Here’s an easy one, I hope:


Suppose $\tau$ is a stopping time and $(M\_t)$ is a martingale which together satisfy the hypotheses of the optional stopping theorem so that $\mathbb{E}[M\_\tau]= \mathbb{E}[M\_0]$.


Will it also be the case that $\mathbb{E}[M\_\sigma]= \mathbb{E}[M\_0]$ for an arbitrary random time $\sigma \leq \tau$, for example $\sigma= \tau-1$?


Thank you very much for your help.



Martingale optional stopping before a stopping time
No. The problem is that $\tau-1$ is not a "stopping time" for the martingale.


Example. Martingale $(X\_n), n=0,1,2$ is the standard random walk on $\mathbb Z$. Our sample space is $[0,1)$ with Lebesgue measure.
\begin{align}
X\_0(\omega) &= 0,\qquad \omega \in [0,1).
\\
X\_1(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/2),
\\
-1,\quad & \omega \in [1/2,1).
\end{cases}
\\
X\_2(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
0,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
-2,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}


The filtration $\mathcal F\_0, \mathcal F\_1, \mathcal F\_2$ for this martingale:  

$\mathcal F\_0$ is the sigma-algebra generated by $X\_0$, so it is the trivial sigma-algebra with atom $[0,1)$.  

$\mathcal F\_1$ is the sigma-algebra generated by $X\_0, X\_1$, so it is the sigma-algebra with atoms $[0,1/2), [1/2,1)$.  

$\mathcal F\_2$ is the sigma-algebra generated by $X\_0, X\_1, X\_2$, so it is the sigma-algebra with atoms $[0,1/4),[1/4,1/2),[1/2,3/4),[3/4,1)$. 


Define stopping time $\tau = 2 \wedge \min\{n : X\_n < 0\}.$ Then
\begin{align}
\tau(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
2,\quad & \omega \in [1/4,1/2),\\
1,\quad & \omega \in [1/2,3/4),\\
1,\quad & \omega \in [3/4,1).
\end{cases}
\\
\sigma(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/4),\\
1,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
0,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
Now $\tau$ is a stopping time, so of course $\mathbb E[X\_\tau] = 0$. But compute $\mathbb E[X\_\sigma] = 1/2 \ne 0$, which shows that $\sigma$ is not a stopping time. Computations:
\begin{align}X\_\tau(\omega) &= \begin{cases}
2,\quad & \omega \in [0,1/4),\\
0,\quad & \omega \in [1/4,1/2),\\
-1,\quad & \omega \in [1/2,3/4),\\
-1,\quad & \omega \in [3/4,1).
\end{cases}
\\
X\_\sigma(\omega) &= \begin{cases}
1,\quad & \omega \in [0,1/4),\\
1,\quad & \omega \in [1/4,1/2),\\
0,\quad & \omega \in [1/2,3/4),\\
0,\quad & \omega \in [3/4,1).
\end{cases}
\end{align}
Calculation that shows $\sigma$ is not a stopping time:
$$
\{\sigma = 0\} = [1/2,1) \notin \mathcal F\_0.
$$


