Machine Learning for Semi Linear PDEs

Abstract

Recent machine learning algorithms dedicated to solving semi-linear PDEs are improved by using different neural network architectures and different parameterizations. These algorithms are compared to a new one that solves a fixed point problem by using deep learning techniques. This new algorithm appears to be competitive in terms of accuracy with the best existing algorithms.

Appendix

1.1 Some Test PDEs

We recall our notations for the semilinear PDEs:

$$\begin{aligned} -\partial _t u(t, x) - \mathcal {L} u(t, x)&= f\left( t, x, u(t, x), \sigma ^\top (t, x) \nabla u(t, x)\right) \\ u(T, x)&= g(x) \end{aligned}$$

where $\mathcal {L}u (t, x) := \frac{1}{2} {{\,\mathrm{{\text {Tr}}}\,}}\left( \sigma ^\top \sigma (t, x) \nabla ^2 u(t, x)\right) + \mu (t, x)^\top \nabla u(t, x)$. For each example, we thus give the corresponding $\mu$, $\sigma$, f and g. In the implementation, we rather use a function

$$\begin{aligned} \tilde{f}(t, x, u(t, x), Du(t, x)) := f(t, x, u(t, x), \sigma ^\top (t, x) Du(t, x)) \end{aligned}$$

for convenience, as this does not influence the results and allows for a more direct formulation for some PDEs.

1.1.1 A Black–Scholes Equation with Default Risk

From [21, 22]. If not otherwise stated, the parameters take values: $\overline{\mu }=0.02$, $\overline{\sigma }=0.2$, $\delta =2/3$, $R=0.02$, $\gamma _h=0.2$, $\gamma _l=0.02$, $v_h=50$, $v_l=70$. We use the initial condition $X_0 = (100, \ldots , 100)$.

$$\begin{aligned} \mu : (t, x)&\mapsto \overline{\mu } x \\ \sigma : (t, x)&\mapsto \overline{\sigma } \; {\mathrm {diag}}(\left\{ x_i\right\} _{i=1..d}) \\ f : (t, x, y, z)&\mapsto -(1-\delta ) \times {\mathrm {min}} \left\{ \gamma _h, \; {\mathrm {max}}\left\{ \gamma _l, \; \frac{\gamma _h - \gamma _l}{v_h - v_l} \left( y - v_h \right) + \gamma _h \right\} \right\} y - R y\\ g : x&\mapsto \min _{i=1..d} x_i \end{aligned}$$

We used the closed formula for the SDE dynamic:

$$\begin{aligned} X_t&= X_s \exp \left[ \left( \left( \bar{\mu }-\frac{\bar{\sigma }^2}{2}\right) \right) (t-s) + \bar{\sigma } (W_t - W_s)\right]&\forall t > s \end{aligned}$$

Baseline (from [21]):

1.1.2 A Black–Scholes–Barenblatt Equation

From [24]. If not otherwise stated, the parameters take values: $\overline{\sigma }=0.4$, $r=0.05$. We use the initial condition $X_0 = (1.0, 0.5, 1.0, \ldots )$.

$$\begin{aligned} \mu : (t, x)&\mapsto 0 \\ \sigma : (t, x)&\mapsto \overline{\sigma } \; {\mathrm {diag}}(\left\{ x_i\right\} _{i=1..d}) \\ f : (t, x, y, z)&\mapsto -r \left( y - \frac{1}{\overline{\sigma }}\sum _{i=1}^{d} z_i\right) \\ g : x&\mapsto \left\| x\right\| ^2 \end{aligned}$$

We used the closed formula for the SDE dynamic:

$$\begin{aligned} X_t&= X_s \exp \left[ -\frac{\bar{\sigma }^2}{2} (t-s) + \bar{\sigma } (W_t - W_s)\right]&\forall t > s \end{aligned}$$

Exact solution:

$$\begin{aligned} u(t, x)&= \exp ((r+\overline{\sigma }^2)(T-t))g(x) \end{aligned}$$

1.1.3 A Hamilton–Jacobi–Bellman Equation

From [21, 22]. If not otherwise stated: $\lambda =1.0$ and $X_0 = (0, \ldots , 0)$.

