# 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

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Does "OU process" stand for the Ornstein-Uhlenbeck process ? The SDE you wrote is that of the Wiener process with drift. – Sasha Oct 27 '11 at 4:34
i mean Wiener process with drift – Cassy Chen Oct 27 '11 at 4:37

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 $$\mathbb{E}(w_t) = \frac{r t}{2} \qquad \operatorname{Var}(w_t) = \frac{\sigma^2 t^2}{12}$$

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You definitely could do simpler to compute $E(w_t)$ and $\text{var}(w_t)$. For example $E(w_t)$ is the integral of $E(u_s)=rs/t$ from $s=0$ to $s=t$, that is $rt/2$. Likewise $\text{var}(w_t)$ is the double integral of $\mathrm{cov}(u_s,u_z)$ on $0\leqslant s,z\leqslant t$ and each $\mathrm{cov}(u_s,u_z)$ is a covariance of the Brownian bridge $W_{0,\sigma}$. – Did Oct 27 '11 at 8:15
@DidierPiau Thank you for the comment. I will work on it – Sasha Oct 27 '11 at 12:49

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

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