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## 2. Processes with Stationary, Independent Increments

### Basic Theory

#### Definitions

Suppose again that $$\bs{X} = \{X_t: t \in T\}$$ is a random process with state space $$S \subseteq \R$$ and where the index set $$T$$ is either $$\N$$ or $$[0, \infty)$$. As usual, we interpret $$T$$ as the time space, so that in the first case we have discrete time and in the second case we have continuous time. Once again, the state space $$S$$ is either countable (often $$\N$$ or $$\Z$$), so that $$X_t$$ has a discrete distribution for $$t \in T$$, or an interval of $$\R$$ (often $$[0, \infty)$$ or $$\R$$) with $$X_t$$ having a continuous distribution for $$t \in T$$. Thus, in terms of the discrete/continuous dichotomy, we have all four cases: discrete time, discrete space; discrete time, continous space; continuous time, discrete space; continuous time, continuous space. We have seen the following definitions before.

Suppose that $$\bs{X} = \{X_t: t \in T\}$$ is a random process as above.

1. $$\bs{X}$$ has stationary increments if for $$s, t \in T$$ with $$s \le t$$, the increment $$X_t - X_s$$ has the same distribution as $$X_{t-s}$$.
2. $$\bs{X}$$ has independent increments if for $$t_1, t_2, \ldots, t_n \in T$$ with $$t_1 \lt t_2 \lt \cdots \lt t_n$$, the increments $$X_{t_1}, X_{t_2} - X_{t_1}, \ldots, X_{t_n} - X_{t_{n-1}}$$ are independent.

In discrete time when $$T = \N$$, the process $$\bs{X}$$ has stationary, independent increments if and only if $$\bs{X}$$ is the partial sum process associated with a sequence of independent, identically distributed variables

The process $$\bs{X} = (X_0, X_1, X_2, \ldots)$$ has stationary, independent increments if and only if there exists an independent, identically distributed process $$\bs{U} = (U_1, U_2, \ldots)$$ such that $X_n = \sum_{i=1}^n U_i, \quad n \in \N$

Proof:

Suppose first that $$(U_1, U_2, \ldots)$$ is a sequence of independent, identically distributed random variables taking values in $$S \subseteq \R$$, and let $$X_n = \sum_{i=1}^n U_i$$. If $$m \le n$$ then $$X_n - X_m = \sum_{i=m+1}^n U_i$$. Hence $$X_n - X_m$$ has the same distribution as $$X_{n-m} = \sum_{i=1}^{n-m} U_i$$ and is independent of $$(U_1, U_2, \ldots, U_m)$$ and hence also $$(X_0, X_1, \ldots, X_m)$$. Conversely, suppose that $$(X_0, X_1, \ldots)$$ has stationary, independent increments. Let $$U_i = X_i - X_{i-1}$$ for $$i \in \N_+$$. Then $$(U_1, U_2, \ldots)$$ is a sequence of independent, identically distributed variables, and $$X_n = \sum_{i=1}^n U_i$$.

It's much harder to characterize proccesses in continuous time with stationary, independent increments. As we have seen before, random processes indexed by an uncountable set are much more complicated in a technical sense than random processes indexed by a countable set. In spite of the technical difficulties, however, many of the underlying ideas are the same. Moreover, we have do have important examples. The non-homogeneous Poisson counting process $$\bs{N} = \{N_t: t \in [0, \infty)\}$$ has independent increments. If the process is in fact homogeneous, then it has stationary increments as well.

#### Distributions and Moments

For a process with stationary, independent increments, if we know the distribution of $$X_t$$ on $$S$$ for each $$t \in T$$, then we can compute all of the finite-dimensional distributions. To state the theorem, suppose that $$X_t$$ has probability density function $$f_t$$ on $$S$$ for $$t \in T$$. Thus, $$f_t$$ is the probability density function for a discrete distribution if $$S$$ is countable, or a probability density function for a continuous distribution if $$S$$ is an interval.