$$\begin{aligned} \mu : (t, x)&\mapsto 0 \\ \sigma : (t, x)&\mapsto \sqrt{2} \; \mathbf {I}_d \\ f : (t, x, y, z)&\mapsto - 0.5 \; \lambda \left\| z\right\| ^2 \\ g : x&\mapsto \log \left( 0.5 \left[ 1 + \left\| x\right\| ^2 \right] \right) \end{aligned}$$

Monte-Carlo solution:

$$\begin{aligned} u(t, X_t) =&-\frac{1}{\lambda } \log \left( \mathbb {E}\left[ \; \exp \left( - \lambda g(X_t + \sqrt{2} B_{T-t}\right) \right] \right) \\ \forall j, \quad \frac{\partial u}{\partial x_j}(t, x) =&\left( \mathbb {E}\left[ \exp \left\{ -\lambda g(X_t + \sqrt{2} B_{T-t})\right\} \right] \right) ^{-1} \\&\times \mathbb {E} \left[ \frac{\partial \, g}{\partial \, x_j}(X_t + \sqrt{2} B_{T-t}) \exp \left\{ -\lambda g(X_t + \sqrt{2} B_{T-t})\right\} \right] \end{aligned}$$

where

$$\begin{aligned} \frac{\partial \, g}{\partial \, x_j}(x)&= \frac{2x_j}{1+\left\| x\right\| ^2} \end{aligned}$$

Baseline (computed using 10 million Monte-Carlo realizations, $d=100$):

1.1.4 An Oscillating Example with a Square Non-linearity

From [15]. If not otherwise stated, the parameters take values: $\mu _0=0.2$, $\sigma _0=1.0$, $a=0.5$, $r=0.1$. The intended effect of the $\min$ and $\max$ in f is to make f Lipschitz. We used the initial condition $X_0=(1.0, 0.5, 1.0, \ldots )$.

$$\begin{aligned} \mu : (t, x)&\mapsto \mu _0 / d \\ \sigma : (t, x)&\mapsto \sigma _0 / \sqrt{d} \; \mathbf {I}_d \\ f : (t, x, y, z)&\mapsto \phi (t, x) \\&\quad + r \left( \max \left[ -\exp (2a(T-t)), \min \left\{ \frac{1}{\sigma _0 \sqrt{d}} y \sum _{i=1}^{d} z_i , \exp (2a(T-t)) \right\} \right] \right) ^2 \end{aligned}$$

where

$$\begin{aligned} \phi : (t, x)&\mapsto \cos \left( \sum _{i=1}^{d} x_i \right) \left( a + \frac{\sigma _0^2}{2} \right) \exp (a(T-t)) + \sin \left( \sum _{i=1}^{d} x_i \right) \mu _0 \exp (a(T-t)) \\&\qquad - r \left( \cos \left( \sum _{i=1}^{d} x_i \right) \sin \left( \sum _{i=1}^{d} x_i \right) \exp (2a(T-t))\right) ^2 \\ g : x&\mapsto \cos \left( \sum _{i=1}^d x_i \right) \end{aligned}$$

Exact solution:

$$\begin{aligned} u(t, x)&= \cos \left( \sum _{i=1}^d x_i \right) \exp (a(T-t)) \\ \forall j, \quad \frac{\partial u}{\partial x_j}(t, x)&= -\sin \left( \sum _{i=1}^d x_i \right) \exp (a(T-t)) \end{aligned}$$

1.1.5 A Non-Lipschitz Terminal Condition

From [35]. If not otherwise stated: $\alpha =0.5$, $X_0 = (0, \ldots , 0)$.

$$\begin{aligned} \mu : (t, x)&\mapsto 0 \\ \sigma : (t, x)&\mapsto \mathbf {I}_d \\ f : (t, x, y, z)&\mapsto - 0.5 \left\| z\right\| ^2 \\ g : x&\mapsto \sum _{i=1}^d \left( \max \left\{ 0, \min \left[ 1, x_i\right] \right\} \right) ^\alpha \end{aligned}$$

