Solution: Partial sums of independent stochastic variables

(a)

We can show that \(M_n\) is a martingale by showing that \(E[M_n|\mathcal{F}_{n-1}] = M_{n-1}\) for all \(n\geq 1\). The process being mean-zero then follows from \((2.4)\) and \((2.5)\) in ABG.

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

(b)

By the same argument as above we get \begin{align*}E[S_n|\mathcal{F}_{n-1}] &= E[X_1+\ldots + X_n|\mathcal{F}_{n-1}] \\ &=E[X_1+\ldots + X_{n-1}|\mathcal{F}_{n-1}] + E[X_n|\mathcal{F}_{n-1}] \\ &= E[X_1+\ldots + X_{n-1}|\mathcal{F}_{n-1}] + E[X_n] \\ &= E[S_{n-1}|\mathcal{F}_{n-1}] +\mu \\ &= S_{n-1} + \mu. \end{align*} Unless \(\mu=0\), this means that \(S_n\) is not martingale.

The transformation \(S'_n = S_n - n\mu\) yields \(S'_n = X_1 + \ldots + X_n - n\mu\ = \sum_{i=1}^n(X_i-\mu) = \sum_{i=1}^n Z_i\), where \(E[Z_i] = E[X_i - \mu] = 0\). By (a) \(S'_n\) is therefore a mean-zero martingale.

(c)

When conditioning on \(\mathcal{F}_{n-1}\), we get that \(X_1, \ldots, X_{n-1}\) are known. This, combined with the independence between \(\mathcal{F}_{n-1}\) and \(X_n\), yields \begin{align*}E[M_n|\mathcal{F}_{n-1}] &= E[X_1\cdot\ldots \cdot X_n|\mathcal{F}_{n-1}] \\ &=X_1\cdot \ldots \cdot X_{n-1}E[X_n|\mathcal{F}_{n-1}] \\ &= M_{n-1}E[X_n] \\ &= M_{n-1}. \end{align*} Hence \(M_n\) is a martingale. For the unconditional expectation we can use directly that the \(X_i\) are independent, and get \begin{align*}E[M_n] &= E[X_1\cdot\ldots \cdot X_n] \\ &=E[X_1]\cdot\ldots \cdot E[X_n] \\ &= 1.\end{align*}