How to compute $\mathbb{E}(\exp(\int_0^t W_s ds)|W_t)$?

I am trying to compute the conditional expectation $$\mathbb{E}\left[\exp\left(\int_0^t W_s ds\right)\middle|\, W_t\right]$$ where $W$ is a standard Wiener process and where $s\le t$. To initially simplify the problem, I have started with the calculations of $\mathbb{E}[W_s|W_t]$ and $\mathbb{E}\left[\int_0^t W_s \,ds\middle|\,W_t\right]$. On the one hand, since $W_t$ and $W_s- \frac{s}{t}W_t$ are independent (having zero covariance and using a gaussian vector argument), we can see that: $$\mathbb{E}\left[W_s\middle | W_t\right]=\mathbb{E}\left[W_s-\frac{s}{t}W_t\middle |\, W_t\right]+\frac{s}{t}W_t=\frac{s}{t}W_t$$ On the other hand, by independence of $W_t$ and $\int_0^t (W_s- \frac{s}{t}W_t)ds$: \begin{align}\mathbb{E}\left[\int_0^t W_s ds\middle|\,W_t\right]&=\mathbb{E}\left[\int_0^t \left(W_s-\frac{s}{t}W_t\right) ds\,\middle |\, W_t\right]+\frac{t}{2}W_t\\[0.3cm]&=\int_0^t \mathbb{E}\left[W_s-\frac{s}{t}W_t\right]ds+\frac{t}{2}W_t=\frac{t}{2}W_t\end{align} Coming back to our initial problem, we thus have: $$\mathbb{E}\left[\exp\left(\int_0^t W_s ds\right)\,\middle|\,W_t\right]=\exp\left(\frac{t}{2}W_t\right)\mathbb{E}\left[\exp\left(\int_0^t \left(W_s-\frac{s}{t}W_t\right) ds\right)\,\middle|\,W_t\right]$$ We also know that $\int_0^t \left(W_s-\frac{s}{t}W_t\right)ds$ is normally distributed with zero mean (easy to see) and variance given by: $$\mathbb{E}\left[\int_0^t\int_0^t \left(W_s- \frac{s}{t}W_t\right)\left(W_u- \frac{u}{t}W_t\right)dsdu\right]=\int_0^t\int_0^t\left(\min(s,u)-\frac{su}{t}\right)dsdu=\frac{t^3}{12}$$ By independence of $W_t$ and $\exp\left(\int_0^t \left(W_s- \frac{s}{t}W_t\right)ds\right)$, we finally obtain ($Z$ being a standard unit normal variable): $$\mathbb{E}\left[\exp\left(\int_0^t W_s ds\right)\,\middle|\,W_t\right]=\exp\left(\frac{t}{2}W_t\right)\mathbb{E}\left[\exp\left(Z\sqrt{\frac{t^3}{12}}\right)\right] =\exp\left(\frac{t}{2}W_t+\frac{t^3}{24}\right)$$ However, I am not sure if this answer and the arguments I have used are correct? Any ideas or comments would be greatly appreciated.

• You rightly show that the result is $$e^{tW_t/2}E(e^X)$$ with $$X=\int_0^tW_sds-\frac{t}2W_t$$ thus, $X$ is normal centered, but your computation of its variance is flawed since it neglects the dependency of $W_s-(s/t)W_t$ and $W_u-(u/t)W_t$ for $u\ne s$. Any idea to compute $E(X^2)$? – Did Oct 13 '16 at 6:05
• Many thanks for your reply. With your notation, I find that $X$ is normally distributed with zero mean and variance $t^3/12$, therefore the final result should be $\exp(\frac{t}{2}W_t+\frac{t^3}{24})$, does this seem correct? – user223935 Oct 13 '16 at 6:15
• Please show how you computed this variance. – Did Oct 13 '16 at 6:18
• I have now edited my post with the computation (general idea, I have obviously used Fubini to interchange expectation and double integral), does it seem correct now? – user223935 Oct 13 '16 at 6:26
• Yes. Well done. – Did Oct 13 '16 at 6:48

A systematic way to do this:

1. Compute the joint distribution of the two Gaussian variables $Y:=\int_0^T W_t dt$ and $X:=W_T$ and then
2. Evaluate the conditional distribution, which we know is obtained from the following least squares regression: $$Y = \alpha + \beta X + Z, \label{LSQ}\tag{1}$$ where $Z$ is zero mean Gaussian variable independent of $X$ with variance $$\sigma^2_Z = \mathrm{Var}(Y-\beta X)=\mathrm{Var}(Y)-\beta^2\mathrm{Var}(X),\label{sigZ}\tag{2}$$ $\beta$ is the least squares slope coefficient, $$\beta = \frac{\mathrm{Cov}(X,Y)}{\mathrm{Var}(X)},\label{beta}\tag{3}$$ and $\alpha$ is the so-called intercept, chosen to make the mean of $Z$ zero, $$\alpha = E[Y]-\beta E[X].\label{alpha}\tag{4}$$ In summary this gives the conditional distribution formula $$Y\mid X \sim N(\alpha +\beta X,\sigma^2_Z).\label{Y|X}\tag{5}$$
3. Finally evaluate the conditional expectation using the moment generating function formula for a Gaussian random variable $$E[\exp(Y)\mid X] =\exp(\alpha +\beta X + \sigma^2_Z/2).\label{MGF}\tag{6}$$

It remains to compute the individual ingredients.

1a) $Y$ is rewritten using (stochastic) integration by parts, $Y = TW_0+\int_0^T (T-t)dW_t.$ This gives $E[Y]=TW_0$ and $\mathrm{Var}(Y)=\int_0^T (T-t)^2dt=T^3/3$.

1b) $E[X] = W_0$ and $\mathrm{Var}(X)=T$ by standard properties of BM.

1c) $\mathrm{Cov}(X,Y) = \mathrm{Cov}(\int_0^T dW_t,\int_0^T (T-t)dW_t) = \int_0^T (T-t)dt =T^2/2$.

Calculate all the parameters of the regression (\ref{LSQ}) using formulae (\ref{sigZ}-\ref{alpha}), starting with (\ref{beta}):

2a) $\beta = T^2/(2T^2)=1/2$, $\alpha = TW_0 - W_0/2$.

2b) $\sigma^2_Z = T^3/3-T^2/4$.

2c) From (\ref{Y|X}) $Y\mid X \sim N(TW_0 + (W_T-W_0)/2, T^3/3-T^2/4)$.

Finally put everything together in (\ref{MGF}):

3a) $E[\exp(Y)\mid X] =\exp(TW_0 + (W_T-W_0)/2 + T^3/6-T^2/8)$.