Comparing Processes Through Their Coordinates
Two stochastic processes can be constructed on different probability spaces and still have the same distribution. To compare them, we do not need to match individual outcomes or sample paths. Instead, we compare the probabilities of seeing each possible list of states at any finite collection of times. These finite-coordinate comparisons are the defining test for equality in distribution of processes.
Throughout, let \((X_n)_{n\ge0}\) be defined on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((Y_n)_{n\ge0}\) be defined on a possibly different probability space \((\widetilde{\Omega},\widetilde{\mathcal{F}},\widetilde{\mathbb{P}})\). Assume both processes take values in the same countable state space \(S\), equipped with the collection of all its subsets. No Markov property, independence, or stationarity is assumed.
\(\mathbb{P}((X_{n_1},\ldots,X_{n_k})\in C)=\widetilde{\mathbb{P}}((Y_{n_1},\ldots,Y_{n_k})\in C)\).
In particular, equality in distribution requires agreement at every finite selection of times, not just at each time considered separately. Since \(S\) is countable, a finite-dimensional distribution can be described by its probabilities on individual tuples: the probability of any subset \(C\subseteq S^k\) is the sum of the tuple probabilities over the countable set \(C\). This gives a practical coordinate-event test.
\(\mathbb{P}(X_{n_1}=i_1,\ldots,X_{n_k}=i_k)=\widetilde{\mathbb{P}}(Y_{n_1}=i_1,\ldots,Y_{n_k}=i_k)\).
Proof. If the processes are equal in distribution, take \(C=\{(i_1,\ldots,i_k)\}\) in the definition. This gives equality of the probabilities of the indicated coordinate events. Conversely, suppose all these coordinate-event probabilities agree. Fix times \(n_1<\cdots<n_k\) and a subset \(C\subseteq S^k\). The events that the coordinate vector equals a particular tuple are disjoint, and their union over tuples in \(C\) is the event that the vector belongs to \(C\). Countable additivity therefore gives \(\mathbb{P}((X_{n_1},\ldots,X_{n_k})\in C)=\sum_{(i_1,\ldots,i_k)\in C}\mathbb{P}(X_{n_1}=i_1,\ldots,X_{n_k}=i_k)\). The corresponding expression for \(Y\) has the same summands by the assumption, so the two probabilities are equal. This holds for every \(k\), every such time selection, and every \(C\), which is equality in distribution. \(\square\)
It is enough to use strictly increasing times in this criterion. If a comparison is initially written with repeated time indices, first list the distinct times in increasing order. If the repeated coordinates demand different states at the same time, the event is empty and has probability zero for both processes. If they demand the same state, the event is exactly the coordinate event at the distinct times. Thus repeated indices add no new test.
Worked Comparisons
Worked Example: Matching Marginals but Not the Process Distribution
Let \(S=\{0,1\}\). On one probability space, let \(Z\) have probabilities \(\mathbb{P}(Z=0)=\mathbb{P}(Z=1)=\tfrac12\), and define \(X_n=Z\) for every \(n\ge0\). On another probability space, let \((Y_n)_{n\ge0}\) be independent fair binary random variables. Compare the processes.
At any one time \(n\), \(\mathbb{P}(X_n=0)=\mathbb{P}(X_n=1)=\tfrac12\). Also \(\widetilde{\mathbb{P}}(Y_n=0)=\widetilde{\mathbb{P}}(Y_n=1)=\tfrac12\). Thus all the one-time distributions agree.
Now use the coordinate event at times \(0\) and \(1\) with assigned states \(0\) and \(1\). For \(X\), both coordinates equal the same \(Z\), so \(\mathbb{P}(X_0=0,X_1=1)=0\). For \(Y\), independence gives \(\widetilde{\mathbb{P}}(Y_0=0,Y_1=1)=\tfrac12\cdot\tfrac12=\tfrac14\). These probabilities differ, so the processes are not equal in distribution.
Check: The pair \((X_0,X_1)\) takes only the values \((0,0)\) and \((1,1)\), each with probability \(\tfrac12\), so its probabilities sum to \(1\). The pair \((Y_0,Y_1)\) has four possible values, each with probability \(\tfrac14\), also summing to \(1\). The disagreement is in the joint distributions, not the marginals.
Worked Example: Equal Distributions Without Equal Sample Paths
Let \((X_n)_{n\ge0}\) be independent fair binary random variables on a probability space, and define \(Y_n=1-X_n\) on that same space. Show that the processes are equal in distribution, even though their paths do not agree.
Take any \(k\ge1\), any distinct increasing times \(n_1<\cdots<n_k\), and any assigned states \(i_1,\ldots,i_k\in\{0,1\}\). Independence and fairness give \(\mathbb{P}(X_{n_1}=i_1,\ldots,X_{n_k}=i_k)=(\tfrac12)^k\). The event \(Y_{n_r}=i_r\) for every \(r\) is the event \(X_{n_r}=1-i_r\) for every \(r\). The values \(1-i_r\) are also a fixed assignment of \(k\) binary coordinates, so \(\mathbb{P}(Y_{n_1}=i_1,\ldots,Y_{n_k}=i_k)=(\tfrac12)^k\). The coordinate-event criterion proves equality in distribution.
For a pathwise comparison, at every outcome and every time, \(X_n\) is either \(0\) or \(1\), and \(Y_n=1-X_n\). Therefore \(Y_n\ne X_n\) for every \(n\), with probability \(1\). The processes have equal distributions but are not equal as random sequences on this probability space.
Check: At two distinct times, each of the four assigned pairs has probability \((\tfrac12)^2=\tfrac14\) for \(X\), and also \(\tfrac14\) for \(Y\). The four pair probabilities sum to \(4\cdot\tfrac14=1\); the same calculation applies to any fixed number \(k\) of distinct coordinates.
