Stochastic Processes and Financial Mathematics
(part one)

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Chapter 7 Stochastic processes and martingale theory

In this section we introduce two important results from the theory of martingales, namely the ‘martingale transform’ and the ‘martingale convergence theorem’. We use these results to analyse the behaviour of stochastic processes, including those from Chapter 4 (random walks, urns, branching processes) and also the gambling game Roulette.

Despite having found a martingale connected to the binomial model, in Proposition 5.5.6, we won’t use martingales to analyse our financial models – yet. That will come in Chapter 15, once we have moved into continuous time and introduced the Black-Scholes model.

7.1 The martingale transform

If $M$ is a stochastic process and $C$ is an adapted process, we define the martingale transform of $C$ by $M$

\[(C\circ M)_n = \sum _{i=1}^n C_{i-1} (M_i-M_{i-1}). \]

Here, by convention, we set $(C\circ M)_0=0$.

If $M$ is a martingale, the process $(C\circ M)_n$ can be thought of as our winnings after $n$ plays of a game. Here, at round $i$, a bet of $C_{i-1}$ is made, and the change to our resulting wealth is $C_{i-1}(M_i-M_{i-1})$. For example, if $C_i\equiv 1$ and $M_n$ is the simple random walk $M_n=\sum _1^n X_i$ then $M_i-M_{i-1}=X_{i}$, so we win/lose each round with even chances; we bet $1$ on each round, if we win we get our money back doubled, if we lose we get nothing back.

We place the bet $C_{i-1}\in m\mc {F}_{i-1}$ on play $i$, meaning that we must place our $i^{th}$ bet using only the information we gained during the first $i-1$ plays. In particular, we don’t yet know the result of the $i^{th}$ play. So our filtration here is $\mc {F}_n=\sigma (C_i,X_i\-i\leq n).$

Theorem 7.1.1 If $M$ is a martingale and $C$ is adapted and bounded, then $(C\circ M)_n$ is also a martingale.

Similarly, if $M$ is a supermartingale (resp. submartingale), and $C$ is adapted, bounded and non-negative, then $(C\circ M)_n$ is also a supermartingale martingale (resp. submartingale).

Proof: Let $M$ be a martingale. Write $Y=C\circ M$. We have $C_n\in m\mc {F}_{n}$ and $X_n\in m\mc {F}_n$, so Proposition 2.2.6 implies that $Y_n\in m\mc {F}_n$. Since $|Cn|\le c$ for some $c$ and all $n$, we have

\[ \E |Y_n| \leq \sum _{k=1}^n \E |C_{k-1} (M_k-M_{k-1})| \leq c \sum _{k=1}^n \E |M_k|+\E |M_{k-1}| < \infty . \]

So $Y_n\in L^1$. Since $C_{n-1}$ is $\F _{n-1}$-measurable, by linearity of conditional expectation, the taking out what is known rule and the martingale property of $M$, we have

\begin{align*} \E [Y_n\|\F _{n-1}] &= \E [Y_{n-1} + C_{n-1} (M_n-M_{n-1}) \| \F _{n-1}]\\ &= Y_{n-1} + C_{n-1} \E [M_n-M_{n-1} \| \F _{n-1}]\\ &= Y_{n-1} + C_{n-1} (\E [M_n|\mc {F}_{n-1}]-M_{n-1})\\ &= Y_{n-1}. \end{align*} Hence $Y$ is a martingale.

The argument is easily adapted to prove the second statement, e.g. for a supermartingale $M$, $\E [M_n-M_{n-1} \| \F _{n-1}] \le 0$. Note that in these cases it is important that $C$ is non-negative. ∎

Stochastic Processes and Financial Mathematics (part one)

Chapter 7 Stochastic processes and martingale theory

7.1 The martingale transform

Stochastic Processes and Financial Mathematics
(part one)