[Math] Proof that a stopped continuous-time martingale is a martingale.

martingalesstopping-times

The proof for a stopped discrete-time martingale is shown as follows.

Let $M=(M_n)_{n\ge0}$ be a discrete-time martinglae w.r.t. the filtration $(\mathcal F_n)_{n\ge0}$, and let $M^T=(M_{n\land T})_{n\ge0}$ be the stopped martingale, where $T$ is a stopping time w.r.t. $(\mathcal F_n)_{n\ge0}$. Since:

  1. We have
    $$\begin{align}
    M_{n\land T}&=1_{T\ge n+1}M_n+1_{T\le n}M_T\\
    &=1_{T\ge n+1}M_n+ \sum_{k=1}^n 1_{T=k}M_k\\
    \end{align}$$
    where $1_{T\ge n+1}, M_n, 1_{T=k}$, and $M_k$ are all $\mathcal F_n-measurable$, hence $M_{n\land T}$ is $\mathcal F_n-measurable$;

  2. ${\Bbb E}|M_{n\land T}| \le \underbrace{{\Bbb E}|M_1|+…+{\Bbb E}|M_n|}_{{\Bbb E}|M_i| \text{ is integrable, }\forall i\ge0}<\infty$, i.e., $M_{n\land T}$ is integrable $\forall n \ge 0$;

  3. We have
    $$\begin{align}
    {\Bbb E}(M_{n+1 \land T}|{\mathcal F_n}) &= {\Bbb E}(M_{n \land T}+1_{T \ge n+1}(M_{n+1}-M_n)|{\mathcal F_n})\\
    &=M_{n \land T}+1_{T \ge n+1}{\Bbb E}(M_{n+1}-M_n)\\
    &=M_{n \land T}\\
    \end{align}$$

Therefore, the stopped martingale satisfies the definition of a discrete martingale. Proof complete.

However, I am not able to extend these three parts of proof to a continuous version, because I cannot devide the time into separate spots with one next to another as the discrete version did. So I really wonder how to give a proof for a continuous-time martingale.

Best Answer

We are going to prove the following proposition:

If $M$ is a continuous martingale and $T$ a stopping time, the stopped process $M^{T}$, i.e., $\{M(t \wedge T), t \geq 0\}$ is a martingale with respect to $\left(\mathcal{F}_{t}\right)$.

Here we go.

The process $M^{T}$ is obviously continuous and adapted. Firstly we use a weak form of optional stopping theorem, saying that a martingale has equal expectation at any bounded stopping time. If $S$ is a bounded stopping time, so is $S \wedge T$; hence $$ E\left[M_{S}^{T}\right]=E\left[M_{S \wedge T}\right]=E\left[M_{0}\right]=E\left[M_{0}^{T}\right]. $$ Then we use this conclusion twice to get our desired equality. If $s<t$ and $A \in \mathcal{F}_{s}$ the r.v. $T=t \mathbf{1}_{A^{c}}+s \mathbf{1}_{A}$ is a stopping time and consequently $$ E\left[X_{0}\right]=E\left[X_{T}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{s} \mathbf{1}_{A}\right] .$$ On the other hand, $t$ itself is a stopping time, and $$ E\left[X_{0}\right]=E\left[X_{t}\right]=E\left[X_{t} \mathbf{1}_{A^{c}}\right]+E\left[X_{t} \mathbf{1}_{A}\right]. $$

Comparing the two equalities yields $X_{s}=E\left[X_{t} | \mathcal{F}_{s}\right].$

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