Problem 12.2 4 from Grimmett (Probability and Random Process):
Let $(M,\mathcal{F})$ be a martingale with $M_0 = 0$ having differences $D_r = M_r - M_{r-1}$. Show that, $X_n = M_n^2 - Q_n$ is a martingale where $Q_n = \sum_{r=1}^n D_r^2$
My attempt,
\begin{align} & X_n = M_n^2 - Q_n \\ &\hspace{10pt} = M_n^2 - \sum_{r=1}^nM_r^2 - \sum_{r=1}^{n-1}M_r^2 + 2\sum_{r=1}^n M_rM_{r-1}; \quad (\text{since $M_0 = 0$ - for second term}) \\ &\hspace{10pt} = M_n^2 - M_n^2 + 2\sum_{r=1}^n M_r M_{r-1} \end{align} As a result, I think, \begin{align} & \mathbb{E}[X_n | \mathcal{F}_{n-1}] = \mathbb{E}[X_{n-1} + 2M_nM_{n-1}|\mathcal{F}_{n-1}] = X_{n-1} + 2M_{n-1}\mathbb{E}[M_n|\mathcal{F}_{n-1}] = X_{n-1}+ 2M_{n-1}^2 \end{align} *martingale property. This confuses me because I don't see if we can directly assume $M_{n-1}^2 = 0$ and/or if I've made some algebraic mistake. On the solutions, it writes: \begin{align} X_n = M_n^2 - Q_n = 2 \sum_{1 \le r < r \le n}D_rD_s. \end{align} However, I can't understand why that previous claim is valid. Any ideas/help or tips is very much appreciated. Thank you :),
Define $D_0$ to be $0$. Then $ \sum\limits_{r=0}^{n} D_r=M_n$. Hence, $X_n=M_n^{2}-Q_n=(\sum\limits_{r=0}^{n} D_r)^{2}-\sum\limits_{r=0}^{n} D_r^{2}=2\sum_{1\leq r<s\leq n}D_rD_s$ by just expanding the square.
To complete the proof you only need the fact that $E(\sum_{1\leq r<s= n}D_rD_s|\mathcal F_{n-1})=0$ which is easy from the martingale property of $(M_n)$.