Stopped process of maximum stopping times 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 $X^{\tau_1 \wedge \tau_2}$.
 A: 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.
