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Stochastic Processes: Definitions and Examples · Tutorial 67 of 994

Finite-Dimensional Distributions

Learn to turn a finite selection of process values into a joint random vector and compute its probability law.

Beginner 10 min read

What You'll Learn

  • Define a finite-dimensional distribution using a finite ordered selection of process indices.
  • Compute joint probabilities for a sequence of independent Bernoulli variables.
  • Find a process’s joint law by partitioning the underlying probability space.
  • Use a shared latent variable to calculate dependent process values.
  • Check consistency under coordinate projection and distinguish joint laws from marginal probabilities.

From a Process to a Finite Random Vector

A stochastic process is a collection of random variables indexed by time or another index set. Its values at any finite selection of indices form an ordinary random vector. The probability law of that vector is called a finite-dimensional distribution. It records how the selected values occur together, not just how often each value occurs separately. This makes it a basic way to describe the joint probabilistic behaviour of a process without needing to specify an entire sample path.

Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \(T\) be an index set, and let \(S\) be a measurable state space. A process \((X_t)_{t\in T}\) consists of measurable random variables \(X_t:\Omega\to S\). For distinct indices \(t_1,\ldots,t_k\), the vector \((X_{t_1},\ldots,X_{t_k})\) takes values in \(S^k\). Its probability law assigns to each measurable set \(B\subseteq S^k\) the probability that the vector lies in \(B\).

Definition: The finite-dimensional distribution of \((X_t)_{t\in T}\) at the ordered indices \(t_1,\ldots,t_k\) is the probability measure on \(S^k\) given by \[ \mathbb{P}\big((X_{t_1},\ldots,X_{t_k})\in B\big),\qquad B\subseteq S^k\ \text{measurable}. \] For a countable state space \(S\), it is equivalently specified by the joint probabilities \(\mathbb{P}(X_{t_1}=x_1,\ldots,X_{t_k}=x_k)\) for all \((x_1,\ldots,x_k)\in S^k\).

The indices are ordered because the coordinates in the vector follow that order. One may list them increasingly when \(T\) is an ordered time set, but the definition also works for any chosen order. Reordering the indices reorders the coordinates. Repeated indices add no new information: for example, \((X_s,X_s)\) is supported only on pairs whose two coordinates are equal. It is usually simpler to use distinct indices.

For real-valued processes, the same joint law can be described by its joint cumulative distribution function \(F_{t_1,\ldots,t_k}(x_1,\ldots,x_k)=\mathbb{P}(X_{t_1}\le x_1,\ldots,X_{t_k}\le x_k)\). For discrete states, a table of joint point probabilities is often more direct. In either case, marginal probabilities alone do not generally determine the finite-dimensional distribution: dependence between the selected values matters.

Two Properties That Help Organize the Laws

A finite-dimensional distribution is a pushforward probability law: it is obtained by applying the map \(\omega\mapsto(X_{t_1}(\omega),\ldots,X_{t_k}(\omega))\) to the original probability space. This observation gives both a practical computation method and a consistency check.

Proposition (Projection consistency): If a finite-dimensional distribution is known at indices \(t_1,\ldots,t_k\), then its distribution on any subcollection of those coordinates is obtained by projecting the joint vector onto those coordinates. In particular, summing a discrete joint probability mass function over the omitted coordinates gives the corresponding lower-dimensional mass function.

Proof. Write \(Y=(X_{t_1},\ldots,X_{t_k})\), and let \(J\) select some of its coordinates in their original order. For any measurable set \(C\) in the space of the selected coordinates, the event that the projection of \(Y\) lies in \(C\) is exactly the event that the selected process values lie in \(C\). Thus its probability is the probability assigned to the inverse image of \(C\) under the projection, which is the projected joint law. In the discrete case, fix values for the selected coordinates. The event that those values occur is the disjoint union of the events obtained by assigning every possible combination of values to the omitted coordinates. Countable additivity says its probability is the sum of the corresponding joint point probabilities. \(\square\)

There is a companion rule for changing coordinate order: if the indices are permuted, the vector’s coordinates are permuted by the same rule, and probabilities are assigned to the correspondingly permuted sets. Both rules follow directly from what the vector records; they do not require independence or stationarity.

