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Stochastic Processes: Definitions and Examples · Tutorial 73 of 994

Stationary Increments

Learn to compare increments over equal-length intervals and verify stationary increments for linear processes and random walks.

Beginner 9 min read

What You'll Learn

  • Define stationary increments using equality in distribution over intervals of equal length.
  • Express increments over discrete and continuous time intervals.
  • Calculate increments for a deterministic linear process.
  • Verify stationary increments for random walks with identically distributed increments.
  • Identify how changing increment distributions can break stationarity of increments.
  • Distinguish equality of increment laws from equality of individual outcomes.

What It Means for Increments to Be Stationary

For a process, an increment records how much its value changes between two times. Stationary increments means that the distribution of this change depends on the length of the time interval, but not on where the interval begins. Thus increments over intervals of equal length have the same law; they need not have the same realized value on a particular sample path.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. For a real-valued discrete-time process \((X_n)_{n\ge0}\), the increment over the interval from \(n\) to \(n+h\) is \(X_{n+h}-X_n\), where \(n,h\) are nonnegative integers. For a real-valued continuous-time process \((X_t)_{t\ge0}\), the increment from \(s\) to \(s+h\) is \(X_{s+h}-X_s\), where \(s,h\ge0\) are real numbers. Equality in distribution is written \(\stackrel{d}{=}\).

Definition: A discrete-time process \((X_n)_{n\ge0}\) has stationary increments if, for every nonnegative integer \(h\), the distribution of \(X_{n+h}-X_n\) is the same for every nonnegative integer \(n\). Equivalently, \(X_{n+h}-X_n\stackrel{d}{=}X_h-X_0\) for all \(n,h\ge0\). A continuous-time process \((X_t)_{t\ge0}\) has stationary increments if, for every \(h\ge0\), \(X_{s+h}-X_s\stackrel{d}{=}X_h-X_0\) for every \(s\ge0\).

The interval of length zero is included: its increment is always zero. The definition compares distributions, not values. In particular, it does not require \(X_{n+h}-X_n=X_h-X_0\) for each outcome. Nor does it assert that increments from different intervals are independent; equality of individual increment distributions alone does not specify how those increments are related to one another.

The starting value cancels from an increment. Consequently, the distribution of \(X_0\) by itself does not determine whether increments are stationary. What matters is the distribution of the changes over intervals of a fixed length. The examples below make this comparison explicit.

Calculating Increments

Worked Example: A Deterministic Linear Process

On any probability space, define the continuous-time process with state space \(\mathbb{R}\) by \(X_t=5-2t\) for \(t\ge0\). For an interval beginning at \(s\) and having length \(h\ge0\), calculate the increment.

Substitute the process formula at both endpoints and subtract:

$$ X_{s+h}-X_s =[5-2(s+h)]-[5-2s] =5-2s-2h-5+2s =-2h. $$

The result is the constant \(-2h\), independent of the start time \(s\). For example, an interval of length \(3\) beginning at \(s=0\) has increment \(X_3-X_0=(-1)-5=-6\). An interval of the same length beginning at \(s=4\) has increment \(X_7-X_4=(-9)-(-3)=-6\). Both increments have the same distribution: a point mass at \(-6\).

Check: Substitution into the general expression gives \(-2(3)=-6\) for both intervals. Since the calculation holds for every \(s,h\ge0\), the process has stationary increments.

This calculation extends to an affine deterministic path \(X_t=b+ct\), where \(b,c\) are fixed real numbers: \(X_{s+h}-X_s=ch\). The intercept \(b\) and the start time \(s\) cancel. In the next result, a random sum gives a broader way to construct processes with stationary increments.

Theorem: Let \((Y_k)_{k\ge1}\) be independent, identically distributed, real-valued random variables on \((\Omega,\mathcal{F},\mathbb{P})\). For a fixed real number \(b\), define the discrete-time process with state space \(\mathbb{R}\) by \(X_0=b\) and \(X_n=b+\sum_{k=1}^{n}Y_k\) for \(n\ge1\). Then \((X_n)_{n\ge0}\) has stationary increments.

