Solution: Property of the predictable variation process

We are to show that \(M_0^2 - \langle M \rangle_0 = 0\) and \(E[M_n^2 - \langle M \rangle_n | \mathcal{F}_{n-1}] = M_{n-1}^2 - \langle M \rangle_{n-1}\), i.e. that \(M_n^2 - \langle M \rangle_n\) is a mean-zero martingale.

First note that we generally assume \(M_0 = 0\) (see section 2.1.1 in ABG), and that by definition we also have \(\langle M \rangle_0 = 0\), and thus \(M_0^2 - \langle M \rangle_0 = 0\) holds. Next we follow an argument similar to the proof of \((2.10)\) in ABG. We use the following \[M_n^2 = (M_{n-1}+M_n-M_{n-1})^2\] and from \((2.7)\) that \[\langle M \rangle_n = \langle M \rangle_{n-1} + E[(M_n-M_{n-1})^2|\mathcal{F}_{n-1}].\]

(i)

Now we first see, keeping in mind that \(E[M_n | \mathcal{F}_m] = M_n\) whenever \(m \geq n\), that \begin{align*}E[M_n^2| \mathcal{F}_{n-1}] &= E[(M_{n-1}+M_n-M_{n-1})^2 | \mathcal{F}_{n-1}] \\ &= E[M_{n-1}^2+2M_{n-1}(M_n-M_{n-1})+(M_n-M_{n-1})^2 | \mathcal{F}_{n-1}] \\ &= M_{n-1}^2 + 2M_{n-1}(E[M_n| \mathcal{F}_{n-1}]-M_{n-1}) +E[(M_n-M_{n-1})^2 | \mathcal{F}_{n-1}]\\ &= 2M_{n-1}E[M_n| \mathcal{F}_{n-1}]-M_{n-1}^2 +E[(M_n-M_{n-1})^2 | \mathcal{F}_{n-1}]\\ &= 2M_{n-1}^2-M_{n-1}^2 +E[(M_n-M_{n-1})^2 | \mathcal{F}_{n-1}]\\ &= M_{n-1}^2 +E[(M_n-M_{n-1})^2 | \mathcal{F}_{n-1}] , \end{align*} where we used that \(E[M_n| \mathcal{F}_{n-1}] = M_{n-1}\), since \(M_n\) is assumed to be a martingale.

(ii)

Next note that \begin{align*}E[\langle M \rangle_n | \mathcal{F}_{n-1}] &= E[\langle M \rangle_{n-1} + E[(M_n-M_{n-1})^2|\mathcal{F}_{n-1}] | \mathcal{F}_{n-1}] \\ &= E[\langle M \rangle_{n-1}|\mathcal{F}_{n-1}] + E[(M_n-M_{n-1})^2|\mathcal{F}_{n-1}] \\ &= \langle M \rangle_{n-1} + E[(M_n-M_{n-1})^2|\mathcal{F}_{n-1}].\end{align*} Note that \(E[\langle M \rangle_{n-1}|\mathcal{F}_{n-1}] = \langle M \rangle_{n-1}\), since \(\langle M \rangle_{n-1}\) is fully determined by \(\{M_0, M_1, \ldots, M_{n-1}\}\).

(iii)

Putting (i) and (ii) together we get \begin{align*}E[M_n^2 - \langle M \rangle_n | \mathcal{F}_{n-1}] &= E[M_n^2| \mathcal{F}_{n-1}] - E[\langle M \rangle_n | \mathcal{F}_{n-1}] \\ &= (M_{n-1}^2 +E[(M_n-M_{n-1})^2 | \mathcal{F}_{n-1}]) - (\langle M \rangle_{n-1} + E[(M_n-M_{n-1})^2|\mathcal{F}_{n-1}]) \\ &= M_{n-1}^2-\langle M \rangle_{n-1},\end{align*} and hence we have shown that \[\underline{\underline{M_0^2 - \langle M \rangle_0 = 0 \text{ and } E[M_n^2 - \langle M \rangle_n | \mathcal{F}_{n-1}] = M_{n-1}^2 - \langle M \rangle_{n-1}}}\]