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6.2 Discrete Time Stationary Processes 171

have the same distribution, then the two processes have the same distribution. The proofs
of these results require rather sophisticated methods of advanced probability theory; see,
for example, Port (1994, Chapter 16) and Billingsley (1995, Chapter 7).

6.2 Discrete Time Stationary Processes
Perhaps the simplest type of discrete time process is one in which X 0 , X 1 , X 2 ,... are
independent, identically distributed random variables so that, for each i and j, X i and X j
have the same distribution. A generalization of this idea is a stationary process in which,
for any n = 1, 2,... and any integers t 1 ,..., t n ,
) and (X t 1 +h ,..., X t n +h )
(X t 1 ,..., X t n
have the same distribution for any h = 1, 2,....

Example 6.1 (Exchangeable random variables). Let X 0 , X 1 ,... denote a sequence of
exchangeable random variables. Then, for any n = 0, 1,..., X 0 , X 1 ,..., X n are exchange-
able. It follows from Theorem 2.8 that any subset of X 0 , X 1 ,... of size m has the same
marginal distribution as any other subset of X 0 , X 1 ,... of size m. Clearly, this implies that
the condition for stationarity is satisfied so that the process {X t : t ∈ Z} is stationary.

The following result gives a necessary and sufficient condition for stationarity that is
often easier to use than the definition.

Theorem 6.1. Let {X t : t ∈ Z} denote a discrete time process and define
Y t = X t+1 , t = 0, 1, 2,....

Then {X t : t ∈ Z} is stationary if and only if {Y t : t ∈ Z} has the same distribution as
{X t : t ∈ Z}.

Proof. Clearly if {X t : t ∈ Z} is stationary then {X t : t ∈ Z} and {Y t : t ∈ Z} have the same
distribution. Hence, assume that {X t : t ∈ Z} and {Y t : t ∈ Z} have the same distribution.
Fix n = 1, 2,... and t 1 ,... t n in Z. Then, by assumption,
D
) = (X t 1 +1 ,..., X t n +1 ).
(X t 1 ,..., X t n
D
Here the symbol = is used to indicate that two random variables have the same distribution.
Hence,
D
{X t : t ∈ T } ={X t+1 , t ∈ Z}.
It then follows that
D
{X t+1 : t ∈ Z} ={Y t+1 : t ∈ Z}.
That is,

D
) = (X t 1 +2 ,..., X t n +2 ).
(X t 1 ,..., X t n