Solution: Predictable and optional variation processes

We can show that \(M_n\) is a martingale by showing that \(E[M_n|M_0,M_1,\ldots M_{n-1}] = M_{n-1}\) for all \(n\geq 1\) (using \((2.1)\) in ABG).

Note that \(X_n\) and \(M_{n-1}\) are independent. Thus we get \begin{align*}E[M_n|M_0,M_1,\ldots M_{n-1}] &= E[X_0+X_1+\ldots + X_n|M_0,M_1,\ldots M_{n-1}] \\ &=E[X_0+X_1+\ldots + X_{n-1}|M_0,M_1,\ldots M_{n-1}] + E[X_n|M_0,M_1,\ldots M_{n-1}] \\ &= E[M_{n-1}|M_0,M_1,\ldots M_{n-1}] + E[X_n] \\ &= M_{n-1}+0 \\ &= M_{n-1}. \end{align*}

Using the definition of the predictable variation process \(\langle M \rangle\), \((2.7)\) in ABG, that \(X_i\) and \(\mathcal{F}_{i-1}\) are independent, and that \[\Delta M_i = M_i-M_{i-1} = \sum_{j=0}^iX_j - \sum_{j=0}^{i-1}X_j = X_i,\] we get \begin{align*}\langle M \rangle_n &= \sum_{i=1}^n Var[\Delta M_i | \mathcal{F}_{i-1}] \\ &= \sum_{i=1}^n Var[X_i | \mathcal{F}_{i-1}] \\ &= \sum_{i=1}^n Var[X_i] \\ &= \sum_{i=1}^n \sigma^2 \\ &= n\sigma^2.\end{align*} Hence the predictable variation process is \[\underline{\underline{\langle M \rangle_n=n\sigma^2 \text{ for } n \geq 1, \quad \langle M \rangle_0=0.}}\]

Similarly, using the definition of the optional variation process \([M]\), \((2.8)\) in ABG, we get \begin{align*}[M]_n &= \sum_{i=1}^n (\Delta M_i)^2 \\ &= \sum_{i=1}^n X_i^2.\end{align*} Hence the optional variation process is \[\underline{\underline{[M]_n=\sum_{i=1}^n X_i^2 \text{ for } n \geq 1, \quad [M]_0=0.}}\]