If $(τ_n)$ is nondecreasing and $M$ is a martingale, then $\text E\left[\sum_n\left|M^{τ_n}_t-M^{τ_{n-1}}_t\right|^2\right]=\text E\left[M_t^2\right]$

Question

Let

• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
• $(\mathcal F_t)_{t\ge0}$ be a right-continuous filtration of $\mathcal A$
• $\tau_n$ be an $\mathcal F$-stopping time for $n\in\mathbb N_0$ with $$\tau_{n-1}\le\tau_n\;\;\;\text{for all }n\in\mathbb N\tag1$$ and $$\tau_n\xrightarrow{n\to\infty}\infty\tag2$$
• $M$ be a right-continuous $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $$M_t\in L^2(\operatorname P)\;\;\;\text{for all }t\ge0\tag3$$

I want to show that $$\operatorname E\left[\sum_{n\in\mathbb N}\left|M^{\tau_n}_t-M^{\tau_{n-1}}_t\right|^2\right]=\operatorname E\left[M_t^2\right]\;\;\;\text{for all }t\ge0\;.\tag4$$

Since $(\tau_n)_{n\in\mathbb N_0}$ is nondecreasing, it's easy to see that $$\operatorname E\left[M_t^{\tau_n}-M_t^{\tau_{n-1}}\mid\mathcal F_{\tau_{n-1}}\right]=0\;\;\;\text{for all }t\ge0\text{ and }n\in\mathbb N\tag5$$ and hence $$\operatorname E\left[M_t^{\tau_n}-M_t^{\tau_{n-1}}\right]=0\;\;\;\text{for all }t\ge0\text{ and }n\in\mathbb N\;,\tag6$$ but I have no idea how we can show $(4)$. So, how can we prove it?

Answer 1

Hints:

1. Using the martingale property show that $$\mathbb{E}((M_t^{\tau_n}-M_t^{\tau_{n-1}})^2) = \mathbb{E}((M_t^{\tau_n})^2)-\mathbb{E}((M_t^{\tau_{n-1}})^2).$$
2. By step 1, we have $$\mathbb{E} \left( \sum_{n=1}^k |M_t^{\tau_n}-M_{t}^{\tau_{n-1}}|^2 \right) = \sum_{n=1}^k \bigg[ \mathbb{E}((M_t^{\tau_n})^2)-\mathbb{E}((M_t^{\tau_{n-1}})^2) \bigg].$$ Show that the right-hand side equals $$\sum_{n=1}^k \mathbb{E}(M_t^2 1_{\{\tau_{n-1} \leq t < \tau_n\}}) + \mathbb{E}(M_{\tau_k}^2 1_{\{t \geq \tau_k\}}).$$
3. Let $k \to \infty$ using the monotone convergence theorem.