conditional expectation of a solution to the SDE Suppose we have an SDE, which is the Wiener process with drift
$dr_t=c dt+\sigma dB_t$, where $B_t$ is brownian
I want to find $\mathbb{E}[e^{-\int_0^t r_s  ds} |r_t=r]$
so my approach is this : write the SDE as : $r_t-r_0=ct+\int\sigma dB_t$
Then I know $r_t$ is distributed as a normal. but then how can i get the distribution of $\int_0^t r_s ds$ and hence the expectation? 
thanks
 A: Since $r_t$ is the Wiener process with drift. Assuming its initial condition is $r_0 = 0$, conditioning on $r_t = r$ is equivalent to considering the Brownian bridge. 
Let $u_t$ be the Brownian bridge process with $u_0 = 0$, and $\lim_{s \to t+0^-} u_s = r$.
Then $u_s = r \frac{s}{t} + \left(1-\frac{s}{t}\right) W_{0,\sigma}\left( \frac{s}{t-s} \right)$ and $\mathrm{d} u_s = \left( \frac{r-u_s}{t-s} \right)\mathrm{d} s + \frac{\sigma}{\sqrt{t}} \mathrm{d} W_s$. 
Variable $w_s = \int_0^s u_\tau \mathrm{d} \tau$ is a Gaussian process as a linear functional of the Brownian bridge, which is Gaussian. Thus to determine its distribution, it suffices to compute moments.
$$
   \mathbb{E}\left( \exp(-w_t) \right) = \exp\left( - \mathbb{E}(w_t)+ \frac{1}{2} \operatorname{Var}(w_t) \right)
$$
The process $w_s$ is also Ito, with SDE $\mathrm{d} w_s = u_s \mathrm{d} s$. To find the moments we use Ito's lemma with $\mathrm{d}X_t = \mu_t \mathrm{d} t + \sigma_t \mathrm{d} B_t$:
$$
   \mathbb{E}( f(X_t)) = \mathbb{E}( f(X_0)) + \int_0^t \mathbb{E}\left( \mu_s f^\prime(X_s) + \frac{1}{2} \sigma_s^2 f^{\prime\prime}(X_s) \right) \mathrm{d} s 
$$
Applying this to polynomials of $u_s$ and $w_s$:
$$
 \begin{eqnarray}
   m_1^\prime(u_s) &=& \frac{r - m_1(u_s)}{t-s} \land m_1(u_0) = 0 \\
   m_2^\prime(u_s) &=& 2 \frac{r m_1(u_s) - m_2(u_s)}{t-s} + 2 \frac{\sigma^2}{2t} \land m_2(u_0) = 0
 \end{eqnarray}
$$
Solving this gives $m_1(u_s) = \frac{r s}{t}$ and $m_2(u_s) = r^2 \frac{s^2}{t^2} + \sigma^2 \frac{s}{t} \frac{t-s}{t}$.
Similarly:
$$
 \begin{eqnarray}
    m_1^\prime(w_s) &=& m_1(u_s) \land m_1(w_0) = 0 \\
    m_{11}^\prime(u_s w_s) &=& m_2(u_s) + \frac{r m_1(w_s) - m_{11}(u_s w_s)}{t-s} \land m_{11}(u_0 w_0) = 0 \\
   m_2^\prime(w_s) &=& 2 m_{11}(u_s w_s) \land m_2(w_0) = 0
 \end{eqnarray} 
$$
Solving these yields:
$$
   m_1(w_s) = \frac{r t}{2} \frac{s^2}{t^2} \qquad
   m_{11}(u_s w_s) = \frac{s^2 \left(r^2 s-s \sigma ^2+\sigma ^2 t\right)}{2 t^2}
  \qquad
   m_2(w_s) = \frac{s^3 \left(3 r^2 s-3 s \sigma ^2+4 \sigma ^2 t\right)}{12 t^2}
$$
Combining moments gives
$$
  \mathbb{E}(w_t) = \frac{r t}{2} \qquad
  \operatorname{Var}(w_t) = \frac{\sigma^2 t^2}{12}
$$
A: This is not too hard once you have noticed that :  
-$(X,Y)=(r_t, \int_0^t r_s.ds)$ is a gaussian vector
-the distribution of $(X,Y)$ conditionnally to $X=x$ is well known 
I am not going further as it is a good exercise to do the calculations once.
Best Regards 
