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.

  • 4
    $\begingroup$ 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 $E(X^2)$? $\endgroup$ – Did Oct 13 '16 at 6:05
  • $\begingroup$ 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? $\endgroup$ – user223935 Oct 13 '16 at 6:15
  • 2
    $\begingroup$ Please show how you computed this variance. $\endgroup$ – Did Oct 13 '16 at 6:18
  • $\begingroup$ 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? $\endgroup$ – user223935 Oct 13 '16 at 6:26
  • 1
    $\begingroup$ Yes. Well done. $\endgroup$ – 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) $E[\exp(Y)\mid X] =\exp(TW_0 + (W_T-W_0)/2 + T^3/6-T^2/8)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.