Worked Example: Agreement at One and Two Times Is Not Enough
Let \((X_n)_{n\ge0}\) be independent fair binary random variables. Construct a process \(Y\) whose distributions at every selection of one or two distinct times match those of \(X\), but whose distribution at three times does not.
Let \(A\) and \(B\) be independent fair binary random variables, and set \(Y_0=A\), \(Y_1=B\), and \(Y_2=A\mathbin{\mathrm{xor}}B\), where “xor” is \(0\) when its inputs agree and \(1\) when they differ. For \(n\ge3\), let \(Y_n\) be independent fair binary random variables, independent also of \(A\) and \(B\).
The pair \((Y_0,Y_1)=(A,B)\) has each of its four possible values with probability \(\tfrac14\). For \((Y_0,Y_2)\), fixing \(Y_0=a\) and \(Y_2=c\) forces the unique value \(B=a\mathbin{\mathrm{xor}}c\); the probability of that pair is therefore \(\tfrac14\). For \((Y_1,Y_2)\), fixing \(Y_1=b\) and \(Y_2=c\) forces the unique value \(A=b\mathbin{\mathrm{xor}}c\), again giving probability \(\tfrac14\). Each of these pairs therefore has the same distribution as two distinct coordinates of \(X\). A pair involving a time \(n\ge3\) has the same fair, independent coordinates as the corresponding pair of \(X\), because those later variables are independent of \(A,B\) and of each other. The one-time distributions are also fair at every time.
At the three times \(0,1,2\), the values of \(Y\) must have even parity: \(Y_0\mathbin{\mathrm{xor}}Y_1\mathbin{\mathrm{xor}}Y_2=0\). The four possible triples are \((0,0,0)\), \((0,1,1)\), \((1,0,1)\), and \((1,1,0)\). Each arises from one of the four equally likely pairs \((A,B)\), so each has probability \(\tfrac14\); the other four triples have probability zero. For \(X\), all eight triples have probability \((\tfrac12)^3=\tfrac18\). For example, \(\mathbb{P}(Y_0=0,Y_1=0,Y_2=1)=0\), whereas \(\mathbb{P}(X_0=0,X_1=0,X_2=1)=\tfrac18\). Thus the processes agree at all selections of at most two distinct times but are not equal in distribution.
Check: The four positive triple probabilities for \(Y\) sum to \(4\cdot\tfrac14=1\); the eight triple probabilities for \(X\) sum to \(8\cdot\tfrac18=1\). This example shows why equality in distribution requires every finite dimension, not merely one- and two-dimensional distributions.
Useful Consequences and a Reliable Test
Proof. Fix \(k\ge1\), times \(n_1<\cdots<n_k\), and states \(t_1,\ldots,t_k\in T\). The event \(g_{n_r}(X_{n_r})=t_r\) for every \(r\) is the event that \(X_{n_r}\in g_{n_r}^{-1}(\{t_r\})\) for every \(r\). Its probability is the sum of the probabilities of the coordinate events \((X_{n_1},\ldots,X_{n_k})=(i_1,\ldots,i_k)\) over all tuples satisfying \(g_{n_r}(i_r)=t_r\) for every \(r\). Equality in distribution makes each such tuple probability equal to the corresponding probability for \(Y\); countable additivity then makes the sums equal. The coordinate-event criterion gives equality in distribution of the transformed processes. \(\square\)
For example, the proposition means that if two processes have the same finite-coordinate distributions, then keeping only whether each coordinate belongs to a chosen subset, or applying a fixed relabeling of the states, cannot make their resulting finite-coordinate distributions differ. The proposition does not say that the transformed processes reveal all information about the original ones; a many-to-one function can discard information.
A practical comparison proceeds in a fixed order. First confirm that both processes use the same state space and time indices. Then choose a finite set of distinct times and calculate the probabilities for assigned states at those times. If one coordinate-event probability differs, the processes are not equal in distribution. To establish equality, every finite selection and every assignment must agree; checking only a few convenient times can disprove equality, but ordinarily cannot prove it.
Common mistake. A tempting but invalid argument is: “At every time \(n\), \(X_n\) and \(Y_n\) have the same distribution, so the processes are equal in distribution.” The first worked example gives the error concretely: at times \(0\) and \(1\), both processes have fair binary marginals, but the event \((0,1)\) has probability \(0\) for \(X\) and \(\tfrac14\) for \(Y\). The correction is to compare joint coordinate events as well as individual marginals, and, for a proof of process equality, to establish agreement at every finite set of times.
Equality in distribution is a statement about probabilities, not a statement that the two processes use the same random variables, the same probability space, or the same paths. Conversely, two processes may be defined on the same probability space and still have the same distribution without agreeing outcome by outcome. Keeping these distinctions explicit prevents a coupling or a particular construction from being mistaken for the definition of the process distribution.
Check Your Understanding
Use coordinate events to decide what can and cannot be concluded in each situation.
- Two processes on \(\{0,1\}\) have the same marginal distribution at every time. What additional probabilities must agree for the processes to be equal in distribution?
- For two processes on a countable state space, explain why agreement on every individual tuple probability implies agreement on every subset of \(S^k\).
- In the complement example, what is the probability that \(Y_0=1\) and \(Y_1=0\)? Explain why this agrees with the corresponding probability for \(X\).
- For the parity construction, identify one triple event that has different probabilities under \(X\) and \(Y\).
- If two processes are equal in distribution and the same state-labeling function is applied at every time, what does the coordinatewise transformation proposition conclude?