Monte-Carlo solution:

$$\begin{aligned} u(t, X_t) =&\log \left( \mathbb {E}\left[ \; \exp \left( g(X_t + B_{T-t})\right) \right] \right) \\ \forall j, \quad \frac{\partial u}{\partial x_j}(t, x) =&\left( \mathbb {E}\left[ \exp \left\{ g(X_t + \sqrt{2} B_{T-t})\right\} \right] \right) ^{-1} \\&\times \mathbb {E} \left[ g'(X_t + \sqrt{2} B_{T-t}) \exp \left\{ g(X_t + \sqrt{2} B_{T-t})\right\} \right] \end{aligned}$$

where

$$\begin{aligned} g'(x)&= \left\{ \begin{array}{ll} 0 &{}\quad {\text {if}}~ x\le 0~ \text {or}~ x\ge 1 \\ \alpha x^{\alpha -1} &{}\quad {\text {else}}\end{array} \right. \end{aligned}$$

Baseline (computed using 10 million Monte-Carlo realizations, $d=10$):

1.1.6 An Oscillating Example with Cox–Ingersoll–Ross Propagation

From [16]. If not otherwise stated: $a=0.1$, $\alpha =0.2$, $T=1.0$, $\hat{k}=0.1$, $\hat{m}=0.3$, $\hat{\sigma }=0.2$. We used the initial condition $X_0 = (0.3, \ldots , 0.3)$. Note that we have $2 \hat{k} \hat{m} > \hat{\sigma }^2$ so that X remains positive.

$$\begin{aligned} \mu : (t, x)&\mapsto \hat{k}\left( \hat{m} - x \right) \\ \sigma : (t, x)&\mapsto \hat{\sigma } {\mathrm {diag}} \left\{ \sqrt{x} \right\} \\ f : (t, x, y, z)&\mapsto \phi (t, x) + a y \left( \sum _{i=1}^d z_i \right) \end{aligned}$$

where

$$\begin{aligned} \phi : (t, x)&\mapsto \cos \left( \sum _{i=1}^{d} x_i \right) \left( -\alpha + \frac{{\hat{\sigma }}^2}{2} \right) \exp (-\alpha (T-t)) \\&\qquad + \sin \left( \sum _{i=1}^{d} x_i \right) \exp (-\alpha (T-t)) \sum _{i=1}^{d} \hat{k}\left( \hat{m} - x_i\right) \\&\qquad + a \cos \left( \sum _{i=1}^{d} x_i \right) \sin \left( \sum _{i=1}^{d} x_i\right) \exp (-2\alpha (T-t)) \sum _{i=1}^d \hat{\sigma } \sqrt{x_i}\\ g : x&\mapsto \cos \left( \sum _{i=1}^d x_i \right) \end{aligned}$$

Exact solution:

$$\begin{aligned} u(t, x)&= \cos \left( \sum _{i=1}^d x_i \right) \exp (-\alpha (T-t)) \\ \forall j, \quad \frac{\partial u}{\partial x_j}(t, x)&= -\sin \left( \sum _{i=1}^d x_i \right) \exp (-\alpha (T-t)) \end{aligned}$$

1.1.7 An Oscillating Example with Inverse Non-linearity

If not stated otherwise, we took parameters $\mu _0 = 0.2$, $\sigma _0 = 1.0$, $a=0.5$, $r=0.1$ and the initial condition $X_0=(1.0, 0.5, 1.0, \ldots )$.

$$\begin{aligned} \mu : (t, x)&\mapsto \mu _0 /d \mathbb {1}_d\\ \sigma : (t, x)&\mapsto \sigma _0 \; \mathbf {I}_d \\ f : (t, x, y, z)&\mapsto \phi (t, x) + r \frac{d y}{\sum _{i=1}^d z_i} \\ g : x&\mapsto 2\sum _{i=1}^d x_i + \cos \left( \sum _{i=1}^d x_i\right) \end{aligned}$$

where

$$\begin{aligned} \phi : (t, x)&\mapsto 2a \sum _{i=1}^d x_i \exp (a(T-t)) + \cos \left( \sum _{i=1}^{d} x_i \right) \left( a + \frac{d \sigma _0^2}{2} \right) \exp (a(T-t)) \\&\quad - \mu _0 \left[ 2- \sin \left( \sum _{i=1}^{d} x_i \right) \right] \exp (a(T-t)) - r \frac{2 \sum _{i=1}^d x_i + \cos \left( \sum _{i=1}^{d} x_i \right) }{\sigma _0 \left[ 2 - \sin \left( \sum _{i=1}^{d} x_i \right) \right] } \end{aligned}$$

