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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

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