Suppose that $$\bs{X} = \{X_t: t \in T\}$$ has stationary, independent increments. If $$t_1, t_2, \ldots, t_n \in T$$ with $$t_1 \lt t_2 \lt \cdots \lt t_n$$ then $$\left(X_{t_1}, X_{t_2}, \ldots, X_{t_n}\right)$$ has probability density function $f_{t_1, t_2, \ldots, t_n}(x_1, x_2, \ldots, x_n) = f_{t_1}(x_1) f_{t_2 - t_1}(x_2 - x_1) \cdots f_{t_n - t_{n-1}}(x_n - x_{n-1}), \quad (x_1, x_2, \ldots, x_n) \in S^n$

Proof:

Let $$t_0 = 0$$ and define $$U_i = X_{t_i} - X_{t_{i-1}}$$ for $$i = 1, 2, \ldots, n$$. Then by the assumption of stationary independent increments, $$\bs{U} = (U_1, U_2, \ldots, U_n)$$ is a sequence of independent variables and $$U_i$$ has PDF $$u \mapsto f_{t_i - t_{i-1}}(u)$$. Let $$V_i = X_{t_i}$$ and $$\bs{V} = (V_1, V_2, \ldots, V_n)$$. Then $$V_i = U_1 + \cdots + U_i$$ for each $$i$$, or in matrix form, $$\bs{V} = A \bs{U}$$ where $$A$$ is the $$n \times n$$ matrix with $$A_{ij} = 1$$ if $$i \ge j$$ and 0 otherwise. The result follows from the change of variables theorem.

Recall that $$\bs{X} = \{X_t: t \in T\}$$ is a second order process if $$\E\left(X_t^2\right) \lt \infty$$ for each $$t \in T$$. For a second order process, the mean and covariance functions are usually of basic importance.

Suppose that $$\bs{X} = \{X_t: t \in T\}$$ is a second order process with stationary, independent increments. Then there exist constants $$\mu \in \R$$ and $$\sigma^2 \in [0, \infty)$$ such that the mean and covariance functions of $$\bs{X}$$ are

1. $$m(t) = \mu t$$ for $$t \in T$$
2. $$c(s, t) = \sigma^2 \min\{s, t\}$$ for $$(s, t) \in T^2$$
Proof:

Let $$v(t) = \var(X_t)$$ for $$t \in T$$. Now for $$s, \; t \in T$$ note that $$X_{s+t} = X_s + (X_{t+s} - X_s)$$. The second term is independent of the first and has the same distribution as $$X_{t}$$. Taking expected values gives $$m(s + t) = m(s) + m(t)$$ and taking variances gives $$v(s + t) = v(s) + v(t)$$. That is, $$m$$ and $$v$$ both satisfy Cauchy's functional equation, named of course, for the ubiquitous Augustin Cauchy. Since the functions $$m$$ and $$v$$ are bounded on bounded subsets of $$T$$, there exists $$\mu \in \R$$ and $$\sigma^2 \in [0, \infty)$$ such that $$m(t) = \mu t$$ and $$v(t) = \sigma^2 t$$ for $$t \in T$$. Finally, let $$s, t \in T$$ with $$s \le t$$. Then $$X_t - X_s$$ is independent of $$X_s$$ and therefore $\cov(X_s, X_t) = \cov\left[X_s, X_s + (X_t - X_s)\right] = \cov(X_s, X_s) + \cov(X_s, X_t - X_s) = \var(X_s) + 0 = \sigma^2 s$

Note from the proof of the last result that $$\mu = \E(X_1)$$ and $$\sigma^2 = \var(X_1)$$.

Recall that two stochastic processes (with the same time and state spaces) are equivalent in distribution if they have the same finite-dimensional distributions. Equivalence in distribution really is an equivalence relation on the class of stochastic processes with given state and time spaces. If a process with stationary independent increments is shifted forward in time and then centered in space, the new process is equivalent to the original.

Suppose that $$\bs{X} = \{X_t: t \in T\}$$ has stationary, independent increments. Fix $$t_0 \in T$$ and define $$Y_t = X_{t_0+t} - X_{t_0}$$ for $$t \in T$$. Then $$\bs{Y} = \{Y_t: t \in T\}$$ is equivalent in distribution to $$\bs{X}$$.

Proof:

Note that for $$s, t \in T$$ with $$s \le t$$, $$Y_t - Y_s = X_{t_0 + t} - X_{t_0 + s}$$. It follows that $$\bs{Y}$$ also has stationary, independent increments and the distribution of $$Y_t$$ is the same as the distribution of $$X_t$$ for each $$t \in T$$. From the PDF result above, $$\bs{Y}$$ has the same finite-dimensional distributions as $$\bs{X}$$.

#### The Markov Property

The last result can be generalized to show that a process with stationary, independent increments is a Markov process. Let $$\mathscr{F}_t = \sigma\{X_s: s \in T, s \le t\}$$ denote the $$\sigma$$-algebra generated by the process up to time $$t$$. Roughly speaking, we can determine if an event $$A \in \mathscr{F}_t$$ occurs by observing the process up to time $$t$$. The collection of $$\sigma$$-algebras $$\mathfrak{F} = \{\mathscr{F}_t: t \in T\}$$ is known as a filtration.

