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[/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
$P(W_{n+1} = -2^{n+1}+1 | W_n = -2^n +1) = \frac{1}{2}$,
then
$\mathbb{E}[W_{n+1} | W_n = -2^n +1] = \frac{1}{2} \times 1 + \frac{1}{2} \times (-2^{n+1} +1) = -2^n + 1 = W_n$
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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