Exact solution:

$$\begin{aligned} u(t, x)&= \left[ 2\sum _{i=1}^d x_i + \cos \left( \sum _{i=1}^d x_i \right) \right] \exp (a(T-t)) \\ \forall j, \quad \frac{\partial u}{\partial x_j}(t, x)&= \left[ 2-\sin \left( \sum _{i=1}^d x_i \right) \right] \exp (a(T-t)) \end{aligned}$$

1.2 Demonstration of Proposition 4.6

In the sequel, C will only depend on u, f, g and may vary from one line to another.

First, note that due to the Lipschitz property in Assumption 4.3, the boundedness of f and g and the regularity of u due to Assumption 4.5, the solution u of (1) and Du satisfy a Feynman–Kac relation (see an adaptation of Proposition 1.7 in [42]):

$$\begin{aligned} u\left( t,x\right)&= \frac{1}{2}\mathbb {E}_{t,x} \left[ \phi \left( t, t+\tau ,X_{t+\tau }^{t,x}, u\left( t+\tau ,X_{t+\tau }^{t,x}\right) , Du\left( t+\tau ,X_{t+\tau }^{t,x}\right) \right) \right. \nonumber \\&\quad \left. +\, \phi \left( t, t+\tau ,\widehat{X}_{t+\tau }^{t,x}, u\left( t+\tau ,\widehat{X}_{t+\tau }^{t,x}\right) , Du\left( t+\tau ,\widehat{X}_{t+\tau }^{t,x}\right) \right) \right] \nonumber \\ Du\left( t,x\right)&= \mathbb {E}_{t,x} \left[ \sigma ^{-\top } \frac{W_{\left( t + \left( \tau \wedge \Delta t \right) \right) \wedge T}-W_{t}}{\tau \wedge \left( T-t\right) \wedge \Delta t} \right. \nonumber \\&\quad \times \frac{1}{2} \left( \phi \left( t,t+\tau , X_{t+\tau }^{t,x}, u\left( t+\tau , X_{t+\tau }^{t,x}\right) , Du\left( t+\tau , X_{t+\tau }^{t,x}\right) \right) \right. \nonumber \\&\quad \left. \left. -\, \phi \left( t,t+\tau , \widehat{X}_{t+\tau }^{t,x}, u\left( t+\tau , \widehat{X}_{t+\tau }^{t,x}\right) , Du\left( t+\tau , \widehat{X}_{t+\tau }^{t,x}\right) \right) \right) \right] . \end{aligned}$$

(39)

First picking $(t,y) \in [0,T] \times \Omega _\epsilon ^{0,x}$, we have using Eq. (39):

