A Process Is a Family of Random Variables
A random variable assigns a value to each outcome in a probability space. A stochastic process makes such assignments repeatedly: for every index, there is a random variable, and all those random variables are defined on the same probability space. The index might represent time, but the definition also allows other ways of organizing the variables.
It helps to keep three roles distinct. An outcome \(\omega\) is an element of the probability space \(\Omega\). An index \(i\) selects one member of the family. The process value \(X_i(\omega)\) is the state assigned to outcome \(\omega\) at index \(i\). The index is not itself a random outcome or a process value.
For a countable state space \(S\), we usually take \(\mathcal{S}\) to contain every subset of \(S\). For a real-valued process, the state space is \(S=\mathbb{R}\), with its Borel sets. The definition requires each fixed-index variable to be measurable. It does not, by itself, require any additional notion of measurability involving the index and outcome together.
The same information can be displayed as a function \(F:I\times\Omega\to S\), where \(F(i,\omega)=X_i(\omega)\). This notation is a convenient way to record all the values, but the defining requirement is that fixing any index produces a random variable.
Proof. Suppose first that \(F\) defines a process by \(X_i(\omega)=F(i,\omega)\). By the definition of a process, each \(X_i\) is measurable, so every fixed-index section of \(F\) is measurable. Conversely, suppose every such section is measurable. Define \(X_i(\omega)=F(i,\omega)\) for each \(i\). Each \(X_i\) is then a random variable with values in \(S\), and all the \(X_i\) are on the given probability space. Thus \((X_i)_{i\in I}\) is a stochastic process. \(\square\)
Finite and Infinite Index Sets
A process need not have infinitely many indices. A finite collection of random variables is a process too; an infinite index set simply supplies values at infinitely many indices. The examples below make the roles of outcome, index, and value explicit.
Worked Example: Counting Heads in Two Tosses
Let \(\Omega=\{HH,HT,TH,TT\}\), let \(\mathcal{F}\) be the set of all subsets of \(\Omega\), and assign probability \(\tfrac14\) to each outcome. Define \(X_n\) to be the number of heads in the first \(n\) tosses, for \(n\in I=\{0,1,2\}\), with \(X_0=0\). The state space is \(S=\{0,1,2\}\).
Each outcome determines a value at each index:
| Outcome \(\omega\) | \(X_0(\omega)\) | \(X_1(\omega)\) | \(X_2(\omega)\) |
|---|---|---|---|
| \(HH\) | 0 | 1 | 2 |
| \(HT\) | 0 | 1 | 1 |
| \(TH\) | 0 | 0 | 1 |
| \(TT\) | 0 | 0 | 0 |
For example, at outcome \(HT\), the process value at index \(2\) is \(X_2(HT)=1\), because one of the first two tosses is a head. The index \(2\) tells us how many tosses to include; it is not the value of the count. As a check, \(X_2=2\) occurs only at \(HH\), so \(\mathbb{P}(X_2=2)=\tfrac14\). Also, \(X_1=1\) occurs at \(HH\) and \(HT\), so \(\mathbb{P}(X_1=1)=\tfrac24=\tfrac12\).
For each fixed \(n\), the assignment from \(\omega\) to \(X_n(\omega)\) is a random variable. If instead we hold \(\omega\) fixed and let \(n\) vary, we see the sequence of values associated with that outcome. That sequence is often called the trajectory or sample path; here, it is simply a way to inspect the process without changing its definition as a family of random variables.
Worked Example: An Infinite Discrete-Time Process
Let \(\Omega=\{1,2,3\}\), with all subsets measurable and each outcome having probability \(\tfrac13\). For every integer \(n\ge0\), define \(X_n(\omega)=\mathbf{1}_{\{n\ge\omega\}}\), where \(\mathbf{1}_A\) equals \(1\) on event \(A\) and \(0\) otherwise. The index set is \(I=\{0,1,2,\ldots\}\), and the state space is \(S=\{0,1\}\).
The values at the first five indices are:
| Outcome \(\omega\) | \(X_0(\omega)\) | \(X_1(\omega)\) | \(X_2(\omega)\) | \(X_3(\omega)\) | \(X_4(\omega)\) |
|---|---|---|---|---|---|
| 1 | 0 | 1 | 1 | 1 | 1 |
| 2 | 0 | 0 | 1 | 1 | 1 |
| 3 | 0 | 0 | 0 | 1 | 1 |
At index \(2\), the event \(X_2=1\) is \(\{\omega:2\ge\omega\}=\{1,2\}\). Therefore \(\mathbb{P}(X_2=1)=\tfrac23\), and \(\mathbb{P}(X_2=0)=\tfrac13\). At index \(0\), no outcome satisfies \(0\ge\omega\), so \(\mathbb{P}(X_0=1)=0\). At every \(n\ge3\), all three outcomes satisfy \(n\ge\omega\), so \(\mathbb{P}(X_n=1)=1\). These calculations use the same three outcomes at each index; only the rule assigning the process value changes with \(n\).
