Let $T$ be stopping time, and $X_n$ be stochastic process. Then the stopped process $X_{n \wedge T}$ can be written as $$X_{n \wedge T} = X_n 1_{T \ge n} + X_T 1_{T < n}$$ where $1_{(\cdot)}$ is indicator function.
So, if we look at $n-1$ case, the above equality should become
$$X_{(n-1) \wedge T} = X_{n-1} 1_{T \ge n-1} + X_T 1_{T < n-1} \ \ \ \ \ (*)$$
But I was wondering is the following equality also true? $$X_{(n-1) \wedge T} = X_{n-1} 1_{T \ge n} + X_T 1_{T < n} \ \ \ \ \ (\star)$$
Since I found one book (Resnick, A probability path, proposition 10.8.2, p.368) uses above gray-colored equality to deduce that a stopped process is still a martingale if the original process is a martingale. But I don't see how this equality $(\star)$ holds and why this equality $(\star)$ seems not match the previous equality $(*)$?
Thank you.
They are the same. Note that
$$1_{T \geq n-1} = 1_{T \geq n} + 1_{T=n-1} \tag{1}$$
and
$$X_{n-1} 1_{T=n-1} = X_{T} 1_{T=n-1}. \tag{2}$$
Therefore,
$$\begin{align*} X_{n-1} 1_{T \geq n-1} + X_{T} 1_{T<n-1} &\stackrel{(1)}{=} X_{n-1} 1_{T \geq n} + X_{n-1} 1_{T=n-1} + X_T 1_{T<n-1} \\ &\stackrel{(2)}{=} X_{n-1} 1_{T \geq n} +\underbrace{X_{T} 1_{T=n-1} + X_T 1_{T<n-1}}_{X_T 1_{T <n}}. \end{align*}$$