Discrete Time Martingale

Discrete Time Filtration

A (discrete time) filtration on the probability space $(\Omega ,\Sigma ,\mathbb{P})$ is a sequence of sigma-algebras on $\Omega$, $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$, with the property that ${\mathcal{F}}_0\subset {\mathcal{F}}_1\subset \cdots \subset \Sigma .$

Discrete Time Stochastic Process

Let $(\Omega ,\Sigma ,\mathbb{P})$ be a probability space with filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$. A sequence of random variables $(X_n{)}_{n\in \mathbb{N}}$ is called a discrete time stochastic process. $(X_n{)}_{n\in \mathbb{N}}$ is said to be adapted to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$ if for each $n\in \mathbb{N}$, $X_n$ is ${\mathcal{F}}_n$-measurable.

Martingale

Let $(\Omega ,\Sigma ,\mathbb{P})$ be a probability space with filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$. Let $(M_n{)}_{n\in \mathbb{N}}$ be an adapted stochastic process with $M_n\in L^2(\Omega ,{\mathcal{F}}_n,\mathbb{P})$ for each $n\in \mathbb{N}$. $(M_n{)}_{n\in \mathbb{N}}$ is called a martingale with respect to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$ if $M_n=E(M_{n+1}\mid {\mathcal{F}}_n)$ for all $n\in \mathbb{N}$. $(M_n{)}_{n\in \mathbb{N}}$ is called a submartingale with respect to $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$ if $M_n\le E(M_{n+1}\mid {\mathcal{F}}_n)$ for all $n\in \mathbb{N}$.

Example 2.5

Let $(X_j{)}_{j\in {\mathbb{N}}^{\ast }}$ be a family of independent random variables on $(\Omega ,\Sigma ,\mathbb{P})$ with$\mathbb{E}(X_j)=0$ for all $j\in {\mathbb{N}}^{\ast }:=\mathbb{N}\setminus \{0\}.$ Equip the probability space with the filtration ${\mathcal{F}}_n:=\sigma (X_1,\dots ,X_n)$ for $n\in {\mathbb{N}}^{\ast }$ and ${\mathcal{F}}_0:=\{\varnothing ,\Omega \}.$ Define, for each $n\in {\mathbb{N}}^{\ast }$, the random variable $S_n:={\sum }_{j=1}^n{X}_j.$ Also set $S_0:=0.$ The family $(S_n{)}_{n\in \mathbb{N}}$ is a martingale with respect to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$. To see this, first note that $\mathbb{E}(X_{n+1}\mid {\mathcal{F}}_n)=\mathbb{E}(X_{n+1}).$ On applying this and by taking out what is known,

$$\begin{array}{rl}\mathbb{E}(S_{n+1}\mid {\mathcal{F}}_n)& =\mathbb{E}(S_n\mid {\mathcal{F}}_n)+\mathbb{E}(X_{n+1}\mid {\mathcal{F}}_n)\\ & =S_n+\mathbb{E}(X_{n+1}\mid {\mathcal{F}}_n)\\ & =S_n+\mathbb{E}(X_{n+1})\\ & =S_n.\end{array}$$

This models your wealth when playing a fair game paying $X_j$ several times. $\mathbb{E}(S_{n+1}-S_n\mid {\mathcal{F}}_n)=0$ means no expected gain based on known information.

Example 2.6

Let $(X_j{)}_{j\in {\mathbb{N}}^{\ast }}$ be a family of independent random variables with $\mathbb{E}(X_j)=1$ for each $j\in {\mathbb{N}}^{\ast }$. Equip the probability space with the filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$ for $n\in {\mathbb{N}}^{\ast }$ and ${\mathcal{F}}_0=\{\varnothing ,\Omega \}.$ Define the random variables $P_0=1,P_n={\prod }_{j=1}^n{X}_j$ for every $n\in {\mathbb{N}}^{\ast }$. As $P_n$ is ${\mathcal{F}}_n$-measurable, Proposition 1.7 can be applied in order to interchange $P_n$ and the expectation. That is

$$\begin{array}{rl}\mathbb{E}(P_{n+1}\mid {\mathcal{F}}_n)& =\mathbb{E}(X_{n+1}{P}_n\mid {\mathcal{F}}_n)\\ & =P_n\cdot \mathbb{E}(X_{n+1}\mid {\mathcal{F}}_n)\\ & =P_n\cdot \mathbb{E}(X_{n+1})\\ & =P_n.\end{array}$$

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Example 2.7

Let $\sigma >0$ and $(X_j{)}_{j\in \mathbb{N}}$ be a family of independent random variables with $\mathbb{E}(X_j)=0$ and $\mathbb{E}(X_j^2)={\sigma }^2$ for each $j\in \mathbb{N}$. Equip the probability space with the filtration ${\mathcal{F}}_n:=\sigma (X_0,\dots ,X_n).$ Define the random variables $M_n:=S_n^2-n{\sigma }^2$ for each $n\in \mathbb{N}$, where $S_n$ is as defined in Example 2.5. Then

