^2_{\lambda_1^\epsilon,\lambda_2^\epsilon}(t)-(1+\epsilon)^2
			\psi^2_{\lambda_1,\lambda_2}(t)\right)
			-\kappa g_\epsilon(t)+\frac{\sigma^2}{2}\left((1+\epsilon)^2-(1+\epsilon)\right)\psi^2_{\lambda_1,\lambda_2}(t)\\
			&=&\frac{\sigma^2}{2}\left(\psi_{\lambda_1^\epsilon,\lambda_2^\epsilon}(t)+(1+\epsilon)
			\psi_{\lambda_1,\lambda_2}(t)\right)g_\epsilon(t)
			-\kappa g_\epsilon(t)+\frac{\sigma^2}{2}\epsilon(1+\epsilon)\psi^2_{\lambda_1,\lambda_2}(t)\\
			&=&f_\epsilon(t) g_\epsilon(t)+\frac{\sigma^2}{2}\epsilon(1+\epsilon)\psi^2_{\lambda_1,\lambda_2}(t),
		\end{eqnarray*}
		where
		\[
		f_\epsilon(t)=\frac{\sigma^2}{2}\left(\psi_{\lambda_1^\epsilon,\lambda_2^\epsilon}(t)+(1+\epsilon)
		\psi_{\lambda_1,\lambda_2}(t)\right)
		-\kappa.
		\]
		Therefore, the function $g_\epsilon(t)e^{-\int_0^t f_\epsilon(s)ds}$ is nondecreasing  and, since $g_\epsilon(0)=0$,
		we have $g_\epsilon(t)\geq 0$.
	\end{remark}    
	
