Processus de diffusion

$\begin{cases}(i)'\displaystyle \lim_{\Delta t \to 0^{+}}\frac{1}{\Delta t}\int_{-\infty}^{\infty}|y-x|^{\delta}p\left(y,t+\Delta t,x,t\right)dy=0 \mbox{ for some } \delta >2 & \\            (ii)'\lim_{\Delta t \to 0^{+}}\frac{1}{\Delta t}\int_{-\infty}^{\infty}(y-x)p\left(y,t+\Delta t,x,t\right)dy=a(x,t)& \\                          (iii)'\lim_{\Delta t \to 0^{+}}\frac{1}{\Delta t}\int_{-\infty}^{\infty}(y-x)^{2}p\left(y,t+\Delta t,x,t\right)dy=b(x,t)& \\                         \forall k\in\{0,1,2\}\mbox{ and } |y-x|>\epsilon: \: \int_{|y-x|>\epsilon}^{}|y-x|^{k}p\left(y,t+\Delta t,x,t\right)dy\leq \frac{1}{\epsilon^{\delta -k}}\int_{-\infty}^{+\infty}|y-x|^{\delta}p\left(y,t+\Delta t,x,t\right)dy &\end{cases}$

alors

$ \begin{cases}   \displaystyle        (i)\lim_{\Delta t \to 0^{+}}\frac{1}{\Delta t}\int_{|y-x|>\epsilon}^{}p\left(y,t+\Delta t,x,t\right)dy=0 & \\          \displaystyle               (ii)\lim_{\Delta t \to 0^{+}}\frac{1}{\Delta t}\int_{|y-x|\leq \epsilon}^{\infty}(y-x)p\left(y,t+\Delta t,x,t\right)dy=a(x,t)& \\        \displaystyle(iii)\lim_{\Delta t \to 0^{+}}\frac{1}{\Delta t}\int_{|y-x|\leq \epsilon}^{}(y-x)^{2}p\left(y,t+\Delta t,x,t\right)dy=b(x,t)&  \end{cases}$