Martingale

Last updated Nov 1, 2022

Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a Probability Space. Let $X: \Omega \to \mathbb{R}^{T}$ be an Adapted Process wrt Filtration $\mathcal{F}{*} := (\mathcal{B}{t})_{t \in T}$ on $\Omega$. $X$ is a Martingale if it satisfies the following properties:

1. $X_{t} \in L^{1}(\Omega)$ $\forall t \in T$,
2. $\mathbb{E}(X_{t} | \mathcal{B}{r}) = X{r}$ Almost Surely for any $r \leq t$.

1. Expectations of a Martingale are Constant

1. Defining property (2) is already true when $r = t$, so really what matters is when $r < t$.

A Martingale is a Supermartingale and a Submartingale

1. This definition is equivalent to the first because property (2) of Supermartingales and Submartingales together implies property (2) of Martingales.

Let $(\Omega, \mathcal{A}, \mathbb{P})$ be a Probability Space. Let $X: \Omega \to \mathbb{R}^{\mathbb{N}}$ be an Adapted Process wrt Filtration $\mathcal{F}{*} := (\mathcal{B}{n})_{n \in \mathbb{N}}$ on $\Omega$. $X$ is a Martingale if it satisfies the following properties:

1. $X_{n} \in L^{1}(\Omega)$ $\forall n \in \mathbb{N}$,
2. $\mathbb{E}(X_{n+1} | \mathcal{B}{n}) = X{n}$ Almost Surely for any $n \in \mathbb{N}$.

1. It is not hard to see that definition (3) is equivalent definition (1) in the case where $T = \mathbb{N}$. First note that $(3)$ is just a special case of $(1)$. We can establish that $(3) \Rightarrow (1)$ by Induction. First observe that for $n = 1$, the only $m \in \mathbb{N}$ s.t. $m \leq n$ is $m = 1$. Therefore, $\forall m \leq n$, we have that $\mathbb{E}(X_{n} | \mathcal{B}{m}) = \mathbb{E}(X{1} | \mathcal{B}{1}) = X{1}$ since Conditioning on known information is Idempotent. For the inductive step, suppose this claim holds for $n \in \mathbb{N}$. Consider $m \leq n+1$. Then we have that $$\mathbb{E}(X_{n+1} | \mathcal{B}{m}) = \mathbb{E}(\mathbb{E}(X{n+1} | \mathcal{B}{n}) | \mathcal{B}{m}) = \mathbb{E}(X_{n} | \mathcal{B}{m}) = X{m}$$ by induction. Note that the first equality holds because of Smoothing. $\blacksquare$