Random Walks with Holding Probabilities

Adding a Hold to a Random Walk

A simple random walk changes location at every step. In some models, however, a step may represent a fixed time interval during which the walker does not move: a patient’s status may remain unchanged, a particle may wait, or a system may have no event in a given time period. A holding probability adds this possibility without changing the basic nearest-neighbor structure. The essential modelling task is to assign the probabilities of moving left, moving right, and staying in place so that they form a valid transition rule.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. Take the state space to be \(S=\mathbb{Z}\) and the time index to be discrete, \(n\ge 0\). Let \(p\) be the probability of a step to the right, \(q\) the probability of a step to the left, and \(r\) the probability of holding. These probabilities must satisfy \(p\ge 0\), \(q\ge 0\), \(r\ge 0\), and \(p+q+r=1\). A genuine walk that permits movement in both directions has \(p>0\) and \(q>0\); zero values are still allowed when modelling one-sided or immobile cases.

The convention is that transition matrices are row-stochastic: entries in row \(i\) describe the next state given current state \(i\), and each row sums to one. Here the row for any state \(i\) has exactly three potentially nonzero entries. Their sum is \(q+r+p=1\), so the transition rule is valid. The fact that the state space is infinite does not change the row-wise check.

Constructing the Walk from Independent Increments

A direct construction makes both the probabilities and the Markov property precise. Let \(Y_1,Y_2,\ldots\) be independent and identically distributed random variables with values in \(\{-1,0,1\}\), where \(\mathbb{P}(Y_k=1)=p\), \(\mathbb{P}(Y_k=-1)=q\), and \(\mathbb{P}(Y_k=0)=r\). Let \(X_0\) be a \(\mathbb{Z}\)-valued random variable independent of these increments, and define \(X_n=X_0+\sum_{k=1}^nY_k\). The information through time \(n\) is \(\mathcal{F}_n=\sigma(X_0,Y_1,\ldots,Y_n)\). The next increment \(Y_{n+1}\) is independent of this information.

Proof. By construction, \(X_{n+1}=X_n+Y_{n+1}\). The increment \(Y_{n+1}\) is independent of \(\mathcal{F}_n\), and \(X_n\) is measurable with respect to \(\mathcal{F}_n\). Therefore, conditional on any positive-probability event \(A\cap\{X_n=i\}\), the next state is \(i+Y_{n+1}\), with probabilities \(p\), \(q\), and \(r\) for values \(i+1\), \(i-1\), and \(i\), respectively. If \(j\) is not one of these three states, the conditional probability is zero. None of these probabilities depends on \(A\) or \(n\), which is the Markov property with time-homogeneous transition probabilities. \(\square\)

The construction also shows what the holding probability does: it gives the increment a positive probability of being zero. The transition law does not require \(p=q\). Equal left and right probabilities describe a symmetric walk; unequal probabilities describe a biased one. In either case, the sum \(p+q\) is the probability that a step changes the state.

Proof. The displacement is \(Y_{n+1}\). Its mean is \(1\cdot p+(-1)\cdot q+0\cdot r=p-q\). Its second moment is \(1^2\cdot p+(-1)^2\cdot q+0^2\cdot r=p+q\). By the definition of variance, \(\operatorname{Var}(Y_{n+1})=\mathbb{E}[Y_{n+1}^2]-(\mathbb{E}[Y_{n+1}])^2=p+q-(p-q)^2\). \(\square\)

Worked Transition Rules

On \(\mathbb{Z}\), every state has a left neighbor, a right neighbor, and itself, so the same rule applies in every row. A finite interval needs additional modelling choices: at an endpoint, a proposed move might be reflected, suppressed, or sent to an absorbing state. Those choices change the endpoint rows and must be specified rather than inferred from the interior rule.

What Holding Probabilities Change—and What They Do Not

A hold is a genuine transition in discrete time: the chain takes one time step even though its state stays the same. It is not the same as deleting that time step or observing the chain only when it moves. For the integer-line model, the probability of a move on a given step is \(p+q=1-r\), while the probability of no change is \(r\). If a problem describes probabilities conditional on a move occurring, those conditional probabilities must be converted to unconditional step probabilities before writing the transition matrix. For example, if the probability of moving right conditional on moving is \(a\), then the unconditional right-step probability is \((1-r)a\), not \(a\).

The state space matters as much as the probabilities. The formula \(p_{i,i+1}=p\), \(p_{i,i-1}=q\), \(p_{ii}=r\) is directly valid on \(\mathbb{Z}\). On a cycle, the neighbors are interpreted modulo the cycle size. On a finite interval, one must replace the formula at the boundary according to an explicitly chosen rule. In every case, the final check is that each possible destination has been included and each row sums to one.

Common mistake. Suppose \(p=\tfrac13\), \(q=\tfrac12\), and \(r=\tfrac16\), as in the first example. Omitting the hold entry and writing only left and right transitions gives a row sum of \(\tfrac12+\tfrac13=\tfrac56\), not \(1\). That row would assign no destination to the remaining probability \(\tfrac16\), so it is not a transition row. The correction is to include \(p_{ii}=r=\tfrac16\), giving \(\tfrac12+\tfrac16+\tfrac13=1\). Another frequent error is to place \(p\) in the left-neighbor column; by definition here \(p\) is the right-step probability and \(q\) is the left-step probability.