Suppose that $$\bs{X} = \{X_t: t \in T\}$$ is a process with stationary independent increments, and that $$X_t$$ has probability density function $$f_t$$ (either in the discrete or continuous cases) for $$t \in T$$. Then $$\bs{X}$$ is a (time homogeneous) Markov process with transition probability density function $$p$$ given by

$p_t(x, y) = f_t(y - x), \quad t \in T, \; x, y \in S$
Proof:

Suppose that $$s, t \in T$$. Note that $$X_{s+t} = X_s + (X_{s+t} - X_s)$$, and the second term is independent of $$\mathscr{F}_s$$ and has the same distribution as $$X_t$$. Hence given $$X_s = x \in S$$, $$X_{s+t}$$ is independent of $$F_s$$ and has the same distribution as $$x + X_t$$, which has the PDF $$y \mapsto f_t(y - x)$$. As usual, we have glossed over some of the technical details that arise when $$T = [0, \infty)$$.

Recall that a stopping time is a random time $$\tau$$ with values in $$T \cup \{\infty\}$$ that satisfies $$\{\tau \le t\} \in \mathscr{F}_t$$ for $$t \in T$$. So roughly speaking, we can determine whether $$\tau \le t$$ by observing the process up to time $$t$$. The $$\sigma$$-algebra associated with $$\tau$$ encodes the information from the random process up to the random time $$\tau$$, analogous to $$\mathscr{F}_t$$ that encodes the information from the process up to the deterministic time $$t$$. The appropriate definition is

$\mathscr{F}_\tau = \{A \in \mathscr{F}: A \cap \{\tau \le t\} \in \mathscr{F}_t \text{ for all } t \in T\}$

The strong Markov property is the Markov property applied to stopping times in addition to deterministic times. A discrete time process with stationary, independent increments is also a strong Markov process. The same is true in continuous time, with the addition of appropriate technical assumptions. Basically, these assumptions require that $$X_t(\omega)$$ be sufficiently nice (in terms of measureability in $$(t, \omega)$$ and continuity in $$t$$) and that we use a filtration that is a slight variation on $$\mathfrak{F}$$. A nice statement of the strong Markov property is a generalization of Theorem 4:

Suppose that $$\bs{X} = \{X_t: t \in T\}$$ is a process with stationary, independent increments, and $$\tau$$ is a stopping time. Then for $$t \in T$$, $$X_{\tau + t} - X_\tau$$ is independent of $$\mathscr{F}_\tau$$ and has the same distribution as $$X_t$$.

### Examples

We have seen several examples of random processes with stationary, independent increments. The following exercises give a quick review.

#### Bernoulli Trials

Let $$\bs{X} = (X_1, X_2, \ldots)$$ be sequence of Bernoulli trials with success parameter $$p \in (0, 1)$$, so that $$X_i = 1$$ if trial $$i$$ is a success, and 0 otherwise. Let $$Y_n = \sum_{i=1}^n X_i$$ for $$n \in \N$$, so $$Y_n$$ is the number of successes in the first $$n$$ trials, and hence has the binomial distribution with parameters $$n$$ and $$p$$.

The sequence $$\bs{Y} = (Y_0, Y_1, \ldots)$$ is a second order process with stationary independent increments. The mean and covariance functions are given by

1. $$m(n) = p n$$ for $$n \in \N$$
2. $$c(m, n) = p (1 - p) \min\{m, n\}$$ for $$(m, n) \in \N^2$$

Now let $$U_i$$ be the number of trials between the $$(i - 1)$$st sucess and the $$i$$th success. Thus, $$\bs{U} = (U_1, U_2, \ldots)$$ is a sequence of independent variables, each with the geometric distribution with parameter $$p$$. Let $$V_n = \sum_{i=1}^n U_i$$ for $$n \in \N$$, so that $$V_n$$ it the trial number of the $$n$$th success and hence has the negative binomial distribution with parameters $$n$$ and $$p$$.