Proof. Fix nonnegative integers \(n\) and \(h\). If \(h=0\), then \(X_{n+h}-X_n=0=X_h-X_0\), so the required equality in distribution holds. If \(h\ge1\), subtract the two process values to obtain \(X_{n+h}-X_n=\sum_{r=1}^{h}Y_{n+r}\). The variables in this sum are independent and each has the same distribution as \(Y_1\). Therefore their sum has the same distribution as \(\sum_{r=1}^{h}Y_r=X_h-X_0\). This holds for every \(n\) and \(h\), proving stationary increments. \(\square\)

The theorem uses identical distribution as well as independence: identical distribution ensures that the summands have the same laws at different positions, and independence ensures that the sum has the same law when its block is shifted. The following examples calculate those laws directly.

Worked Example: A Simple Symmetric Random Walk

Let \((\Omega,\mathcal{F},\mathbb{P})\) support independent random variables \(Y_1,Y_2,\ldots\), each equal to \(1\) or \(-1\) with probability \(\tfrac12\). Define the discrete-time process on state space \(\mathbb{Z}\) by \(X_0=0\) and \(X_n=\sum_{k=1}^{n}Y_k\). Find the increment distribution over intervals of length \(2\), and compare two such intervals.

For any \(n\ge0\), \(X_{n+2}-X_n=Y_{n+1}+Y_{n+2}\). The sum is \(2\) if both steps are \(1\), is \(-2\) if both are \(-1\), and is \(0\) if the steps have opposite signs. Independence gives

$$ \begin{aligned} \mathbb{P}(X_{n+2}-X_n=2)&=\tfrac12\cdot\tfrac12=\tfrac14,\\ \mathbb{P}(X_{n+2}-X_n=-2)&=\tfrac12\cdot\tfrac12=\tfrac14,\\ \mathbb{P}(X_{n+2}-X_n=0)&=\tfrac12\cdot\tfrac12+\tfrac12\cdot\tfrac12=\tfrac12. \end{aligned} $$

For the interval from \(0\) to \(2\), the increment is \(X_2-X_0=Y_1+Y_2\), so its probabilities at \(2,-2,0\) are respectively \(\tfrac14,\tfrac14,\tfrac12\). For the interval from \(3\) to \(5\), it is \(X_5-X_3=Y_4+Y_5\), with exactly the same probabilities.

Check: The probabilities sum to \(\tfrac14+\tfrac14+\tfrac12=1\). The increment has mean \(2(\tfrac14)+(-2)(\tfrac14)+0(\tfrac12)=0\), as also follows from adding two zero-mean steps. The process has stationary increments by the theorem, not because these two realized increments must match.

Worked Example: A Random Walk with Nonzero Mean Steps

Suppose \((Y_k)_{k\ge1}\) are independent and identically distributed on a probability space, with \(\mathbb{P}(Y_k=0)=\tfrac13\) and \(\mathbb{P}(Y_k=2)=\tfrac23\). Let \(X_0=0\) and \(X_n=\sum_{k=1}^{n}Y_k\), a discrete-time process with state space \(\mathbb{Z}\). Find the distribution of an increment of length \(2\) and check it on two intervals.