Continuous Indices and State Values
An index set can also be continuous. The definition still asks for one random variable for each fixed index; it does not change the roles of \(\omega\), the index, and the process value.
Worked Example: A Threshold Process
Take \(\Omega=[0,1]\), \(\mathcal{F}\) to be the Borel sets, and \(\mathbb{P}\) to be the uniform probability measure on \([0,1]\). Let \(U(\omega)=\omega\). For every \(t\ge0\), set \(X_t(\omega)=\mathbf{1}_{\{U(\omega)\le t\}}\). The index set is \(I=[0,\infty)\), and the state space is \(S=\{0,1\}\).
At a fixed index \(t=\tfrac14\), the event \(X_{1/4}=1\) is \(\{U\le\tfrac14\}=[0,\tfrac14]\), which has probability \(\tfrac14\). Thus \(\mathbb{P}(X_{1/4}=1)=\tfrac14\) and \(\mathbb{P}(X_{1/4}=0)=1-\tfrac14=\tfrac34\). At \(t=1\), every \(\omega\in[0,1]\) satisfies \(U(\omega)\le1\), so \(\mathbb{P}(X_1=1)=1\). At \(t=0\), the event is the singleton \(\{0\}\), whose uniform probability is \(0\), so \(\mathbb{P}(X_0=1)=0\).
Fixing an outcome gives another perspective: if \(\omega=\tfrac34\), then \(X_t(\tfrac34)=0\) for \(0\le t<\tfrac34\) and \(X_t(\tfrac34)=1\) for \(t\ge\tfrac34\). The outcome is the threshold \(\tfrac34\); \(t\) is the index being checked; and the resulting value is either \(0\) or \(1\).
Equality at Each Index and Equality as Families
Because a process contains many random variables, “equal with probability one” can refer to two different statements. Two variables may agree with probability one at each fixed index, while there may be no single probability-one event on which they agree at every index. For a countable index set, the stronger statement does follow.
Proof. For each \(i\in I\), let \(A_i=\{\omega:X_i(\omega)\ne Y_i(\omega)\}\). The assumption gives \(\mathbb{P}(A_i)=0\). Since \(I\) is countable, its elements can be listed as \(i_1,i_2,\ldots\) (with a finite list if \(I\) is finite). The event that the processes disagree at some index is \(\bigcup_{i\in I}A_i\). Countable subadditivity gives \(\mathbb{P}(\bigcup_{i\in I}A_i)\le\sum_{i\in I}\mathbb{P}(A_i)=0\). Therefore the complement, the event that they agree at every index, has probability one. \(\square\)
The countability assumption matters. For example, use the uniform probability space \([0,1]\), take \(I=[0,1]\), and define \(X_t(\omega)=\mathbf{1}_{\{\omega=t\}}\) and \(Y_t(\omega)=0\). For each fixed \(t\), the event \(X_t\ne Y_t\) is the singleton \(\{\omega=t\}\), which has probability zero. But for every \(\omega\in[0,1]\), choosing \(t=\omega\) gives \(X_t(\omega)=1\ne0=Y_t(\omega)\). Thus the event of agreement at every index is empty, not a probability-one event. This example illustrates why a statement about each fixed random variable need not automatically hold simultaneously over an uncountable index set.
A Common Mix-Up
A frequent mistake is to treat the index as if it were the value of the process. In the two-toss example, the incorrect calculation “at index \(2\), the value is \(2\)” confuses how many tosses are examined with how many heads occurred. For the outcome \(HT\), the correct calculation is \(X_2(HT)=1\). For the outcome \(HH\), it is \(X_2(HH)=2\). The index is fixed at \(2\) in both cases; the outcome determines the value.
When reading a process, ask three separate questions: Which probability space contains the outcome? Which index selects the random variable? Which state does that variable assign to the outcome? Keeping those questions separate is the basic discipline for working with stochastic processes.
Check Your Understanding
For each question, identify whether it concerns an outcome, an index, or a process value.
- In the two-toss example, what is \(X_1(TH)\), and which outcome and index are being used?
- For the infinite discrete-time process, compute \(\mathbb{P}(X_1=1)\) and \(\mathbb{P}(X_3=1)\).
- In the threshold process, describe \(X_{1/2}(\omega)\) for \(\omega=\tfrac14\) and for \(\omega=\tfrac34\).
- State the measurability requirement in the definition of a process. Does it require a sigma-algebra on the index set?
- Why does coordinate-wise almost-sure equality imply agreement at every index with probability one for a countable index set?