The sequence $$\bs{V} = (V_0, V_1, \ldots)$$ is a seond order process with stationary, independent incrmements. The mean and covariance functions are given by

1. $$m(n) = \frac{1}{p} n$$ for $$n \in \N$$
2. $$c(m, n) = \frac{1 - p}{p^2} \min\{m, n\}$$ for $$(m, n) \in \N^2$$

#### Simple Random Walk

Closely related to the Bernoulli trials process is the simple random walk. Suppose that $$\bs{U} = (U_1, U_2, \ldots)$$ is a sequence of independent random variables with $$\P(U_i = 1) = p$$ and $$\P(U_i = -1) = 1 - p$$ for each $$i \in \N_+$$, where $$p \in (0, 1)$$ is a parameter. Let $$X_n = \sum_{i=1}^n U_i$$ for $$n \in \N$$, the position of the walker at time $$n$$.

The simple random walk $$\bs{X} = (X_0, X_1, X_2, \ldots)$$ is a second order process with stationary, independent increments. The mean and covariance functions are given by

1. $$m(n) = (2 p - 1)n$$ for $$n \in \N$$
2. $$c(m, n) = 4 p (1 - p) \min\{m, n\}$$ for $$(m, n) \in \N^2$$

#### Renewal Processes

Consider a renewal process with interarrival times $$\bs{X} = (X_1, X_2, \ldots)$$. By definition, this is a sequence of independent, identically distributed variables with values in $$[0, \infty)$$. Let $$\mu$$ and $$\sigma^2$$ denote the mean and variance of the interarrival times (assumed finite). Let $$T_n = \sum_{i=1}^n X_i$$ for $$n \in \N$$, so that $$T_n$$ is the time of the $$n$$th arrival.

The sequence of arrival times $$\bs{T} = (T_0, T_1, \ldots)$$ is a second order process with stationary, independent increments. The mean and covariance functions are given by

1. $$m(n) = \mu n$$ for $$n \in \N$$
2. $$c(m, n) = \sigma^2 \min\{m, n\}$$

#### Poisson Processes

Consider the Poisson model of random points in $$[0, \infty)$$, with rate parameter $$r \in (0, \infty)$$. This is a special renewal process with interarrival times $$\bs{X} = (X_1, X_2, \ldots)$$ that have the exponential distribution with rate parameter $$r$$. The time of the $$n$$th arrival $$T_n = \sum_{i=1}^n X_i$$ has the gamma distribution with parameters $$n$$ and $$r$$. The results of the previous exercise apply with $$\mu = 1 / r$$ and $$\sigma^2 = 1 / r^2$$.

The sequence of arrival times $$\bs{T} = (T_0, T_1, \ldots)$$ is a second order process with stationary, independent increments. The mean and covariance functions given by

1. $$m(n) = \frac{1}{r} n$$ for $$n \in \N$$
2. $$c(m, n) = \frac{1}{r^2} \min\{m, n\}$$ for $$(m, n) \in \N^2$$

Now let $$\bs{N} = \{N_t: t \in [0, \infty)\}$$ denote the counting process. Thus $$N_t$$ is the number of random points in $$[0, t]$$, which has the Poisson distribution with parameter $$r t$$.

The Poisson counting process $$\bs{N}$$ is a second order process with stationary, independent increments. The mean and covaraince functions are given by

1. $$m(t) = \lambda t$$ for $$t \in [0, \infty)$$
2. $$c(s, t) = \lambda \min\{s, t\}$$ for $$(s, t) \in [0, t)^2$$

Suppose now that $$r: [0, \infty) \to (0, \infty)$$ is measurable and let $m(t) = \int_{(0, t]} r(s) \, d\lambda(s), \quad t \in [0, \infty)$ where as usual $$\lambda$$ denotes Lebesgue measure on $$\R$$. Consider the hon-homogeneous Poisson process with rate function $$r$$ and hence mean function $$m$$. As before, let $$T_n$$ denote the time of the $$n$$th arrival for $$n \in \N$$. Unless $$r$$ is contant (so that the Poisson process is in fact homogeneous), the arrival time sequence $$\bs{T} = (T_0, T_1, \ldots)$$ has independent increments, but not stationary increments. Similarly, if $$N_t$$ denotes the number of arrivals in $$(0, t]$$ for $$t \in [0, \infty)$$, then the counting process $$\bs{N} = \{N_t: t \in [0, \infty)\}$$ has independent, but not stationary increments.

#### Summary

The examples involving Bernoulli trials and the simple random walk have discrete time and state spaces. In the Poisson model, the arrival time process has a discrete time space and a continuous state space, while the counting process has a continuous time space and a discrete state space. We are missing an example of a process with stationary, independent increments and with continuous time and state spaces. Brownian motion, one of the most important random processes, will fill the gap.