compact. $B$ has open interior with $b_{0}\in B^{\circ}$. The domain of $\lambda$, denoted as $L$, has $\lambda_0\in L$ and is a set of uniformly bounded right-continuous functions satisfying that $\lambda(0)=1$ for all $\lambda\in L$ (which means $L\subset L^{\infty}\left([0,\infty)\right)$ and under the weak-$\ast$ topology, the closure of $L$ is compact).

$C2$. $E_{\Omega_{0}}\left(\left|\log\left(V_{\Omega}\right)\right|\right)$, $E_{\Omega_{0}}\left(\left|\nabla_{b_{i}}\log\left(V_{\Omega}\right)\nabla_{b_{j}}\log\left(V_{\Omega}\right)\right|\right)$, $E_{\Omega_{0}}\left(\left|\nabla_{b_{i}b_{j}}\log\left(V_{\Omega}\right)\right|\right)$ are finite for all $i,j=1,\dots,p$ and all $\Omega\in A\times B\times L$; and the matrix $I=\left\{ E_{\Omega_{0}}\left(\nabla_{b_{i}}\log\left(V_{\Omega}\right)\nabla_{b_{j}}\log\left(V_{\Omega}\right)\right)\right\} _{1\leq i,j\leq p}$ and $H=\left\{ E_{\Omega_{0}}\left(\nabla_{b_{i}b_{j}}\log\left(V_{\Omega}\right)\right)\right\} _{1\leq i,j\leq p}$ are positive definite.

$C3$. There exists a positive function $d\in L^{1}(pdf_{\Omega_0})-$ such that for all $\Omega\in A\times B\times L$, $\left|\log V_{\Omega}\right|\leq d\left(Z_T,T\right)$ almost surely with respect to the probability measure $pdf_{\Omega_0}$. $C4$. For all $a\in A$, $q\left(z,t\mid a\right)\in C^{2}\left(\mathbb{R}^{p}\times\mathbb{R}_{+}\right)$ and the map given through $q\left(.\mid.\right):A\rightarrow C^{2}\left(\mathbb{R}^{p}\times\mathbb{R}_{+}\right)$ is continuous with respect to the $C^{2}$ topology.

$C5$. For every $a\in A$, there is an $p\times p$ matrix $M_a$, such that $\frac{\partial q\left(z,0\mid a\right)}{\partial z}\rightarrow M_a$ as $\left\Vert z\right\Vert\rightarrow \infty$ ($\left\Vert.\right\Vert$ is the Euclidean Norm of a vector). And for different $a$ and $a'$, $M_a-M_{a'}$ has at least one eigenvalue with non-zero real part.

$C6$. $pdf_{\Omega_0}$ has the full support $\mathbb{R}^{p}\times \mathbb{R}_{+}$. The true initial $p(.,0)$ satisfies that $\int_{\mathbb{R}^{p}}\exp(c\cdot z)p(z,0)dz\not=1$ for every $c\in \mathbb{R}^{p}$. Condition 