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
$latex \mathbb{E}^P [|X_n|] < \infty$ for all $latex n \ge 0$ $latex \mathbb{E}^P[X_{n+m}|F_n] = X_n$ for all $latex n, m \ge 0$ 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. [caption id="attachment_2845" align="alignnone" width="300"]Martingales. Source: NYU Martingales. Source: NYU[/caption]

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, X_i = 1
– player lose $1 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 the n^{th} play is 2^{n-1} assuming we are still playing at time n. And we can let W_n denote total winnings after n coin tosses, assuming W_0 = 0. Then W_n is a martingale!

To see this, let us prove that W_n \in \{ 1, -2^n +1 \} for all n.
Suppose we win for first time on n^{th} bet. Then
W_n = -(1 + 2 + \ldots + 2^{n-2}) + 2^{n-1}
= -(2^{n-1} - 1) + 2^{n-1} = 1
If we have not yet won after n bets then,
W_n = -(1 + 2 + \ldots + 2^{n-1}) = -2^n +1
Finally, to show W_n is a martingale, we just need to show \mathbb{E}[W_{n+1} | W_n] = W_n which can be easily prove using iterated expectations.
For case 1, W_n = 1, then P(W_{n+1} = 1 | W_n = 1) = 1 so \mathbb{E}[W_{n+1} | W_n = 1] = 1 = W_n
For case 2, W_n = -2^n +1, meaning we bet 2^n on (n+1)^{th} toss so W_{n+1} \in \{ 1 , -2^{n+1} + 1 \}. Since
P(W_{n+1} = 1) | W_n = -2^n +1) = \frac{1}{2}, and
P(W_{n+1} = -2^{n+1}+1 | W_n = -2^n +1) = \frac{1}{2},
\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_n
Thus, we showed that \mathbb{E} [W_{n+1} | W_n] = W_n

To 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.
Let X_n denote the number of red balls in the urn after n draws. Then
P(X_{n+1} = k+1 | X_n = k) = \frac{k}{n + 2}
P(X_{n+1} = k | X_n = k) = \frac{n + 2 - k}{n + 2}
We can show that M_n := \frac{X_n}{n+2} is a martingale.

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