Introduction to Martingales

This is a rather important topics for anyone interested in doing Finance.
Lets look at their definition first.
A Martingale is a random process \{X_n: 0 \le n \le \infty \} with respect to the information filtration F_n and the probability distribution P, if
\mathbb{E}^P [|X_n|] < \infty for all n \ge 0
\mathbb{E}^P[X_{n+m}|F_n] = X_n for all n, m \ge 0

Martingales are used widely and one example is to model fair games, thus it has a rich history in modelling of gambling problems. If you google Martingale, you will get an image related to a Horse, because it started with Horse-betting.

Martingales. Source: NYU

Martingales. Source: NYU

We define a submartigale by replacing the above condition 2 with
\mathbb{E}^P [X_{n+M}| F_n] \ge X_n for all n, m \ge 0
and a supermartingale with
\mathbb{E}^P [X_{n+M}| F_n] \le X_n for all n, m \ge 0 .
Take note that a martingale is both a submartingale and a supermartingale. Submartingale in layman terms, refers to the player expecting more as time progresses, and vice versa for supermartingale.

Let us try to construct a Martingale from a Random Walk now.
Let S_n := \sum_{i=1}^n X_i be a random walk where the X_i’s are IID with mean \mu.
Let M_n := S_n -n \mu. Then M_n is a martingale because:
\mathbb{E}_n [M_{n+m}]
= \mathbb{E}_n [\sum_{i=1}^{n+m} X_i - (n+m) \mu]
= \mathbb{E}_n [\sum_{i=1}^{n+m} X_i] - (n+m) \mu since expectation distributes linearly
= \sum_{i=1}^n X_i + \mathbb{E}_n [\sum_{i=n+1}^{n+m} X_i] - (n+m) \mu
= \sum_{i=1}^n X_i + m \mu - (n+m) \mu = M_n

So how will a martingale betting strategy be like?
Here, we let X_1, X_2, \ldots be IID random variables with P(X_i = 1) = P(X_i = -1) = \frac{1}{2}. We can imagine X_i to represent the result of a coin-flipping game where,
– player win 1 if the coin comes up heads, that is,latex X_i = 1- player lose1 if the coin comes up tails, that is, X_i = -1

Consider further now a doubling strategy where we keep doubling the bet until we eventually win. Once we win, we stop and our initial bet is 1.  The first thing we note is that the size of bet on thelatex n^{th}play islatex 2^{n-1}assuming we are still playing at time n. And we can letlatex W_ndenote total winnings after n coin tosses, assuminglatex W_0 = 0. Thenlatex W_nis a martingale!  To see this, let us prove thatlatex W_n \in \{ 1, -2^n +1 \}for all n.  Suppose we win for first time onlatex n^{th}bet. Thenlatex W_n = -(1 + 2 + \ldots + 2^{n-2}) + 2^{n-1}latex = -(2^{n-1} – 1) + 2^{n-1} = 1If we have not yet won after n bets then,latex W_n = -(1 + 2 + \ldots + 2^{n-1}) = -2^n +1Finally, to showlatex W_nis a martingale, we just need to showlatex \mathbb{E}[W_{n+1} | W_n] = W_nwhich can be easily prove using iterated expectations. For case 1,latex W_n = 1, thenlatex P(W_{n+1} = 1 | W_n = 1) = 1solatex \mathbb{E}[W_{n+1} | W_n = 1] = 1 = W_nFor case 2,latex W_n = -2^n +1, meaning we betlatex 2^nonlatex (n+1)^{th}toss solatex W_{n+1} \in \{ 1 , -2^{n+1} + 1 \}. Sincelatex P(W_{n+1} = 1) | W_n = -2^n +1) = \frac{1}{2}, andlatex P(W_{n+1} = -2^{n+1}+1 | W_n = -2^n +1) = \frac{1}{2}, thenlatex \mathbb{E}[W_{n+1} | W_n = -2^n +1] = \frac{1}{2} \times 1 + \frac{1}{2} \times (-2^{n+1} +1) = -2^n + 1 = W_nThus, we showed thatlatex \mathbb{E} [W_{n+1} | W_n] = W_nTo bring what we learnt a further step, lets look at Polya's Urn briefly. Consider an urn which contains red balls and green balls. Initially there is just one green ball and one red ball in the urn. At each time step, a ball is chosen randomly from the urn: If ball is red, the its returned to the urn with an additional red ball. If ball is green, then its returned to the urn with an additional green ball. Letlatex X_ndenote the number of red balls in the urn after n draws. Thenlatex P(X_{n+1} = k+1 | X_n = k) = \frac{k}{n + 2}latex P(X_{n+1} = k | X_n = k) = \frac{n + 2 – k}{n + 2}We can show thatlatex M_n := \frac{X_n}{n+2}$ is a martingale.

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