Random Walks

Random Walk Probability of Returning to Origin after n steps

Random Walk Probability of Returning to Origin

Assume that the walk starts at $x=0$ with steps to the right or left occurring with probabilities $p$ and $q=1-p$. We can write the position $X_n$ after $n$ steps as
\[X_n=R_n-L_n \tag{1}\]
where $R_n$ is the number of right or positive steps (+1) and $L_n$ is the number of left or negative steps ($-1$).

Therefore the Total steps can be calculated as:  \[n=R_n+L_n \tag{2}\]
Hence
\begin{align*}
L_n&=n-R_n\\
\Rightarrow X_n&=R_n-n+R_n\\
R_n&=\frac{1}{2}(n+X_n) \tag{3}
\end{align*}
The equation (3) will be an integer only when $n$ and $X_n$ are both even or both odd (eg. To go from $x=0$ to $x=7$ we must take an odd number of steps).

Now, let $v_{n,x}$ be the probability that the walk is at state $x$ after $n$ steps assuming that $x$ is a positive integer. Then
\begin{align*}
v_{n,x}&=P(X_n=x)\\
&=P(R_n=\frac{1}{2}(n+x))
\end{align*}
$R_n$ is a binomial random variable with index $n$ having probability $p$, since the walker either moves to the right or not at every step, and the steps are independent, then
\begin{align*}
v_{n,x}&=\binom{n}{\frac{1}{2}(n+x)}p^{\frac{1}{2}(n+x)}q^{n-\frac{1}{2}(n+x)}\\
&=\binom{n}{\frac{1}{2}(n+x)}p^{\frac{1}{2}(n+x)}q^{\frac{1}{2}(n-x)} \tag{4}
\end{align*}
where $(n,x)$ are both even or both odd and $-n \leq x \leq n$. Note that a similar argument can be constructed if $x$ is a negative integer.

Example: For a total number of steps is 2, the net displacement must be one of the three possibilities: (1) two steps to the left, (2) back to the start, (3) or two steps to the right. These correspond to values of $x = -2, 0,+2$. It is impossible to get more than two units away from the origin if you take only two steps and it is equally impossible to end up exactly one unit from the origin if you take two steps.

For a symmetric case ($p=\tfrac{1}{2}$), starting from the origin, there are $2^n$ different paths of length $n$ since there is a choice of right or left move at each step. Since the number of steps in the right direction must be $\tfrac{1}{2}(n+x)$ and the total number of paths must be the number of ways in which $\frac{1}{2}(n+x)$ can be chosen from $n$: that is
\[N_{n,x}=\binom{n}{\tfrac{1}{2}(n+x)}\]
provided that $\tfrac{1}{2}(n+x)$ is an integer.

By counting rule, the probability that the walk ends at $x$ after $n$ steps is given by the ratio of this number and the total number of paths (since all paths are equally likely). Hence
\[v_{n,x}=\frac{N_{n,x}}{2^n}=\binom{n}{\tfrac{1}{2}(n+x)}\frac{1}{2^n}\]
The probability $v_{n,x}$ is the probability that the walk ends at state $x$ after $n$ steps: the walk could have overshot x before returning there.

Related Probability/ First Passage through x

A related probability is the probability that the first visit to position x occurs at the $n$th step. The following is a descriptive derivation of the associated probability-generating function of the symmetric random walk in which the walk starts at the origin, and we consider the probability that it returns to the origin.

From equation (4), the probability that a walk is at the origin at step $n$ is
\begin{align*}
v_{n,x}&=\binom{n}{\frac{1}{2}(n+x)}p^{\frac{1}{2}(n+x)}q^{n-\frac{1}{2}(n+x)}\\
&=\binom{n}{\tfrac{1}{2}(n+0)} \left(\frac{1}{2}\right)^{\tfrac{1}{2}n} \left(\frac{1}{2}\right)^{\tfrac{1}{2}n}\\
&=\binom{n}{\tfrac{1}{2}n}\frac{1}{2^n}=p_n \,\,\,\text{(say)}, \quad (n=2,4,6,\cdots) \tag{5}
\end{align*}
Here $p_n$ is the probability that after $n$ steps the position of the walker is at the origin. We also assume that $p_n=0$ if $n$ is odd. From equation (5) we can construct a generating function.
\begin{align*}
H(s)&=\sum_{n=0}^\infty p_n s^n\\
&=\sum_{n=0}^\infty p_{2n}s^{2n}=\sum_{n=0}^\infty \frac{1}{2^{2n}}\binom{2n}{n}s^{2n} \tag{6}
\end{align*}
Note that $p_0=1$, and H(s) is not a probability generating function since $H(1)\neq1$.

