$S_t$ and $X_t$ stochastic processes; $S_t$ martingale. Why $E[\int_{0}^T S_tdX_t]=-S_0X_0$?

Question

Given $S_t$ and $X_t$ two stochastic processes, with $S_t$ martingale, why does the following hold true? $$E[\int_{0}^T S_tdX_t]=E[S_T X_T - S_0 X_0 - \int_0^T X_t dS_t] = -S_0X_0$$ The first equality comes from integration by parts and is crystal-clear. My doubts arise as to the second equality. I thought that $\int_0^T X_t dS_t=X_t[S_T-S_0]$ but I guess I am completely wrong. Why does the second equality holds true? Is this due to exploitation of martingale nature of $S_t$ (exploited by conditioning with respect to filtration at time 0, that is with respect to $\mathcal{F}_0$) or to a more trivial fact?

Answer 1

In general, the above identity is wrong: Let $T>0$, $(S_t)_{t\ge 0}$ be the exponential martingale of the Brownian motion, and let $X_t = t$ for all $t\ge 0$. Then $S_0=1$ and thus by Fubini, $$ E[\int_0^T S_t d X_t] = E[\int_0^T S_t d t] = \int_0^T \underbrace{E[S_t]}_{=1} d t = T \not= -1 \cdot 0 = -S_0 X_0. $$ Do you have any assumptions on $S$ and $X$?

For example, one can easily show by a Riemann-approximation that $$ E[\int_0^T S_t d X_t] = E[S_T X_T], $$ if e.g. $(X_t)$ is a continuous, increasing, integrable and $(\mathcal{F}_t)_{t\ge 0}$-adapted process with $X_0=0$ and $(S_t)$ is a bounded continuous a $(\mathcal{F}_t)_{t\ge 0}$-martingale. (Different assumptions might be possible...)