Computation rule: If the process is a function of a random element \(Z\) with known distribution, say \(X_t=g_t(Z)\), then the joint event at \(t_1,\ldots,t_k\) is found by evaluating all the functions \(g_{t_1},\ldots,g_{t_k}\) at the same value of \(Z\). Partition the possible values of \(Z\) according to the resulting vector, then add the probabilities within each part.

Worked Distributions

Worked Example: Two Coordinates of Independent Tosses

Let \((\Omega,\mathcal{F},\mathbb{P})\) describe three independent fair coin tosses, and let \(X_n=1\) if toss \(n+1\) is heads and \(X_n=0\) if it is tails, for \(n=0,1,2\). This is a discrete-time process with index set \(T=\{0,1,2\}\) and state space \(S=\{0,1\}\). Find the finite-dimensional distribution at indices \(0\) and \(2\).

There are \(2^3=8\) equally likely toss sequences. The first coordinate records toss 1 and the second records toss 3. For each pair \((x_0,x_2)\), toss 2 can be either 0 or 1, so there are two sequences giving that pair. The joint probabilities are therefore

$$ \mathbb{P}(X_0=x_0,X_2=x_2)=\frac{2}{8}=\frac14, \qquad (x_0,x_2)\in\{0,1\}^2. $$

Thus each of \((0,0),(0,1),(1,0),(1,1)\) has probability \(\tfrac14\). The probabilities sum to \(4\cdot\tfrac14=1\). Summing over the second coordinate gives \(\mathbb{P}(X_0=0)=\tfrac14+\tfrac14=\tfrac12\) and \(\mathbb{P}(X_0=1)=\tfrac12\). Summing over the first coordinate gives the same two probabilities for \(X_2\), as required for fair tosses. The full law also records that these two selected tosses can take any of their four pairs of values.

Worked Example: Thresholds of One Uniform Random Variable

Let \(U\) be uniform on \([0,1]\) on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Define a discrete-time process with index set \(T=\{0,1,2\}\), state space \(S=\{0,1\}\), and \(X_n=\mathbf{1}_{\{U\le a_n\}}\), where \(a_0=\tfrac14\), \(a_1=\tfrac12\), and \(a_2=\tfrac34\). Find the joint law at all three indices.

All three coordinates use the same \(U\). As \(U\) moves through the indicated intervals, the process vector takes these values:

$$ \begin{array}{c|c|c} \text{Range of }U &(X_0,X_1,X_2)&\text{Probability}\\ [0,\tfrac14] &(1,1,1)&\tfrac14\\ (\tfrac14,\tfrac12] &(0,1,1)&\tfrac14\\ (\tfrac12,\tfrac34] &(0,0,1)&\tfrac14\\ (\tfrac34,1] &(0,0,0)&\tfrac14 \end{array} $$

The listed ranges are disjoint and cover \([0,1]\); each has length \(\tfrac14\), hence probability \(\tfrac14\). The endpoint convention in \(X_n=\mathbf{1}_{\{U\le a_n\}}\) assigns each threshold to the interval on its left. The joint masses sum to \(4\cdot\tfrac14=1\).

As a check, the probability that \(X_0=1\) is the mass of \((1,1,1)\), namely \(\tfrac14\). The probability that \(X_1=1\) is the sum of the first two masses, \(\tfrac14+\tfrac14=\tfrac12\), and the probability that \(X_2=1\) is the sum of the first three masses, \(3\cdot\tfrac14=\tfrac34\). These agree with the threshold probabilities. Projecting onto indices \(0\) and \(2\), the masses are \(\tfrac14\) at \((1,1)\), \(\tfrac12\) at \((0,1)\), and \(\tfrac14\) at \((0,0)\); \((1,0)\) has probability zero.

Worked Example: A Process Specified by a Random Pattern

On a probability space let \(J\) take values \(1,2,3\) with probabilities \(\tfrac12,\tfrac13,\tfrac16\), respectively. Define a discrete-time process with index set \(T=\{0,1,2,3\}\) and state space \(S=\{0,1\}\) by the following paths: when \(J=1\), the path is \((0,0,1,1)\); when \(J=2\), it is \((0,1,0,1)\); and when \(J=3\), it is \((1,0,0,1)\). Find the finite-dimensional distribution at indices \(0,1,3\).

Evaluate the three selected coordinates for each possible value of \(J\). If \(J=1\), the vector is \((0,0,1)\), with probability \(\tfrac12\). If \(J=2\), it is \((0,1,1)\), with probability \(\tfrac13\). If \(J=3\), it is \((1,0,1)\), with probability \(\tfrac16\). These vectors are distinct, so no probabilities need to be combined. All other vectors in \(\{0,1\}^3\) have probability zero.

