is a stochastic process, integrable and adapted to filtration . Then is a martingale if almost surely whenever .
It is natural to think about a martingale as defined in the context of a process evolving in time. Then this definition is very reasonable:
- Integrable: the entire construction is about taking expectations. So these need to exist.
- Adapted: can’t be affected by what happens after time n. It must be defined by what has happened up to time n.
- Expectation condition: if you look at the process at a given time m, the best estimate for what will be in the future is in fact what it is now. As you might expect, it is sufficient that almost surely for each n.
There are many situations where the expected change in a variable over a time period is zero, whatever the value of the variable is at the start. For example, gambling. For illustration, assume we are speculating on the outcomes of tossing a coin repeatedly. You might have a complicated strategy, for example ‘double your stake when you lose’ (the so-called martingale strategy), or anything else. But ultimately, you can’t see into the future. So before every coin toss, you have to decide your stake, based on what’s happened up until now, and you will win or lose with equal probability, so your expected gain is 0, regardless of how you make your stake choice. Thus under any strategy determined without looking into the future (called previsible), the process recording your winnings is a martingale.
The double-if-you-lose-then-stop strategy has the following property. Suppose your initial stake is £1. Then with probability 1/2, you win £1 after one time toss. With probability 1/4, you lose the first toss, then stake £2 on the second and win, making a profit of “£1. And so forth. So the probability that you win £1 is: 1/2+1/4+1/8+…=1. So you have a guaranteed profit of £1? But the expected profit after time n is 0 for all times. How can this work?
There are obvious practical objections. The player only has finite money £N in his bank account, so the game must terminate after log N tosses. He also has at most about 100 years to play the game. And, in more theoretical terms, there is the event that he loses all the coin tosses. This has probability 0, but the associated profit is . So the expected overall profit looks as if it might yet be 0, or at least, not necessarily defined. So we need some theory to deal with questions like these.
Optional Stopping Theorem
So in the above example, we don’t care hugely about the entire process, just about the final profit. Effectively, we have a random variable T, which records when we first make a profit. Appropriately, this is called a stopping time, because this first profit time is determined by the past and present, rather than the future, so we could stop the process at the stopping time. Consider expectations of for these stopping times is the most important initial question in the theory.
OST: in the following circumstances:
- When T is bounded by a positive constant.
- When is bounded by an integrable RV, (including Y a constant), and T is finite almost surely.
- When the increments are bounded a.s., and .
It is clear why none of these conditions apply to the above example.
- T is not bounded: with positive probability it will be greater than any bound.
- T is finite almost surely, but can’t be bounded, because for large n with positive probability it is .
- The increments are not bounded, and =1/2 + 1/4 x 2 + 1/8 x 4 + … = 1/2 + 1/2 + 1/2 + … = . But this does give the result we want in almost every situation without these problems.
- Can write . If this sum is finite, can apply linearity of expectation.
- This has been set up with conditions just right to use Dominated Convergence.
- By writing as a sum of indicator functions in T, can bound and then use Dominated Convergence.