Solution: The martingale property is preserved under optional stopping

We are considering the process \(M^T\) defined by \(M_n^T = M_{\min \{n,T\}}\), where \(M\) is a martingale, and are to show that \(M^T\) itself is a martingale. By the martingale preservation property it is sufficient to find a predictable process \(H\) such that \(M^T = H \bullet M\).

Let \(H_n = I(T > n - 1 ))\). Since \(T\) is an (optional) stopping time, we know that whether or not \(T>n-1\) is only a function of what has happened up to (and including) time \(n-1.\) Thereby the value of \(H_n\) is know at time \(n-1\), so the \(\{H_n\}\) process is predictable. In the following we show that with this \(\{H_n\}\) process we get \(H\bullet M = M^T.\)

First note that we for \(n=0\) have \((H\bullet M)_0 = H_0M_0 = I(T > -1) M_0 = M_0 = 0 = M_{\min \{0,T\}}\).

For \(n\geq 1\) we get \begin{align*}(H\bullet M)_n &= \sum_{i=1}^nH_i(M_i-M_{i-1}) \\ &= \sum_{i=1}^nI(T > i-1) (M_i-M_{i-1}) \\ &= \sum_{i=1}^n I(i\leq T) (M_i-M_{i-1}) \\ &= \sum_{i=1}^{\min \{n,T\}}(M_i-M_{i-1}) \\ &= M_{\min \{n,T\}}.\end{align*} Thus we have shown that \[\underline{\underline{M^T = H \bullet M.}}\]