The binomial coefficient can be re-arranged as follows:
\begin{align*}
\binom{2n}{n}&=\frac{(2n)!}{n!n!}=\frac{2n(2n-1)(2n-2)\cdots3.2.1}{n!n!}\\
&=\frac{2^nn!(2n-1)(2n-3)\cdots3.2.1}{n!n!}\\
&=\frac{2^{2n}}{n!}\frac{1}{2}\frac{3}{2}\cdots(n-\tfrac{1}{2})\\
&=(-1)^n \binom{-\tfrac{1}{2}}{n}2^{2n} \tag{7}
\end{align*}
Using equation (6) in (7)
\[H(s)=\sum_{n=0}^\infty \frac{1}{2^{2n}}(-1)^n \binom{-\tfrac{1}{2}}{n}s^{2n}2^{2n}=(1-s^2)^{-\tfrac{1}{2}} \tag{8}\]
by binomial theorem, provided $|s|<1$. Note that this expansion guarantees that $p_n=0$ if $n$ is odd.

Note that the equation (8) does not sum to one. This is called defective distribution which still gives the probability that the walk is at the origin at step n.

We can estimate the behavior of $p_n$ for large n by using Stirling’s Formula (asymptotic estimate for $n!$ for large $n$), $n!\approx\sqrt{2\pi} n^{n+\tfrac{1}{2}}e^{-n}$

From equation (5)
\begin{align*}
p_{2n}&=\frac{1}{2^{2n}}\binom{2n}{n}=\frac{1}{2^{2n}}\frac{(2n)!}{n!n!}\\
&\approx\frac{1}{2^{2n}}\frac{\sqrt{2\pi}(2n)^{2n+\tfrac{1}{2}}e^{-2n}}{[\sqrt{2\pi}(n^{n+\tfrac{1}{2}}e^{-n})]^2}\\
&=\frac{1}{\sqrt{\pi n}}; \qquad \text{for large $n$}
\end{align*}
Hence $np_n\rightarrow \infty$ confirming that the series $\sum\limits_{n=0}^\infty p_n$ must diverge.

Random Walk Probability of Returning to Origin after $n$ Steps Some EXAMPLES

Example: Consider a random walk starting from $x_0=0$ and find the probability that after 5 steps the position is 3. i.e. $X_5=3$, $p=0.6$.

Solution: Here the number of steps is $n=5$ and the position is $x=3$. Therefore positive and negative steps are

$R_n= \frac{1}{2}(n+x)=\frac{1}{2}(5+3)=4$ and $X_n=R_n-L_n \Rightarrow 3=4+L_n=1$
The probability that the event $X_5=3$ will occur in a random walk with $p=0.6$ is
\[P(X_5=3)=\binom{5}{\frac{1}{2}(5+3)}(0.6)^{\tfrac{1}{2}(3+5)}(0.4)^{\tfrac{1}{2}(5-3)}=0.2592\]

Click the links to learn Stochastic Processes Introduction, Random Walk Models, Simple Random Walk

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Simple Random Walk: Unrestricted Random Walk

A simple random (or unrestricted random walk) walk on a line or in one dimension occurs with probability $p$ when walker step forward (+1) and/or has probability $q=1-p$ if walker steps back ($-1$). For ith step, the modified Bernoulli random variable $W_i$ (takes the value $+1$ or $-1$ instead of {0,1}) is observed and the position of the walk at the nth step can be found by
\begin{align}
X_n&=X_0+W_1+W_2+\cdots+W_n\nonumber\\
&=X_0+\sum_{i=1}^nW_i\nonumber\\
&=X_{n-1}+W_n
\end{align}
In the gambler’s ruin problems X0=k, but here we assume (without loss of generality) that walks start from the origin so that Xo=0.

Several derived results for random walks are restricted by boundaries. We consider here random walks without boundaries called unrestricted random walks. We are interested in

  1. The position of the walk after a number of steps and
  2. The probability of a return to the origin, the start of the walker.

From equation (1) the position of the walker at step n simply depends on the position at (n-1)th step, because the simple random walk possesses the Markov property (the current state of the walk depends on its immediate previous state, not on the history of the walks up to the present state)

Furthermore Xn=Xn-1 ±1

and the transition probabilities from one position to another is
P(Xn=j|Xn-1=j-1)=p, and P(Xn=j|Xn-1=j+1)=q is independent of the number of plays in the game or steps is represented by n.

Mean and Variance of Xn can be calculated as:
\begin{align*}
E(X_n)&=E\left(X_0+\sum_{i=1}^n W_i\right)\\
&=E\left(\sum_{i=1}^n W_i\right)=nW_n\\
V(X_n)&=V\left(\sum_{i=1}^n W_i\right)=nV(W)
\end{align*}
Since $W_i$ are independent and identically distributed (iid) random variables and where $W$ is the common or typical Bernoulli random variable in the sequence {Wi}. Thus
\begin{align*}
E(W)&=1.p+(-1)q=p-q\\
V(W)&=E(W^2)-[E(W)]^2\\
&=1^2p+(-1)^2q-(p-q)^2\\
&=p+q-(p^2+q^2-2pq)\\
&=1-p^2-q^2+2pq\\
&=1-p^2-(1-p)^2+2pq\\
&=1-p^2-(1+p^2-2p)+2pq\\
&=1-p^2-1-p^2+2p+2pq\\
&=-2p^2+2p+2pq\\
&=2p(1-p)+2pq=4pq
\end{align*}
So the probability distribution of the position of the random walk at stage n has to mean E(Xn)=n(p-q) and variance V(Xn)=4npq.

