_{3,\ell }\right) \right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( \Delta t_{\ell }^{3 \delta +1} \right) e^{4(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta a_{\ell ,n^*} \right) } \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*}\right) \left( 1+h_{4,\ell }\right) \right] }\\&\le \frac{d_1 \Delta t_{\ell }^{1-\delta } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{3(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} \Delta a_{\ell ,n^*}\left( 1+ h_{3,\ell }\right) \right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \Delta a_{\ell ,n^*} \right] }\\&\le C \Delta t_{\ell }^{1-\delta } \underset{\Delta t_{\ell }\rightarrow 0}{\longrightarrow } 0. \end{aligned}$$

Fig. 20

Example 5.1 with IS (with $\delta =0.5$): Histogram of $\# \{n \in {\mathcal {S}}: k_n>0\}$, for number of samples $M_{\ell }=10^5$. a $\ell =6$. b $\ell =10$

Finally, for $I_3$, using Lemma 4.1 and Assumptions 4.2 and 4.3, we obtain

$$\begin{aligned}&0 \le I_3\le \frac{d_1^2 \Delta t_{\ell }^2 \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left|Y_{\ell }\right|^2\right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }\\&\quad = \frac{d_1^2 \Delta t_{\ell }^2 \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] }\\&\quad =\frac{d_1^2 \Delta t_{\ell }^2 \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( \Delta t_{\ell }^{ \delta +1} \right) e^{2(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta a_{\ell ,n^*}\right) \left( 1+ h_{2,\ell }\right) \right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( \Delta t_{\ell }^{3 \delta +1} \right) e^{4(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta a_{\ell ,n^*}\right) \left( 1+ h_{4,\ell }\right) \right] }\\&\quad \le \frac{d_1^2 \Delta t_{\ell }^{2(1-\delta )} \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{2(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} \Delta a_{\ell ,n^*} \left( 1+ h_{2,\ell }\right) \right] }{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \Delta a_{\ell ,n^*} \right] }\\&\quad \le {\widetilde{C}} \Delta t_{\ell }^{2(1-\delta )} \underset{\Delta t_{\ell }\rightarrow 0}{\longrightarrow } 0. \end{aligned}$$

Fig. 21

Example 5.2 with IS (with $\delta =0.5$): Histogram of $\# \{ \text {IS steps : s.t.}\, {\overline{K}}:=\sum _{j \in {\mathcal {J}}_1} \sum _{n \in {\mathcal {S}}_j}k^j_n>0\}$, for number of samples $M_{\ell }=10^5$. a $\ell =4$. b $\ell =8$

Fig. 22

Example 5.3 with IS (with $\delta =0.5$): Histogram of $\# \{ \text {IS steps : s.t.}\, {\overline{K}}:=\sum _{j \in {\mathcal {J}}_1} \sum _{n \in {\mathcal {S}}_j}k^j_n>0\}$, for number of samples $M_{\ell }=10^5$. a) $\ell =6$. b) $\ell =10$

Therefore, we conclude that

$$\begin{aligned} \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( Y_{\ell }-\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^4\right] \underset{\Delta t_{\ell } \rightarrow 0}{\sim } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] . \end{aligned}$$

(A.5)

Finally, using (A.4), (A.5), Lemma 4.1 and Assumptions 4.2 and 4.3, we obtain

$$\begin{aligned}&\kappa _{\ell }:=\frac{ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \left( Y_{\ell }-\mathrm {E}\left[ Y_{\ell }\right] \right) ^4\right] }{\left( \text {Var}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }\right] \right) ^2} \underset{\Delta t_{\ell } \rightarrow 0}{\sim } \frac{ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4\right] }{\left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] \right) ^2}\nonumber \\&\quad =\frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^4(T) ; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] }{\left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2(T) ; \left( {\mathcal {F}}_{N_{\ell }}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] \right) ^2} \nonumber \\&\quad = \Delta t_{\ell }^{\delta -1} \frac{\mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{4(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \Delta a_{\ell ,n^*} \left( 1+ h_{4,\ell }\right) \right] }{\left( \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ e^{2(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \Delta a_{\ell ,n^*} \left( 1+ h_{2,\ell }\right) \right] \right) ^2} \nonumber \\&\quad ={\mathcal {O}}\left( \Delta t_{\ell }^{\delta -1}\right) . \end{aligned}$$

(A.6)

$\square$

Proof of Theorem 4.2

Let $0< \delta < 1$. Then, using (A.4), Lemma 4.1 and Assumptions 4.2 and 4.3, we obtain

$$\begin{aligned}&\text {Var}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell } \right] \underset{\Delta t_{\ell } \rightarrow 0}{\sim } \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2\right] \nonumber \\&\quad = \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ \mathrm {E}_{{\overline{\pi }}_{\ell }}\left[ Y_{\ell }^2; \left( {\mathcal {F}}_{N_{\ell }-1}, {\mathcal {I}}^s_{\ell }={\mathcal {S}}\right) \right] \right] \nonumber \\&\quad =\left[ \left( \Delta t_{\ell }^{\delta +1} \right) e^{2(\Delta t_{\ell }^{1-\delta } -\Delta t_{\ell }) \sum _{n \in {\mathcal {S}}} \Delta a_{\ell ,n}} \right. \nonumber \\&\qquad \left. e^{- \left( \Delta t_{\ell }^{1-\delta } \Delta a_{\ell ,n^*} \right) } \left( \Delta a_{\ell ,n^*}\right) \left( 1+ h_{2,\ell }\right) \right] \nonumber \\&\quad ={\mathcal {O}}\left( \Delta t_{\ell }^{1+\delta }\right) . \end{aligned}$$

(A.7)

$\square$

Appendix B: Numerical evidence of Assumption 4.1

In Figs. 20, 21 and 22, we plot the histograms, for Examples 5.1, 5.2 and 5.3, with $\delta =0.5$, corresponding to $\# \{ \text {IS steps : s.t.}\, {\overline{K}}:=\sum _{j \in {\mathcal {J}}_1} \sum _{n \in {\mathcal {S}}_j}k^j_n>0\}$, that is the number of times where we perform IS and succeeded to separate the two paths. These Figs. show that our assumption 4.1 (c) is valid since for small values of $\Delta t_{\ell }$, we have at most one jump created by IS such that it separates the two paths.