# how to show that the price process is a martingale

Suppose I have an $d$-dimensional semimartingale $S=\{S_t\}$ with $t\in[0,T]$ under $P$. $S$ need not to be continuous (RCLL can be assumed). Suppose $Q$ is an equivalent measure w.r.t. $P$ such that

$$E_Q[\int_0^T\theta_u dS_u]\le 0$$

for all admissbale $\theta$'s, i.e. a process $\theta$ is called admissable if it is predictable, $S$-integrable such that $\int_0^t\theta_udS_u\ge -a$ for $a>0$ and for all $t$ (the integral is uniformly bounded from below). No I want to show, if $S$ is bounded then $S$ is a martingale under $Q$.

What I did so far:

2. Since we assume that $S$ is bounded, it is also integrable

All we have to prove is therefore the martingale property. Since we have the above properties, I'm quite sure to prove the martingale property in its elementary form:

Let $A\in\mathcal{F}_s$, $s\le t$, we want to show

$$E_Q[S_t\mathbf1_A]=E_Q[S_s\mathbf1_A]$$

This is equivalent to $E_Q[\mathbf1_A(S_t-S_s)]=0$. Now my idea is to use nice integrands, such as $\theta=\mathbf1_{]]s,t]]}$, which is admissable hence we have

$$-a\le E_Q[\int_s^tdS]=E_Q[S_t-S_s]\le 0$$

Somehow I have to find an integrand $\theta$, such that $\int\theta_udS_u$=$\mathbf1_A(S_t-S_s)$ and show that we even have $E_Q[\int_0^T\theta_u dS_u]=0$, for this $\theta$. Unfortunately here I got stuck. Some hints would be appreciated. Thanks in advance!

math

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choose $\theta_u = \pm 1_{(0,\tau)}, \tau$ a bounded stopping time. – mike Oct 16 '12 at 11:10
@mike Stupid me! I also used this approach, but I don't realized to use $-\mathbf1_{(0,\tau)}$ to get an equality instead of an inequality! You can turn your comment into an answer and I will accept it – math Oct 16 '12 at 12:33