For the symmetric random walk (where p=½) after n steps, the expected position is the origin, and it yields the maximum value of V(Xn)=4npq=4np(1-p).

If p>½ then drift is expected away from the origin in a positive direction and if p<½ it would be expected that the drift would be in the negative direction.

Since V(Xn) is proportional to $n$, it grows with increasing n, and we would be increasingly uncertain about the position of the walker as n increases.
i.e.
\begin{align*}
\frac{\partial V(X_n)}{\partial p}&=\frac{\partial}{\partial p} {4npq}\\
&=\frac{\partial}{\partial p} \{4np-4np^2 \}=4n-8np \quad \Rightarrow p=\frac{1}{2}
\end{align*}
Just knowing the mean and standard deviation of a random variable does not enable us to identify its probability distribution. But for large n, we can apply the CLT.
\[Z_n=\frac{X_n-n(p-q)}{\sqrt{4npq}}\thickapprox N(0,1)\]
Applying continuity correction, approximate probabilities may be obtained for the position of the walk.

Example : Consider unrestricted random walk with n=100, p=0.6 then
\begin{align*}
E(X_n)&=E(X_{100})=nE(W)=n(p-q)\\
&=100(0.6-0.4)=20\\
V(X_n)&=4npq=4\times 100\times 0.6 \times 0.4=96
\end{align*}
The position of walk at the 100th step between 15 and 25 pace/step from the origin is
\[P(15\leq X_{100}\leq30)\thickapprox P(14.5<X_{100}<25.5)\]
\[-\frac{5.5}{\sqrt{96}}<Z_{100}=\frac{X_{100}-20}{\sqrt{96}}<\frac{5.5}{96}\]
hence
\[P(-0.5613<Z_{100}<0.5613)=\phi(0.5613)-\phi(-0.5613)=0.43\]
where Φ(Z) is the standard normal distribution function.

Read more about Simple Random walk: Random Walks Model

References

  1. http://www.encyclopediaofmath.org/index.php/Random_walk
  2. http://mathworld.wolfram.com/RandomWalk1-Dimensional.html

Random Walks Model: A Mathematical Formalization of Path

A random walk (first introduced by Karl Pearson in 1905) is a mathematical formalization of a path consisting series of random steps.

Random Walks Example

  1. The path traced by a molecule as it travels in a liquid or gas,
  2. The search path of a foraging animal,
  3. The price of a fluctuating stock, and (iv) the financial status of a gambler.
    All these random steps in the example can be modeled as random walks, although they may not be truly random in reality.

Suppose there are $a+1$ positions marked out on a straight line and numbered 0,1,2,…, a. A person starts at $k$ where $0<k<a$. The walk proceeds in such a way that, at each step, there is probability p that the walker goes forward one step to $k+1$ and a probability $q=1-p$ that the walker goes back one step to $k-1$. The walk continues until either $0$ or $a$ is reached and then ends.

In a random walk, the position of a walker after having moved $n$ times is known as the state of the walk after $n$ steps or after covering $n$ stages. Thus the walk described above starts at stage $k$ at step $0$ and moves to either stage $k-1$ or stage $k+1$ after 1 step and so on.

If the walk is bounded, then the ends of the walk are known as barriers and they may have various properties. In this case, the barriers are said to be absorbing implying that the walk must end once a barrier is reached since there is no escape.

A useful diagrammatic way of representing random walk is by a transition or process diagram. In a transition diagram, the possible states of the walker can be represented by points on a line. If a transition between two points can occur in one step then those points are joined by a curve or edge as shown with an arrow indicating the direction of the walk and a weighting denoting the probability of the step occurring. A transition diagram is also known as a direct graph.

For small Markov processes the simplest way to represent the process is often in terms of its state transition diagram. In-state transition diagram each state (outcome) represents the process as a node in a graph. The arcs in the graph represent possible transitions between states of the process. The arcs are labeled by the transition rates between the states.

Example:  Suppose a meteorologist notices that the weather on a given day seems to depend on the weather conditions of the previous day. He/ She observes that if it is raining one day, then the next day is sunny 60% of the time and rainy 40% of the time; on the other hand, if it is sunny, the next day is sunny with probability 30% and rainy with probability 70%. Note that there are two outcomes (i) sunny and (ii) rainy in this Markov process.

The transition probability between sunny and rainy is 70%, between sunny and sunny is 30%, between rainy and sunny is 60%, and between rainy and rainy is 40%. The simple weather forecasting Markov Process in the transition diagram is

Random Walks https://itfeature.com
Random Walks

Random walks model are widely used in many fields such as Ecology, Economics, Psychology, Computer Science, Physics, Chemistry, Biology, etc. Random walks explain the observed behavior of processes in all these fields, serving as a fundamental model for the recorded stochastic activity.

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