$$\begin{aligned} B(\theta ,t, y) = \,&| {\bar{u}}^{\epsilon }(\theta , t, y) - u(\theta ,t,y)|^2 \\ \le \,&2 | u(t,y)-u(\theta ,t,y)|^2 + 2 \left( \mathbb {E}_{t,y}\left[ 1_{X^{t,y}_{t+\tau } \in \Omega ^{0,x}_\epsilon } \right. \right. \\&\times \frac{1}{2}\left( \frac{f(t,X^{t,y}_{t+\tau },u(\theta , t+\tau , X^{t,y}_{t+\tau }), v(\theta ,t+\tau , X^{t,y}_{t+\tau }))}{ \rho (\tau )} \right. \\&- \frac{f(t,X^{t,y}_{t+\tau },u( t+\tau , X^{t,y}_{t+\tau }), Du(t+\tau , X^{t,y}_{t+\tau }))}{ \rho (\tau )} \\&+ \frac{f(t,{\hat{X}}^{t,y}_{t+\tau },u(\theta , t+\tau , {\hat{X}}^{t,y}_{t+\tau }), v(\theta ,t+\tau , {\hat{X}}^{t,y}_{t+\tau }))}{ \rho (\tau )} \\&\left. - \left. \frac{f(t,{\hat{X}}^{t,y}_{t+\tau },u( t+\tau , {\hat{X}} ^{t,y}_{t+\tau }), Du(t+\tau , {\hat{X}}^{t,y}_{t+\tau }))}{ \rho (\tau )} \right) \right] \\&+ \frac{1}{2}\mathbb {E}_{t,y} \big [ 1_{X^{t,y}_{t+\tau } \notin \Omega ^{0,x}_\epsilon } \big (\phi \big (t, t+\tau ,X_{t+\tau }^{t,y}, u(t+\tau ,X_{t+\tau }^{t,y}), Du(t+\tau ,X_{t+\tau }^{t,y}) \big ) \\&+ \left. \phi \big (t, t+\tau ,\widehat{X}_{t+\tau }^{t,y}, u(t+\tau ,\widehat{X}_{t+\tau }^{t,y}), Du(t+\tau ,\widehat{X}_{t+\tau }^{t,y}) \big ) \big )\big ] \right) ^2 \end{aligned}$$

Using Jensen equality, the boundedness of g and f, the fact that $\rho$ is bounded by below and Assumption 4.3, we get:

$$\begin{aligned} B(\theta ,t,y) \le \,&2 | u(t,y)-u(\theta ,t,y)|^2 \\&+ C \left( K^2 \mathbb {E}\left[ 1_{X^{t,y}_{t+\tau } \in \Omega ^{0,x}_\epsilon } \left( (u(t+\tau ,X^{t,y}_{t+\tau })-u(\theta ,t+\tau ,X^{t,y}_{t+\tau }))^2 \right. \right. \right. \\&+ \left. \left. \left. || Du(t+\tau ,X^{t,y}_{t+\tau }) -v(\theta ,t+\tau ,X^{t,y}_{t+\tau })||^2_2 \right) \right] + \mathbb {E}[ 1_{X^{t,y}_{t+\tau } \notin \Omega ^{0,x}_{\epsilon }}]\right) \end{aligned}$$

Then using the results in [43, 44], we know that there exists $(u(\theta ^{*},t,x), v(\theta ^*,t,x )) \in \kappa$ such that:

$$\begin{aligned} \sup _{({\hat{t}}, y) \in [t,T]\times \Omega _\epsilon ^{0,x}} ( | u({\hat{t}}, y) - u(\theta ^*, {\hat{t}}, y)| + || Du({\hat{t}}, y)- v(\theta ^{*}, {\hat{t}},y)||_2) \le \epsilon \end{aligned}$$

(40)

Then we have that

$$\begin{aligned} B(\theta ^{*},t,y) \le&C \left( \epsilon + \mathbb {E}_{t,y}\left[ 1_{X^{t,y}_{t+\tau } \notin \Omega ^{0,x}_{\epsilon }}\right] \right) \nonumber \\ \le&C \left( \epsilon + \mathbb {E}_{t,y}\left[ 1_{\exists s \in [t,T] / X^{t,y}_{s} \notin \Omega ^{0,x}_{\epsilon }}\right] \right) \end{aligned}$$

(41)

Similarly introducing ${\hat{W}}^{t}_\tau = \sigma ^{-\top } \frac{W_{(t+ (\tau \wedge \Delta t )) \wedge T}-W_{t}}{\tau \wedge (T-t) \wedge \Delta t}$,

