# Martingale Convergence

Last quarter, I did a DRP  on martingales, and at the start of this quarter, gave a short talk on what I learned. Here are the notes from that talk.

### 1. Conditional Expectation

Definition 1. Suppose ${(\Omega, \mathcal{F}_o, \mathbb{P})}$ is a probability space, ${\mathcal{F}\subset \mathcal{F}_o}$ is a sub ${\sigma}$-algebra, and ${X}$ is a random variable.

1. The conditional expectation of ${X}$ given ${\mathcal{F}}$, ${E(X\mid \mathcal{F})}$, is any ${\mathcal{F}}$-measurable random variable ${Y}$ such that

$\displaystyle \int_A X\, d\mathbb{P}=\int_A Y\, d\mathbb{P}\text{ for all } A\in \mathcal{F}.$

2. If ${Y}$ is another random variable, then ${E(X\mid Y)}$ is defined as ${E(X\mid \mathcal{F})}$ where ${\mathcal{F}}$ is the ${\sigma}$-algebra generated by ${Y}$.

Fact. The conditional expectation exists, is unique, and is integrable.

Intuitively, we can think of ${E(X\mid \mathcal{F})}$ as the best guess of the value of ${X}$ given the information available in ${\mathcal{F}}$.

Example 1.

1. If ${X}$ is ${\mathcal{F}}$-measurable, then ${E(X\mid \mathcal{F})=X}$.
2. If ${X}$ and ${Y}$ are independent, then ${E(X\mid Y)=E(X)}$.
3. Let ${X_1,\cdots}$ be random variables with mean ${\mu}$, and let ${S_n=X_1+\cdots+X_n}$ be the partial sums. Then, if ${m>n}$,

$\displaystyle E(S_m\mid S_n)=E(X_m+\cdots+X_{n+1}+S_n\mid S_n)=E(X_m\mid S_n)+\cdots+E(X_{n+1}\mid S_n)+E(S_n\mid S_n)=\mu(m-n)+S_n.$

The idea is that you know your position at ${S_n}$, and you take ${m-n}$ steps whose sizes are, on average, ${\mu}$, so your best guess for your position is ${S_n+\mu(m-n)}$.

### 2. Martingales

A martingale is a model of a fair game.

Definition 2. Consider a filtration (increasing sequence of ${\sigma}$-algebras) ${\mathcal{F}_n}$ and a sequence of random variables ${X_n}$, each measurable with respect to ${\mathcal{F}_n}$ and integrable. Then, if ${E(X_{n+1}\mid \mathcal{F}_n)=X_n}$ for all ${n}$, we say ${S_n}$ is a martingale.

Example 2. Let ${X_1,\cdots}$ be random variables that take only the values ${-1}$ and ${1}$ with probability ${1/2}$ each. Then ${S_n=X_1+\cdots+ X_n}$ is a martingale, because

$\displaystyle E(S_{n+1}\mid S_n)=E(X_{n+1}+S_n\mid S_n)=E(X_{n+1}\mid S_n)+E(S_n\mid S_n)=E(X_{n+1})+S_n=S_n.$

Theorem 3. If ${X_n}$ is a martingale with ${\sup E(X_n^+)<\infty}$, then ${X_n}$ converges a.s. to a limit ${X}$ with ${E(|X|)<\infty}$.

I’ll only sketch this proof, because even though the idea is nice, the details are a little annoying. The idea is to set up a way of betting on a martingale, show that you can’t make money off such a betting system, and then use this to draw conclusions about the martingales behavior.

Definition 4. Let ${\mathcal{F}_n}$ be a filtration. A sequence ${H_m}$ of random variables is said to be predictable if ${H_m}$ is ${\mathcal{F}_{m-1}}$ measurable for all ${m}$.

That is, the value of ${H_m}$ can always be predicted with certainty at time ${m}$. Then, if we “bet” an amount ${H_m}$ on the martingale ${X_m}$, our total winnings will be

$\displaystyle (H\cdot X)_n=\sum_{m=1}^n H_m(X_m-X_{m-1}).$

If ${X_n}$ is a supermartingale and ${H_n}$ is bounded, then ${(H\cdot X)_n}$ is a supermartingale as well.

Now, fix an interval ${(a,b)}$ and choose ${H_n}$ in the following way: if ${X_n, then bet 1, and continue to do so until ${X_n>b}$. Then bet zero until you fall back below ${a}$, and repeat. Every time you go from below ${a}$ to above ${b}$, you will make a profit of at least ${(b-a)}$. Let ${U_n}$ be the number of these upcrossings that occur. Then, you can use the previous fact, together with the bound on ${\sup E(X_n^+)}$, to show that the number of these upcrossings is finite with probability 1. Since this is true for arbitrary ${(a,b)}$, the ${X_n}$ must converge.

Finally, we give a brief application of the martingale convergence theorem.

Proposition 5. Let ${X_n}$ be a sequence of random variables such that ${P(X_n=1)=P(X_n=-1)=\frac{1}{2}.}$ Then ${{\sum_{j=1}^{\infty} \frac{X_j}{j}}}$ converges with probability 1.

Proof: Let ${{S_n=\sum_{j=1}^n\frac{X_j}{j}}.}$ It’s easy to check that ${S_n}$ is a martingale, and that ${E(S_n)}$ is bounded (in fact, it’s always 0), so the ${S_n}$ converge almost surely. Thus the random harmonic series converges almost surely. $\Box$

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