	We can now prove the following Lemma, which will be useful in  Section \ref{sect-estimjointlaw} to  prove suitable estimates on the joint law of the process $(X,Y)$.
	\begin{lemma}\label{RemarqueMomentsNeg}
		For every $q>0$  there exists $C>0$ such that for all $y_0\geq 0$,
		\begin{equation}
		\mathbb{E}_{y_0}\left(\int_0^t Y_v dv\right)^{-q}\leq \frac{C}{t^{2q}}.
		\end{equation}
	\end{lemma}
	\begin{proof}
		If we take $\lambda_1=0$ and $\lambda_2=-s$ with $s>0$  in Proposition \ref{Laplace2},
		we get
		\[
		\mathbb{E}_{y_0}\left( e^{-s \int_0^t Y_v dv}\right)=e^{y_0\psi_{0,-s}(t) +\theta\kappa\phi_{0,-s}(t)}.
		\]
		Since $\psi'_{0,-s}(0)=-s<0$, we can deduce by the  proof of Proposition \ref{Laplace2} that $\psi'_{0,-s}(t)=-se^{\int_0^t(\sigma^2\psi(u)-\kappa)du}$. Therefore, since $\psi_{0,-s}=0$, we have  
		\begin{equation}\label{psi0-s}
		\psi_{0,-s}(t)=-s\int_0^te^{\int_0^u(\sigma^2\psi(v)-\kappa)dv}du.
		\end{equation}
		Again from the proof of Proposition \ref{Laplace2},
		\[
		\psi_{0,-s}(t)\geq 
		\frac{\kappa}{\sigma^2}-\sqrt{\left(\frac{\kappa}{\sigma^2}\right)^2+2\frac{s}{\sigma^2}}
		\geq -\sqrt{2s/\sigma^2},
		\]
		so, by using \eqref{psi0-s}, we deduce that
		$$
		\psi_{0,-s}(t)\leq 
		-s\int_0^te^{\int_0^u-(\sigma \sqrt{2s}+\kappa)dv}du=-s\int_0^te^{-\lambda_su}du=-\frac{s}{\lambda_s}(1-e^{-t\lambda_s}).
		$$	
		where $\lambda_s=\sigma\sqrt{2s}+\kappa$. Since $\phi_{0,-s}(t)=\int_0^t \psi_{0,-s}(u)du$, we have
		$$
		\phi_{0,-s}(t)\leq-\frac{s}{\lambda^2_s} \left( t\lambda_s   -1+e^{-t\lambda_s}\right).
		$$
		Therefore, since $\psi_{0,-s}(t)\leq 0$,  for any $y_0\geq 0$ we get
		\begin{align*}
		\mathbb{E}_{y_0}\left(e^{-s\int_0^tY_vdv}\right)&\leq e^{\kappa\theta\phi_{0,-s}(t)}\leq e^{ -\frac{\kappa\theta s}{\lambda_s^2} (t\lambda_s-1+e^{-t\lambda_s})    }.
		\end{align*}
		Now, recall that for every $q>0$ we can write
		$$
		\frac{1}{y^q}=\frac{1}{\Gamma(q)}\int_0^\infty s^{q-1}e^{-sy}ds.
		$$
		Therefore
		\begin{align*}
		\mathbb{E}_{y_0}\left(\int_0^t Y_v dv\right)^{-q}&=\mathbb{E}_{y_0}\left(\frac{1}{\Gamma(q)}\int_0^\infty s^{q-1}
		e^{-s\int_0^t Y_v dv}ds\right)\\
		&\leq\frac{1}{\Gamma(q)}\int_0^1s^{q-1}
		e^{  -\frac{\kappa\theta s}{\lambda_s^2} (t\lambda_s-1+e^{-t\lambda_s})} ds+ \frac{1}{\Gamma(q)}\int_1^\infty s^{q-1}
		e^{  -\frac{\kappa\theta s}{\lambda_s^2} (t\lambda_s-1+e^{-t\lambda_s})} ds.
		\end{align*}
		Recall that $\lambda_s=\sigma\sqrt{2s}+\kappa$, so the first terms in the right hand side is finite. Moreover, for $s>1$, we have $\frac{\kappa\theta s}{\lambda_s^2}\leq C$.  Then, by noting that the function $u\mapsto tu-1+e^{-tu}$ is nondecreasing, we have
		\begin{align*}
		\mathbb{E}_{y_0}\left(\int_0^t Y_v dv\right)^{-q}&\leq C+ \frac{1}{\Gamma(q)}\int_1^\infty s^{q-1}
		e^{  -C (t\sigma\sqrt{2s}-1+e^{-t\sigma\sqrt{2s}})} ds\\
		&\leq C+ \frac{1}{t^{2q}\Gamma(q)}\int_0^\infty v^{q-1}
		e^{  -C (\sigma\sqrt{2v}-1+e^{-\sigma\sqrt{2v}})} dv\\
		&\leq \frac{C}{t^{2q}},
		\end{align*}
		which concludes  the proof.
	\end{proof}
	
	Now recall that the diffusion $(X,Y)$ evolves according to the following stochastic differential system
	
	\begin{equation*}
	\begin{cases}
	dX_t=\left( \frac{\rho\kappa\theta}{\sigma}-\frac{Y_t}{2}\right)dt + \sqrt{Y_t}dB_t,\\
	dY_t=\kappa(\theta-Y_t)dt+\sigma\sqrt{Y_t}dW_t.
	\end{cases}
	\end{equation*}	 	
	If we set $\tilde{X}_t=X_t-\frac{\rho}{\sigma}Y_t$, we have 
	\begin{equation}
	\label{model2}
	\begin{cases}
	d\tilde{X}_t=\left( \frac{\rho\kappa}{\sigma}-\frac{1}{2}\right)Y_tdt+\sqrt{1-\rho^2}\sqrt{Y_t} d\tilde{B}_t,\\
	dY_t=\kappa(\theta-Y_t)dt+\sigma\sqrt{Y_t}dW_t.
	\end{cases}
	\end{equation}
	where $\tilde{B}_t=(1-\rho^2)^{-1/2}\left(B_t-\rho W_t\right)$. Note that $\tilde B $ is a standard Brownian motion with $\langle \tilde{B},W\rangle_t=0$.

	\begin{