$$\begin{aligned} C(\theta ^{*},t,y)&= || {\bar{v}}^\epsilon ( \theta ^{*} ,t, y) - v(\theta ^{*} ,t,y)||^2_2 \nonumber \\&\le 2 || Du(t,y)-v(\theta ^{*} ,t,y)||^2_2 + 2 \left( \frac{1}{2} \mathbb {E}_{t,y}\left[ 1_{X^{t,y}_{t+\tau } \in \Omega ^{0,x}_\epsilon } {\hat{W}}^{t}_\tau \right. \right. \nonumber \\&\qquad \times \left( \frac{f(t,X^{t,y}_{t+\tau },u(\theta ^{*} , t+\tau , X^{t,y}_{t+\tau }), v(\theta ^{*},t+\tau , X^{t,y}_{t+\tau }))}{ \rho (\tau )} \right. \nonumber \\&\qquad - \frac{f(t,X^{t,y}_{t+\tau },u( t+\tau , X^{t,y}_{t+\tau }), Du(t+\tau , X^{t,y}_{t+\tau }))}{ \rho (\tau )} \nonumber \\&\qquad - \frac{f(t,{\hat{X}}^{t,y}_{t+\tau },u(\theta ^{*} , t+\tau , {\hat{X}}^{t,y}_{t+\tau }), v(\theta ^{*},t+\tau , {\hat{X}}^{t,y}_{t+\tau }))}{ \rho (\tau )} \nonumber \\&\qquad \left. \left. + \frac{f(t,{\hat{X}}^{t,y}_{t+\tau },u( t+\tau , {\hat{X}} ^{t,y}_{t+\tau }), Du(t+\tau , {\hat{X}}^{t,y}_{t+\tau }))}{ \rho (\tau )} \right) \right] \nonumber \\&\qquad + \frac{1}{2}\mathbb {E}_{t,y} \big [ 1_{X^{t,y}_{t+\tau } \notin \Omega ^{0,x}_\epsilon } {\hat{W}}^{t}_\tau \big (\phi \big (t, t+\tau ,X_{t+\tau }^{t,y}, u(t+\tau ,X_{t+\tau }^{t,y}), Du(t+\tau ,X_{t+\tau }^{t,y}) \big ) \nonumber \\&\qquad - \left. \phi \big (t, t+\tau ,\widehat{X}_{t+\tau }^{t,y}, u(t+\tau ,\widehat{X}_{t+\tau }^{t,y}), Du(t+\tau ,\widehat{X}_{t+\tau }^{t,y}) \big ) \big )\Bigg ] \right) ^2 \nonumber \\&\le 2 || Du(t,y)-v(\theta ^{*} ,t,y)||^2 \nonumber \\&\qquad + \,C \mathbb {E}_{t,y}\left[ ({\hat{W}}^{t}_\tau )^\top {\hat{W}}^{t}_\tau \left( \frac{||f||_\infty ^2 }{\rho (\tau )^2} + \frac{|g(X^{t,y}_{t+\tau }) - g( {\hat{X}}^{t,y}_{t+\tau })|^2}{{\bar{F}}(T)^2}\right) \right] \mathbb {E}_{t,y}\left[ 1_{X^{t,y}_{t+\tau } \notin \Omega ^{0,x}_{\epsilon }} \right] \nonumber \\&\qquad + C K \mathbb {E}_{t,y}\left[ 1_{X^{t,y}_{t+\tau } \in \Omega ^{0,x}_\epsilon } \frac{\left( {\hat{W}}^{t}_\tau \right) ^\top {\hat{W}}^{t}_\tau }{\rho (\tau )^2} \left( u(t+\tau ,X^{t,y}_{t+\tau }) - u\left( \theta ^{*},t+\tau ,X^{t,y}_{t+\tau }\right) \right) ^2 \right] \nonumber \\&\qquad + \,C K \mathbb {E}_{t,y}\left[ 1_{X^{t,y}_{t+\tau } \in \Omega ^{0,x}_\epsilon } \frac{({\hat{W}}^{t}_\tau )^\top {\hat{W}}^{t}_\tau }{\rho (\tau )^2} || Du\left( t+\tau ,X^{t,y}_{t+\tau }\right) \right. \nonumber \\&\qquad \left. - \,v(\theta ^{*},t+\tau ,X^{t,y}_{t+\tau }) ||^2_2 \right] , \end{aligned}$$

(42)

where we have used Jensen and Cauchy Schwarz.

Introducing $G\in {{\,\mathrm{\mathbb {R}}\,}}^d$ composed of centered unitary independent Gaussian random variables

$$\begin{aligned} \mathbb {E}_{t,x}\left[ \frac{({\hat{W}}^{t}_\tau )^\top {\hat{W}}^{t}_\tau }{\rho (\tau )^2}\right] = \mathbb {E}\left[ \frac{1}{\tau \rho (\tau )^2}\right] \mathbb {E}[G^\top \sigma ^{-1} \sigma ^{-\top } G] < \infty \end{aligned}$$

when $u<1$ in Eq. (11).

