Suppose $X$ is an adapted process and $\tau_1, \ldots , \tau_k$ are stopping times such that $X^{\tau_1}, \ldots , X^{\tau_k}$ are all martingales. I want to show that then $X^{\tau_1 \vee \ldots \vee \tau_k}$ is a martingale.

Therefore, we have that \begin{align*} &X^{\tau_i} = \{X_{\tau_i \wedge t}: t \in \mathbb{R}_+ \} \qquad &&\text{w.r.t. the filtration $\{\mathcal{F}_t\}$.} \\ \text{and} \qquad &\mathbb{E}[X^{\tau_i}_t \mid \mathcal{F}_s] = X^{\tau_i}_s \qquad &&\text{for all $s < t$ and $1 \leq i \leq k$.} \end{align*} So, \begin{align*} X^{\tau_1 \vee \ldots \vee \tau_k} &= \{X_{(\tau_1 \vee \ldots \vee \tau_k) \wedge t}: t \in \mathbb{R}_+ \}\\ &= \{X_{(\tau_1 \vee t) \wedge \ldots \wedge (\tau_k\vee t)}: t \in \mathbb{R}_+ \}. \end{align*} In general, suppose $M$ is a right-continuous martingale and $\sigma$ is a stopping time. Then the stopped process $\{M_{\sigma \wedge t}: t \in \mathbb{R}_+ \}$ is a martingale w.r.t. the filtration $\{\mathcal{F}_t\}$. $(*)$

Hence, it suffices to elaborate the case $k=2$ and verify that $X^{\tau_1 \vee \tau_2}$ is indeed a martingale and then apply $(*)$ for all $k$.

However, I do not see how to write $X^{\tau_1 \vee \tau_2}$ in terms of $X^{\tau_1}$, $X^{\tau_2}$ and $X^{\tau_1 \wedge \tau_2}$.


1 Answer 1


We have $$ (\tau_1\vee\tau_2)\wedge t = (\tau_1\wedge t)\vee(\tau_2\wedge t)\\ $$ and
\begin{align} (\tau_1\wedge t)+(\tau_2\wedge t)&=(\tau_1\wedge t)\vee(\tau_2\wedge t)+(\tau_1\wedge t)\wedge(\tau_2\wedge t)\\ &=(\tau_1\wedge t)\vee(\tau_2\wedge t) + (\tau_1\wedge\tau_2\wedge t), \end{align} from which $$(\tau_1\wedge t)\vee(\tau_2\wedge t) = (\tau_1\wedge t)+(\tau_2\wedge t)-(\tau_1\wedge\tau_2\wedge t). $$ It follows that $$X_{(\tau_1\vee\tau_2)\wedge t} = X_{\tau_1\wedge t}+X_{\tau_2\wedge t}- X_{\tau_1\wedge\tau_2\wedge t}$$ and hence for $s<t$, \begin{align} \mathbb E\left[X_{(\tau_1\vee\tau_2)\wedge t}\mid\mathcal F_s\right] &= \mathbb E\left[X_{\tau_1\wedge t}\mid\mathcal F_s \right]+\mathbb E\left[X_{\tau_2\wedge t}\mid\mathcal F_s \right]-\mathbb E\left[X_{\tau_1\wedge\tau_2\wedge t}\mid\mathcal F_s \right]\\ &= X_{\tau_1\wedge s}+X_{\tau_2\wedge s}-X_{\tau_1\wedge\tau_2\wedge s}\\ &= X_{(\tau_1\vee\tau_2)\wedge s}. \end{align} This shows that $X^{\tau_1\vee\tau_2}$ is a martingale, from which the general result follows by induction.


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