I have a question regarding the conditional expectation of two integrals of random variables Here are my hypothesis
Let $W_t$ be a Brownian motion. $\Gamma_t$, $\mu_t$ et $\sigma_t$ three deterministic parameters such as:
$\lambda_t = \Gamma_t\exp(X_t)$
$dX_t=-\mu_tdt+\sigma_tdW_t$
$X_0$
$T_0, T_1 \in [0,T]$
I would like to calculate: $\mathbb{E}\Bigr(\exp \bigr(-\int_{0}^{T_1}{\lambda_sds\bigr)}\Bigr)$
Conditionning the above by $X_{T_0}$ I have:
$\mathbb{E}\Bigr(\exp(-\int_{0}^{T_1}{\lambda_sds}\bigr)\Bigr)=\mathbb{E}\Bigr(\mathbb{E}\bigr(\exp(-\int_{0}^{T_1}{\lambda_sds)}|X_{T_0}\bigr)\Bigr).$
But
\begin{align*} \mathbb{E}\biggr(\exp\Bigr(-\int_{0}^{T_1}{\lambda_sds\Bigr)}|X_{T_0}\biggr)&=\mathbb{E}\biggr(\exp\Bigr(-\int_{0}^{T_0}{\lambda_sds\Bigr)}\exp\Bigr(-\int_{T_0}^{T_1}{\lambda_sds\Bigr)}|X_{T_0}\biggr) \\ &=\mathbb{E}\biggr(\exp\Bigr(-\int_{0}^{T_0}{\Gamma_s\exp(X_s)ds\Bigr)}\exp\Bigr(-\int_{T_0}^{T_1}{\Gamma_s\exp(X_s)ds\Bigr)}|X_{T_0}\biggr) \end{align*}
My question is the following: conditioning to $X_{T_0}$, do we have:
$\exp\Bigr(-\int_{0}^{T_0}{\Gamma_s\exp(X_s)ds}\Bigr)$ and $\exp\Bigr(-\int_{T_0}^{T_1}{\Gamma_s\exp(X_s)ds}\Bigr)$ independent ?
In other words can I write:
\begin{align*} \mathbb{E}\biggr(\exp\Bigr(-\int_{0}^{T_0}{\Gamma_s\exp(X_s)ds\Bigr)}\exp\Bigr(-\int_{T_0}^{T_1}{\Gamma_s\exp(X_s)ds\Bigr)}|X_{T_0}\biggr) \\ &=\mathbb{E}\biggr(\exp\Bigr(-\int_{0}^{T_0}{\Gamma_s\exp(X_s)ds\Bigr)}|X_{T_0}\biggr).\mathbb{E}\biggr(\exp\Bigr(-\int_{T_0}^{T_1}{\Gamma_s\exp(X_s)ds\Bigr)}|X_{T_0}\biggr) \end{align*}
?
Many thanks for your help.