Similarly using the fact that g is Lipschitz,

$$\begin{aligned} \mathbb {E}_{t,y}\left[ ({\hat{W}}^{t}_\tau )^\top {\hat{W}}^{t}_\tau \left( \frac{||f||_\infty ^2 }{\rho (\tau )^2} + \frac{|g(X^{t,y}_{t+\tau }) - g( {\hat{X}}^{t,y}_{t+\tau })|^2}{{\bar{F}}(T)^2}\right) \right]< C < \infty \end{aligned}$$

Then using Eq. (40) in Eq. (42), we get

$$\begin{aligned} C(\theta ^{*}, t,y) \le C ( \epsilon + \mathbb {E}_{t,y}[ 1_{\exists s \in [t,T] / X^{t,y}_{s} \notin \Omega ^{0,x}_{\epsilon }}]) \end{aligned}$$

(43)

Injecting (41) and (43) in Eq. (31) and using the definition of $\Omega ^{0,x}_{\epsilon }$ complete the proof.

1.3 Demonstration of Proposition 4.7

Let $\bar{K}$ be a compact. It is always possible to find $n_0$ such that for $n> n_0$, $\bar{K} \subset \Omega ^{0,x}_{\epsilon _n}$.

Let u be a function from ${{\,\mathrm{\mathbb {R}}\,}}\times {{\,\mathrm{\mathbb {R}}\,}}^d$ to ${{\,\mathrm{\mathbb {R}}\,}}$ and v function from ${{\,\mathrm{\mathbb {R}}\,}}\times {{\,\mathrm{\mathbb {R}}\,}}^d$ to ${{\,\mathrm{\mathbb {R}}\,}}^d$. Let

$$\begin{aligned} \psi (t,t+\tau ,x,u) =&1_{X^{t,x}_{t+\tau } \in \Omega ^{0,x}_{\epsilon _n}} \frac{1}{2} \left[ \phi \left( t, t+\tau ,X_{t+\tau }^{t,x}, u(t+\tau ,X_{t+\tau }^{t,x}) \right) \right. \\&\left. +\, \phi \left( t, t+\tau ,\widehat{X}_{t+\tau }^{t,x}, u\left( t+\tau ,\widehat{X}_{t+\tau }^{t,x}\right) \right) \right] . \end{aligned}$$

We note that for $\zeta$ an independent uniform random variable in [0, T]

$$\begin{aligned} D_n&:= \mathbb {E} \left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} \Bigg ( \mathbb {E}_{t,X^{0,x}_t} \left[ \psi (t,t+\tau ,X^{0,x}_t,u(\theta _n, \cdot ,\cdot ),v(\theta _n, \cdot ,\cdot )) \right] \right. \\&\qquad \left. - \,u(\theta _n,t ,X^{0,x}_{t}) \Bigg )^2 {{\,\mathrm{{\text {d}}}\,}}t \right] \\&= T \mathbb {E} \left[ 1_{X^{0,x}_{\zeta } \in \Omega ^{0,x}_\epsilon } \Bigg ( \mathbb {E}_{\zeta ,X^{0,x}_\zeta } \left[ \psi (\zeta ,\zeta +\tau ,X^{0,x}_\zeta ,u(\theta _n, \cdot ,\cdot ),v(\theta _n, \cdot ,\cdot )) \right] \right. \\&\qquad \left. -\, u(\theta _n, \zeta ,X^{0,x}_{\zeta }) \Bigg )^2 \right] \\&= T \ell (\theta _n) \end{aligned}$$