For every \(n\ge0\), \(X_{n+2}-X_n=Y_{n+1}+Y_{n+2}\). The sum is \(0\) when both steps are zero, \(2\) when exactly one step is \(2\), and \(4\) when both steps are \(2\). Thus

$$ \begin{aligned} \mathbb{P}(X_{n+2}-X_n=0)&=\tfrac13\cdot\tfrac13=\tfrac19,\\ \mathbb{P}(X_{n+2}-X_n=2)&=\tfrac23\cdot\tfrac13+\tfrac13\cdot\tfrac23=\tfrac49,\\ \mathbb{P}(X_{n+2}-X_n=4)&=\tfrac23\cdot\tfrac23=\tfrac49. \end{aligned} $$

In particular, \(X_2-X_0=Y_1+Y_2\) has these probabilities, and \(X_6-X_4=Y_5+Y_6\) has the same probabilities because the steps have the same independent laws. The three probabilities sum to \(\tfrac19+\tfrac49+\tfrac49=1\). The expected increment is \(0(\tfrac19)+2(\tfrac49)+4(\tfrac49)=\tfrac83\), which agrees with twice the one-step mean \(2[0(\tfrac13)+2(\tfrac23)]=\tfrac83\).

Check: The formula for the increment uses exactly two steps, regardless of the starting time. The distribution therefore depends on the interval length \(2\), not on whether the interval starts at \(0\) or \(4\).

When the Increment Law Changes with Time

Worked Example: A Process Without Stationary Increments

Let \(\Omega=\{0,1\}\), let \(\mathcal{F}\) contain all subsets of \(\Omega\), and assign probability \(\tfrac12\) to each outcome. Define \(Y_1(\omega)=\omega\), set \(Y_k=0\) for every \(k\ge2\), and define the discrete-time process on state space \(\mathbb{R}\) by \(X_0=0\) and \(X_n=\sum_{k=1}^{n}Y_k\). Test whether its increments are stationary.

For the interval from \(0\) to \(1\), the increment is \(X_1-X_0=Y_1\). It equals \(0\) with probability \(\tfrac12\) and \(1\) with probability \(\tfrac12\). For the interval from \(1\) to \(2\), the increment is \(X_2-X_1=Y_2=0\), so it equals \(0\) with probability \(1\).

The two intervals both have length \(1\), but their increment distributions differ: the first assigns probability \(\tfrac12\) to \(1\), while the second assigns probability \(0\) to \(1\). Hence the process does not have stationary increments.

Check: Directly, \(X_1\) is equally likely to be \(0\) or \(1\), while \(X_2=X_1\). Thus \(X_2-X_1=0\) always, confirming the second increment calculation.

A common mistake is to replace equality in distribution with equality on each outcome. For instance, in the symmetric random walk example one might claim \(X_5-X_3=X_2-X_0\), because both are sums of two steps. That calculation is not justified: the left side is \(Y_4+Y_5\), while the right side is \(Y_1+Y_2\); these random variables need not be equal on a given outcome. The correct conclusion is that they have the same distribution, since both are sums of two independent steps with the same one-step law.

Another mistake is to check only one interval length. Stationary increments requires the comparison for every length, and, for each length, for every allowed starting time. In the nonstationary example, the length-one comparison already disproves the property. To establish the property, as for the random walks above, the argument must cover arbitrary \(n\) and \(h\). This is why writing the increment as a sum over the interval is useful: it identifies precisely which assumptions make its law independent of the starting time.

Stationary increments is useful whenever a model describes changes over time through a common increment law. A linear deterministic trend has a fixed change for each interval length; a random walk with identically distributed steps has a common increment distribution for each number of steps. In either case, the start time can be changed without changing the law of an increment of the same length. The nonstationary example shows why this property must be checked rather than inferred merely from the fact that a process is built by adding increments.

Check Your Understanding

Use the definition to answer each question, distinguishing distributional equality from pathwise equality.

  1. For \(X_t=7+4t\), what is the increment over an interval of length \(h\), and why does it not depend on the start time?
  2. For a random walk with independent, identically distributed steps, write \(X_{n+h}-X_n\) as a sum and state why its distribution is independent of \(n\).
  3. In the symmetric random walk, calculate the probabilities that an increment of length \(1\) equals \(1\) and \(-1\).
  4. Why does equality in distribution of two increments not imply that their realized values are equal on every outcome?
  5. In the final worked example, which two equal-length intervals have different increment laws, and what value distinguishes the laws?

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