For $t \in [0,T]$. consider two stochastic integrals with a nonnegative constant integrand $c$
- $$\mathbb{E} \left[ \int_0^{t(\omega)^* \wedge T} c \cdot dW_t \right]$$
where $t^*$ is random stopping time, and $W_t$ is Wiener process.
In this simple situation, "can I just interchange the expectation and the integral sign" (By Fubini's result?) and get
$$\mathbb{E}\left [\int_0^{t(\omega)^* \wedge T} c \cdot dW_t \right] = ? =c \cdot \int_0^{t(\omega)^* \wedge T} \mathbb{E}[ dW_t ]= 0?$$
2.Consider another stochastic integral
$$\mathbb{E}\left[ \int_0^{t(\omega)^* \wedge T} dt \right]$$
I know $\mathbb{E}\left[ \int_0^{t(\omega)^* \wedge T} c \cdot dt \right] = c \cdot \mathbb{E}\left[ t(\omega)^* \wedge T \right]$, but can I compute this expectation more explicitly? (I attempt to get rid of the random stopping time and get a deterministic-like result)
Thanks you
For 1), we have $$E[\int_0^{t(\omega)^*\wedge T}c dW_s]=cE[W_{t(\omega)^*\wedge T}]$$.
$t(\omega)^*\wedge T$ is a stopping time for which is bounded by $T$. Hence by the optinoal stopping theorem $$E[W_{t(\omega)^*\wedge T}]=E[W_0]=0$$
For 2). This depends on the exact definition of your $t(\omega)^*$. For a general stopping time this can not be simplified further. You can get an upper bound by $T$ since $t(\omega)^*\wedge T\le T$ for any $T\in\mathbb{R}$