Integration with respect to two different Brownian motions

Question

Let $B$ be the standard Brownian motion. The process $W_s=B_{s+a}-B_a$ is also a Brownian motion. I just want an example of a process $X_s$ such that $$E\int_0^tX_sdB_s\neq E\int_0^tX_sdW_s.$$

Answer 1

Such a process does not exist. Two possible argumentations:

• Since a stochastic integral with respect to Brownian motion is a martingale, the expectation is constant, i.e. $$\mathbb{E} \left( \int_0^t X_s \, dB_s \right) = \mathbb{E} \left( \int_0^t X_s \, dW_s \right) = 0.$$ (To be more precise: Here, we need some integrability assumptions on $(X_t)_{t \geq 0}$ to ensure that the stochastic integral is not only a local martingale, but a martingale.)
• Since both $(W_t)_{t \geq 0}$ and $(B_t)_{t \geq 0}$ are Brownian motions, the stochastic integrals $$\int_0^t X_s \, dB_s \quad \text{and} \quad \int_0^t X_s \, dW_s$$ are equal in distribution. In particular, the expectation value coincides.