The three nonzero probabilities sum to \(\tfrac12+\tfrac13+\tfrac16=\tfrac36+\tfrac26+\tfrac16=1\). Projecting onto index \(0\), the probability of 1 is \(\tfrac16\); projecting onto index \(1\), it is \(\tfrac13\); and at index \(3\), it is 1. These also follow directly from the three possible paths: only the third has value 1 at index 0, only the second has value 1 at index 1, and all end in 1 at index 3.

Why Joint Probabilities Matter

The first example has independent selected coordinates, but that is not automatic for a process. In the threshold example, \(X_0=1\) implies \(X_2=1\), because \(U\le\tfrac14\) also implies \(U\le\tfrac34\). A joint law makes such dependence visible. It also supports calculations of events involving several times, such as the probability that a process visits a specified set of states at two selected indices.

A useful general result underlies the examples: probabilities of joint values are found by identifying the underlying outcomes that produce those values. For a discrete state space, this gives a point-mass formula.

Proposition (Joint point-mass formula): For a process with countable state space \(S\) and distinct indices \(t_1,\ldots,t_k\), the finite-dimensional distribution has point masses \[ p(x_1,\ldots,x_k)=\mathbb{P}(X_{t_1}=x_1,\ldots,X_{t_k}=x_k),\qquad (x_1,\ldots,x_k)\in S^k. \] The probability of any set \(B\subseteq S^k\) is the sum of these masses over the points in \(B\).

Proof. For each \(x=(x_1,\ldots,x_k)\in S^k\), let \(A_x=\{\omega:(X_{t_1}(\omega),\ldots,X_{t_k}(\omega))=x\}\). If \(x\ne y\), then at least one coordinate differs, so \(A_x\cap A_y=\varnothing\). The event that the vector belongs to \(B\) is exactly the union of \(A_x\) over \(x\in B\). The union is countable because \(S\) is countable and \(k\) is finite. Countable additivity therefore gives \(\mathbb{P}((X_{t_1},\ldots,X_{t_k})\in B)=\sum_{x\in B}\mathbb{P}(A_x)\), which is the stated formula. Taking \(B=S^k\) also shows that all point masses sum to 1. \(\square\)

In practice, the formula can be implemented by a table: list the possible underlying outcomes, record the process vector for the selected indices, and add probabilities whenever different underlying outcomes give the same vector. If the state space is countable, there may be infinitely many rows, but the same disjoint-event argument applies.

Common Mistake: Multiplying Marginals Without Independence

A frequent error is to compute a joint probability by multiplying two marginal probabilities without first establishing independence. For the threshold process above, the marginal probabilities are \(\mathbb{P}(X_0=1)=\tfrac14\) and \(\mathbb{P}(X_2=1)=\tfrac34\). The incorrect calculation would be \(\mathbb{P}(X_0=1,X_2=1)=\tfrac14\cdot\tfrac34=\tfrac{3}{16}\). But the joint event is exactly \(\{U\le\tfrac14\}\), so its probability is \(\tfrac14\), not \(\tfrac{3}{16}\). The product rule for joint probabilities requires independence of the two random variables; here they are dependent because \(X_0=1\) forces \(X_2=1\).

When computing a finite-dimensional distribution, keep the selected values together as a vector and work with the event that the whole vector takes each value. Then check that the probabilities sum to one and that summing over omitted coordinates recovers the appropriate lower-dimensional distributions. Those checks can expose a missing outcome, a duplicated case, or an incorrect independence assumption.

Check Your Understanding

For each question, identify the process values that form the joint vector before calculating probabilities.

  1. A process has state space \(\{0,1\}\) and three equally likely paths \((0,0,1)\), \((0,1,1)\), and \((1,1,0)\). What is its finite-dimensional distribution at indices \(0\) and \(1\)?
  2. In the threshold example, what is the probability of \((X_0,X_1)=(0,1)\)? Verify your answer by summing the appropriate three-coordinate masses.
  3. Why must the probabilities in a discrete finite-dimensional probability mass function sum to 1?
  4. If a joint law at three indices is known, how do you obtain the law at just the first and third indices?
  5. Give a reason that multiplying two marginal probabilities may give the wrong joint probability for process values at different indices.