$$\begin{array}{rl}\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)& =\mathbb{E}\left((S_n+X_{n+1}{)}^2-n{\sigma }^2-{\sigma }^2\mid {\mathcal{F}}_n\right)\\ & =\mathbb{E}\left(S_n^2-n{\sigma }^2+2S_n{X}_{n+1}+X_{n+1}^2-{\sigma }^2\mid {\mathcal{F}}_n\right)\\ & =M_n-{\sigma }^2+2\mathbb{E}(S_n{X}_{n+1}\mid {\mathcal{F}}_n)+\mathbb{E}(X_{n+1}^2\mid {\mathcal{F}}_n)\\ & =M_n-{\sigma }^2+2S_n\mathbb{E}(X_{n+1}\mid {\mathcal{F}}_n)+\mathbb{E}(X_{n+1}^2)\\ & =M_n-{\sigma }^2+2S_n\mathbb{E}(X_{n+1})+{\sigma }^2\\ & =M_n,\end{array}$$

for each $n\in \mathbb{N}$. This demonstrates that $M_n$ is a martingale with respect to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$.

Predictable Random Variables

A sequence of random variables $(A_n{)}_{n\in \mathbb{N}}$ is called predictable with respect to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$ if $A_n$ is ${\mathcal{F}}_{n-1}$-measurable for all $n\in {\mathbb{N}}^{\ast }$.

Martingale Transform

Let $(M_n{)}_{n\in \mathbb{N}}$ be a martingale and $(A_n{)}_{n\in \mathbb{N}}$ a predictable process with respect to the filtration $({\mathcal{F}}_n{)}_{n\in \mathbb{N}}$. Define ${\tilde{M}}_n:={\sum }_{j=1}^n{A}_j(M_j-M_{j-1}),$ for each $j\in \mathbb{N}$. $({\tilde{M}}_n{)}_{n\in \mathbb{N}}$ is called the martingale transform of $(M_n{)}_{n\in \mathbb{N}}$ by $(A_n{)}_{n\in \mathbb{N}}$.

Theorem

Let $(M_n{)}_{n\in \mathbb{N}}$ be a martingale and $(A_n{)}_{n\in \mathbb{N}}$ a predictable process. If $\Vert A_n{\Vert }_{\infty }<\infty$ for each $n\in \mathbb{N}$, then $({\tilde{M}}_n{)}_{n\in \mathbb{N}}$ is a martingale.

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Proof

Let $n\in {\mathbb{N}}^{\ast }$. As $A_n$, $M_n$ and $M_{n-1}$ are all ${\mathcal{F}}_n$-measurable, ${\tilde{M}}_n$ must also be ${\mathcal{F}}_n$-measurable. Let us check that ${\tilde{M}}_n\in L^2(\Omega ,{\mathcal{F}}_n,\mathbb{P}).$ On applying Cauchy-Schwarz,

$$\begin{array}{rl}\mathbb{E}\left({\tilde{M}}_n^2\right)& \le \mathbb{E}\left[\left(\sum _{i=1}^n|A_i{|}^2\right)\left(\sum _{j=1}^n|M_j-M_{j-1}{|}^2\right)\right]\\ & \le n{\left(\underset{1\le i\le n}{\sup }\Vert A_i{\Vert }_{\infty }\right)}^2\sum _{j=1}^n\left(\mathbb{E}(M_j^2)+\mathbb{E}(M_{j-1}^2)+2\mathbb{E}(|M_j{M}_{j-1}|)\right)\\ & \le n{\left(\underset{1\le i\le n}{\sup }\Vert A_i{\Vert }_{\infty }\right)}^2\sum _{j=1}^n\left(\mathbb{E}(M_j^2)+\mathbb{E}(M_{j-1}^2)+2\mathbb{E}(M_j^2{)}^{1/2}\mathbb{E}(M_{j-1}^2{)}^{1/2}\right)\\ & <\infty .\end{array}$$

To show the martingale property for ${\tilde{M}}_n$, observe that

$$\begin{array}{rl}\mathbb{E}\left({\tilde{M}}_n-{\tilde{M}}_{n-1}\mid {\mathcal{F}}_{n-1}\right)& =\mathbb{E}\left(A_n(M_n-M_{n-1})\mid {\mathcal{F}}_{n-1}\right)\\ & =A_n\mathbb{E}\left(M_n-M_{n-1}\mid {\mathcal{F}}_{n-1}\right)\\ & =0.\end{array}$$

Hence,

\mathbb{E} \left( \widetilde{M}_n \mid \mathcal{F}_{n-1} \right) = \widetilde{M}_{n-1}, $$so $(\widetilde{M}_n)_{n\in\mathbb{N}}$ is a martingale. >[!remark] > >>Consider once again our gambler playing cards. Let $M_n$ be the wealth of the gambler at time $n$ given that they have only bet a unit stake at each round. Suppose that the process $(M_n)_{n \in \mathbb{N}}$ is a martingale. This means that the gambler's total wealth is not expected to change. Suppose instead that the gambler decides on the strategy of betting $A_n$ units for the $n$th round of cards. Then their total wealth will be given by the martingale transform of $(M_n)_{n \in \mathbb{N}}$ by $(A_n)_{n \in \mathbb{N}}$. Will the gambler's strategy improve their fortune? The above theorem states that no matter what strategy is used by the gambler, the total wealth will remain a martingale. This is a rigorous formulation of what is intuitively obvious; you can't beat a fair game.