Using the Eq. (39), the expression of $D_n$, the Lipschitz property of f, Jensen inequality, the boundedness of f, g, and the fact that $\rho$ is bounded by below by $\hat{\rho }(T)$:

$$\begin{aligned} F_n&= \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in {\bar{K}}} (u( t,X^{0,x}_t) - u( \theta _n, t,X^{0,x}_t)) ^2 {{\,\mathrm{{\text {d}}}\,}}t \right] \\&\le \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} (u( t,X^{0,x}_t) - u( \theta _n, t,X^{0,x}_t)) ^2 {{\,\mathrm{{\text {d}}}\,}}t \right] \\&\le 3 \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} \left( \mathbb {E}_{t,X^{0,x}_t} (\psi (t,t+\tau ,X^{0,x}_{t+\tau },u) \right. \right. \\&\quad \left. \left. - \psi (t,t+\tau ,X^{0,x}_{t+\tau }, u(\theta _n,\cdot ,\cdot ))) \right) ^2 {{\,\mathrm{{\text {d}}}\,}}t \right] \\&\quad +\, 3 \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} \left( \mathbb {E}_{t,X^{0,x}_t} ( \psi (t,t+\tau ,X^{0,x}_{t+\tau }, u(\theta _n,\cdot ,\cdot ))) - u( \theta _n, t,X^{0,x}_t) \right) ^2 {{\,\mathrm{{\text {d}}}\,}}t \right] \\&\quad +\, 3 \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} \mathbb {E}_{t,X^{0,x}_t} \left( 1_{X^{0,x}_{t+ \tau } \notin \Omega ^{0,x}_{\epsilon _n}} \frac{1}{4} \big [ \phi \big (t, t+\tau ,X_{t+\tau }^{t,x}, u(t+\tau ,X_{t+\tau }^{t,x}) \big ) \right. \right. \\&\quad + \left. \left. \phi \big (t, t+\tau ,\widehat{X}_{t+\tau }^{t,x}, u(t+\tau ,\widehat{X}_{t+\tau }^{t,x}) \big ) \big ]^2 \right) {{\,\mathrm{{\text {d}}}\,}}t \right] \\&\le 3 K^2 \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} \mathbb {E}_{t,X^{0,x}_t} \left[ 1_{\tau <T -t} 1_{X^{0,x}_{t+\tau } \in \Omega ^{0,x}_{\epsilon _n}}\right. \right. \\&\quad \left. \left. \times \frac{(u( \theta _n, t+\tau ,X^{0,x}_{t+\tau }) - u(t+\tau ,X^{0,x}_{t+\tau }))^2}{\rho (\tau )^2 }\right] {{\,\mathrm{{\text {d}}}\,}}t \right] \\&\quad +\, 3 T \ell (\theta _n) + 3 \left( \frac{ ||f||_\infty ^2}{{\hat{\rho }}(T)^2} + \frac{ ||g||_\infty ^2}{{\bar{F}}(T)^2}\right) \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} \mathbb {E}_{t,X^{0,x}_t} \left( 1_{X^{0,x}_{t+ \tau } \notin \Omega ^{0,x}_{\epsilon _n}} \right) {{\,\mathrm{{\text {d}}}\,}}t\right] \\&\le 3 K^2 \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} \mathbb {E}_{t,X^{0,x}_t} \left[ \int _t^T ds \right. \right. \\&\quad \left. \left. \times \,\left( 1_{X^{0,x}_{s} \in \Omega ^{0,x}_{\epsilon _n}} \frac{(u( \theta _n, s,X^{0,x}_{s}) - u(s,X^{0,x}_{s}))^2}{\rho (s-t) } \right) ds \right] {{\,\mathrm{{\text {d}}}\,}}t \right] + 3 T \ell (\theta _n) + C \epsilon \\&\le 3 \frac{K^2 T}{{\hat{\rho }}(T)} \mathbb {E}\left[ \int _0^T 1_{X^{0,x}_{t} \in \Omega ^{0,x}_{\epsilon _n}} (u( \theta _n, t,X^{0,x}_{t}) - u(t,X^{0,x}_{t}))^2 {{\,\mathrm{{\text {d}}}\,}}t \right] + 3T \ell (\theta _n) + C \epsilon \end{aligned}$$

Then, if K is small enough

$$\begin{aligned} F_n \le {\hat{\rho }}(T) \frac{T \ell (\theta _n) + C \epsilon }{ {\hat{\rho }}(T) - 3 K^2 T}, \end{aligned}$$

which completes the proof.