1
$\begingroup$

Right now, I want to figure out the covariance of a stochastic integral and a riemann integral in the following form:

$$E\left(\int_{0}^{t}\exp[B(t)-B(s)]ds \cdot \int_{0}^{t}\exp[B(t)-B(s)]dW(s)\right).$$ where $B(\cdot)$ and $W(\cdot)$ are independent Brownian motions, or $d\langle B, W \rangle_t = \gamma dt$.

My guess to this is 0 and my calculation is

$$E\left(\int_{0}^{t}\exp[B(t)-B(s)]ds \cdot \int_{0}^{t}\exp[B(t)-B(s)]dW(s)\right) = E\left(\int_{0}^{t}\exp[B(t)-B(s)]\int_{0}^{t}\exp[B(t)-B(r)]dr dW(s)\right) = 0 $$ by virtue of $E(\int \cdot dW) = 0$. Am I correct? If not, what tool can I use to derive the correct result?

$\endgroup$
  • $\begingroup$ What is your filtration $\{\mathcal{F}_t, t\ge 0\}$, the stochastic integrals and covariation $\langle B,W\rangle$ need it. $\endgroup$ – JGWang Mar 24 '17 at 3:04
  • $\begingroup$ The filtration $\mathcal{F}_{t}:=\sigma (B(t),W(t):t\geq 0)$ $\endgroup$ – Tom Mar 25 '17 at 8:49
1
$\begingroup$

First of all, $\int_0^t\exp[B(t)-B(s)]\,dW(s)$ should be understood as $\exp[B(t)]\int_0^t\exp[-B(s)]\,dW(s)$, otherwise, the integral couldn't be defined as an Ito's integral.

Let \begin{align} Z(t)&\Bigl(=\int_0^t\exp[B(t)-B(s)]\,ds\int_0^t\exp[B(t)-B(s)]\,dW(s)\Bigr)\\ &=e^{2B(t)}\int_0^t\exp[-B(s)]\,ds\int_0^t\exp[-B(s)]\,dW(s) \end{align} Using Ito's formula, we have \begin{align} dZ(t)&=2Z(t)\,dB(t)+2Z(t)\,dt+\Bigl(e^{B(t)}\int_0^t\exp[-B(s)]\,dW(s)\Bigr)\,dt\\ &\quad +\Bigl(e^{B(t)}\int_0^t\exp[-B(s)]\,ds\Bigr)\,dW(t) +2\gamma \Bigl(e^{B(t)}\int_0^t\exp[-B(s)]\,ds\Bigr)\,dt \tag{1} \end{align} Denote $m(t)=\mathsf{E}[Z(t)]$, then $m(0)=0$ and from (1) we have \begin{align} dm(t)&=2m(t)\,dt+\mathsf{E}\Bigl[e^{B(t)}\int_0^t e^{-B(s)}\,dW(s)\Bigr] +2\gamma\int_0^t\mathsf{E}[e^{B(t)-B(s)}]\,dt\\ &=2m(t)\,dt+6\gamma(e^{t/2}-1).\tag{2} \end{align} where we use following equalities: \begin{align} \mathsf{E}\Bigl[e^{B(t)}\int_0^t e^{-B(s)}\,dW(s)\Bigr]&=2\gamma(e^{t/2}-1),\tag{3}\\ \mathsf{E}[e^{B(t)-B(s)}]&=e^{(t-s)/2} \end{align} Now from (2) and $m(0)=0$ we could get the expression of \begin{align} m(t)&=\mathsf{E}\Bigl[\int_0^t\exp[B(t)-B(s)]\,ds\int_0^t\exp[B(t)-B(s)]\,dW(s)\Bigr]\\ &=\gamma e^{2t}-4\gamma e^{t/2}+3\gamma.\end{align}

To get (3) using same way as above(Ito's formula and solving a ODE). Let $X(t)=e^{B(t)}\int_0^te^{B(s)}\,dW(s)$ and $m_X(t)=\mathsf{E}[X(t)]$, then \begin{align} dX(t)&=X(t)\,dB(t)+\frac12X(t)\,dt+dW(t)+\gamma dt,\\ dm_X(t)&=\frac12m_X(t)+\gamma dt, \qquad m_X(0)=0\\ m_x(t)&=\gamma\int_0^te^{(t-s)/2}\,ds=2\gamma(e^{t/2}-1). \end{align} Please excuse me that above deduction may be included some errors and typos, since the process includes many details be checked.

$\endgroup$
  • $\begingroup$ Can you briefly explain equation (3)? I am not quite familiar with how to find $E(\int_{0}^{t} B(s) dW(s))$ and $E(\int_{0}^{t} B(s) dW(s))^2$ when $d\langle B, W \rangle_t = \gamma dt$. Does ito isometry still applies? $\endgroup$ – Tom Mar 28 '17 at 10:39
  • $\begingroup$ I have posted a new question here: mathoverflow.net/questions/265752/… $\endgroup$ – Tom Mar 28 '17 at 11:30
  • $\begingroup$ @Tom, thank you for your replication. I add the proof of (3). $\endgroup$ – JGWang Mar 29 '17 at 3:59

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.