How to compute the expectation of $e^{W_t}\int_0^t e^{-W_s} dW_s$

Question

I am having some trouble solving $\mathbb{E}[e^{W_t}\int_0^t e^{-W_s} dW_s]$. My idea would be to apply itos formular on $e^{W_t}$ to get \begin{align}\mathbb{E}[e^{W_t}\int_0^t e^{-W_s} dW_s] = \mathbb{E}[(1+\int_0^t e^{W_s} dW_s+1/2\int_0^t e^{W_s} ds)\int_0^t e^{-W_s} dW_s] \\ = \mathbb{E}[\int_0^t e^{-W_s} dW_s]+\mathbb{E}[\int_0^t e^{W_s-W_s} ds]+1/2\mathbb{E}[\int_0^t e^{W_s} ds\int_0^t e^{-W_s} dW_s] \end{align} My assumption would be that the first term is a martingale, thus having expectation $0$. The second expectation is deterministic and should equal to $t$. Unfortunately, I have no idea how to solve the third expectation.

Answer 1

It is well-known and easy to see by Ito's lemma that $$ M_t=e^{W_t-t/2} $$ is a martingale that satsifes the SDE $$ dM_t=M_t\,dW_t\,. $$ Equivalently, $$ M_t=1+\int_0^tM_s\,dW_s\,. $$ The calculation of $\mathbb E[e^{W_t}\int_0^te^{-W_s}\,dW_s]$ then boils down to \begin{align} &e^{t/2}\,\mathbb E\Big[M_t\int_0^te^{-W_s}\,dW_s\Big]= e^{t/2}\,\mathbb E\Big[\underbrace{\Big(\int_0^te^{W_s-s/2}\,dW_s\Big)}_{\textstyle=:X_t}\underbrace{\Big(\int_0^te^{-W_s}\,dW_s\Big)}_{\textstyle=:Y_t}\Big]\\ &=e^{t/2}\mathbb E\big[\big\langle X,Y\big\rangle_t\big]=e^{t/2}\mathbb E\Big[\int_0^te^{W_s-s/2}\,e^{-W_s}\,ds\Big]=e^{t/2}\int_0^te^{-s/2}\,ds\,. \end{align} Can you finish?

Answer 2

Itô's lemma is indeed useful in this situation, but it is easier to apply it in its local form to the process $X_t = f(W_t)$, where $f(x) = \int_0^x e^{x-y} \,\mathrm{d}y$, hence $$ \mathrm{d}X_t = \frac{1}{2}\left(X_t - 2e^{W_t} + 1\right)\mathrm{d}t + \left(X_t - e^{W_t} + 1\right)\mathrm{d}W_t. $$ Now, recalling that $e^{W_t}$ follows a log-normal distribution, whose mean is given by $e^{t/2}$ in the present case, we find by averaging the previous SDE : $$ \frac{\mathrm{d}}{\mathrm{d}t}\langle X_t\rangle = \frac{1}{2}\langle X_t\rangle - e^{t/2} + \frac{1}{2} $$ This is an inhomogeneous first-order linear ODE for $\langle X_t \rangle$, which is solved by $\langle X_t \rangle = (A-t)e^{t/2} - 1$, with $A$ a constant, which is determined thanks to the initial condition $\langle X_t \rangle = A-1 = 1$ (itself derived from $X_0 = f(W_0) = f(0) = 1$, hence $A = 2$.

Answer 3

Heuristically, $\int_{0}^{t} e^{-W_s} \, \mathrm{d}W_s$ is the sum of infinitesimal elements $e^{-W_s} (W_{s+\mathrm{d}s}-W_s)$ over an infinitely fine subpartition of $[0, t]$. So,

\begin{align*} \mathbf{E}\left[ e^{W_t} \int_{0}^{t} e^{-W_s} \, \mathrm{d}W_s \right] &= \int_{0}^{t} \mathbf{E}\left[ e^{W_t} e^{-W_s} (W_{s+\mathrm{d}s}-W_s) \right] \\ &= \int_{0}^{t} \mathbf{E}[e^{W_t-W_{s+\mathrm{d}s}}]\mathbf{E}\left[ e^{W_{s+\mathrm{d}s}-W_s} (W_{s+\mathrm{d}s}-W_s) \right]. \end{align*}

In the last line, we utilized the fact that $W_t-W_{s+\mathrm{d}s}$ and $W_{s+\mathrm{d}s}-W_s$ are independent. To simplify the last integral, we invoke the following formulas:

$$ \mathbf{E}[e^{X}] = e^{v/2} \quad\text{and}\quad \mathbf{E}[Xe^{X}] = ve^{v/2} \qquad\text{for}\quad X \sim \mathcal{N}(0, v). $$

Since $W_b - W_a \sim \mathcal{N}(0, b-a)$ for $0 \leq a \leq b$, it therefore follows that

\begin{align*} \mathbf{E}\left[ e^{W_t} \int_{0}^{t} e^{-W_s} \, \mathrm{d}W_s \right] &= \int_{0}^{t} e^{\frac{1}{2}(t-s)} \, \mathrm{d}s. \end{align*}

Although the above computation is only heuristic, it can be justified either by reducing the stochastic integral to the limit of Riemann sums (where all the above steps can be justified with due modifications) or by invoking an alternative argument (such as using martingale as in @Kurt G.'s answer).

Answer 4

Another way to do this is to express $$e^{W_t}\int_0^t e^{-W_s}dW_s$$ as an Ito process. Denote $U_t = e^{W_t}$ and $V_t = \int_0^t e^{-W_s}dW_s$, then by the integration by parts formula: $$ d(U_t V_t) = \left(\dfrac{1}{2}U_t V_t + 1\right)dt + (U_t V_t + 1)dW_t $$ which means $$ U_t V_t = \int_0^t \left(\dfrac{1}{2}U_s V_s + 1\right)ds + \int_0^t (U_s V_s + 1)dW_s $$ By the property of the Ito integral and Fubini theorem, taking expectation we have $$ \mathbb{E}(U_t V_t) = \int_0^t \left(\dfrac{1}{2}\mathbb{E}(U_s V_s) + 1\right)ds $$ The only left to do is to solve the ODE: $$ \dfrac{df}{dt} = \dfrac{1}{2}f + 1, f(0) = 0 $$ where $f(t) = \mathbb{E}(U_t V_t) = \mathbb{E}\left(\displaystyle e^{W_t}\int_0^t e^{-W_s}dW_s\right)$