This image is from http://blog.motheyes.com/2010/03/ministry-of-stochastic-walks/ .
Here is a short guide to the random walk found in probability. Note that this random walk will be a symmetrical random walk with equal probabilities of \(\dfrac{1}{2}\) for each of the two outcomes.
Let us first consider a unbiased/even coin. We have a probability of 0.5 for heads and 0.5 for tails.
Denote the outcome of tosses as \(\omega = \omega_{1} \omega_{2} \omega_{3}...\omega_{n}\) where \(\omega\)) is omega for \(n\) coin tosses. \(\omega\) is an infinite sequence of outcomes \(\omega_{1}, \omega_{2}\) up to \(\omega_{n}\).
Define the random variable \(X_j\) for the \(j^{th}\) coin toss for \(j\) from 1 to n where:
\[ X_j = \begin{cases} +1, & \text{if }\omega_j = \text{Heads } \\ -1, & \text{if }\omega_j =\text{Tails}\end{cases}\]
Each coin toss is either +1 for a heads or a -1 for a tails with a probability of 0.5 each.
Note that this \(X_j\) random variable does not have to relate to coin tosses. Once can define \(X_j\) to be dependent on up/down movements, even/odd numbers, etc.
\(E[X_j] = 0\) since \(E[X_j] = 1 \times \dfrac{1}{2} - 1 \times \dfrac{1}{2}= 0\).
\(Var(X_j) = 1\) since
\[Var(X_j) = E[X_j^2] - (E[X_j])^2 = E[X_j^2] - 0^{2} = 1^2 \times \dfrac{1}{2} + (-1)^2 \times \dfrac{1}{2} = 1\]
Now we have the random variable \(X_j\) for j from 1 to n. But, what if we want a running total of these +1 and -1 outcomes for 1 to n?
Let us define \(M_0 = 0\) and this “running total” as \(M_k\) where:
\[M_k = \displaystyle\sum_{j=1}^{k} X_j\] for k = 1, 2, …
This stochastic (random) process \(M_k\) is a symmetric random walk.
Mean is zero:
\[E[M_k] = E [\displaystyle\sum_{j=1}^{k} X_j] = \displaystyle\sum_{j=1}^{k} E[X_j] = \displaystyle\sum_{j=1}^{k} (1 * 0.5 + (- 1) * 0.5) = 0\]
Variance is just \(k\) for \(M_k\).
\[Var(M_k) = Var( \displaystyle\sum_{j=1}^{k} X_j ) = \displaystyle\sum_{j=1}^{k} Var(X_j) = \displaystyle\sum_{j=1}^{k} 1 = k\]
Note that the independence of coin tosses was assumed such that the covariance in the double sum is zero.
The symmetric random walk is a martingale. That is \(\displaystyle E[M_l | F_k ] = M_k\). (The conditional expectation given the filtration at time \(k < l\) is just the symmetric random walk at time k.) We don’t expect the symmetric random walk to change from time \(k\) to \(l\).
The quadratic variation of the symmetric random walk is just time k. This is because:
\[[M, M]_k = \sum_{j=1}^{k} (M_j - M_j-1)^2 = \sum_{j=1}^{k} (X_j)^2 = \sum_{j=1}^{k}(\pm 1)^2 = k\] (Add 1 k times).
The increments of the symmetric random walk are independent. For example, \((M_1 - 0 = M_1 - M_0) , M_2 - M_1 , M_3 - M_2, ... ,\) and \(M_k - M_{k-1}\) are independent increments. This means that increments over non-overlapping intervals are independent since the intervals depend on different coin tosses.
The symmetric random walk is a “running total” on the random variable \(X_j\). This random variable \(X_j\) is either +1 or -1 with equal probabilities from one of two outcomes (heads / tails for example). The symmetric random walk has a lot useful properties and is useful for understanding Brownian Motion.