Solution: Equivalent definitions of a martingale

We are to show that \(\left\{E[M_n | \mathcal{F}_{n−1}] = M_{n−1} \quad \forall n \geq 1 \quad (2.3)\right\} \Leftrightarrow \left\{E[M_n | \mathcal{F}_{m}] = M_{m} \quad \forall n,m \colon \quad n > m \geq 0 \quad (2.4)\right\}.\)

Proving \(\Leftarrow\):
Assume \((2.4)\) holds. Let \(m = n-1\). The implication immediately follows.

Proving \(\Rightarrow\):

Assume \((2.3)\) holds. The case \(m = n-1\) holds trivially, but we need to show the remaining cases with \(m\) such that \(0 \leq m < n-1\). We use the hint provided in the exercise, namely \(E[M_n | \mathcal{F}_{m_1} ] = E[E[M_n | \mathcal{F}_{m_2} ]|\mathcal{F}_{m_1} ]\) whenever \(0 \leq m_1 < m_2 < n\). With an arbitrary \(m\) such that \(0 \leq m < n-1 = m_2\) we thus get \begin{align*}E[M_n | \mathcal{F}_{m} ] &= E[E[M_n | \mathcal{F}_{n-1} ]|\mathcal{F}_{m} ] \\ &= E[M_{n-1}|\mathcal{F}_{m} ].\end{align*} Now, if \(n-1 = m + 1\) we get \(E[M_{n-1}|\mathcal{F}_{m} = M_m]\) by \((2.3)\). If not, the same argument as before yields \begin{align*}E[M_n | \mathcal{F}_{m} ] &= E[M_{n-1}|\mathcal{F}_{m} ] \\ &= E[M_{n-2}|\mathcal{F}_{m} ] \\&= \ldots \\ &= E[M_{m+1}|\mathcal{F}_{m} ] \\ &= M_m,\end{align*} and so the desired implication holds.