From a Path-Space Measure to a Process
A probability measure on a space of paths gives a single object from which all the finite-dimensional laws of a process can be read. The points of the space are paths, while the coordinate maps observe the path at individual times. This viewpoint is useful both when a measure is constructed by the Kolmogorov Extension Theorem and when a process is first given on some other probability space and then represented by its paths.
Let \(T\) be an index set, let \((S,\mathcal{A})\) be a measurable state space, and write \(S^T\) for the set of all functions \(\omega:T\to S\). Equip \(S^T\) with the product sigma-algebra \(\mathcal{A}^{\otimes T}\), generated by sets that restrict finitely many coordinates to measurable subsets of \(S\). Suppose \(P\) is a probability measure on \((S^T,\mathcal{A}^{\otimes T})\). This is a finite, and therefore sigma-finite, measure space; it is not assumed complete. For each \(t\in T\), define the coordinate map \(X_t:S^T\to S\) by \(X_t(\omega)=\omega(t)\). The maps \(X_t\) are the coordinate process.
The product sigma-algebra is defined so that coordinate observations are measurable. Indeed, for \(A\in\mathcal{A}\), \(X_t^{-1}(A)=\{\omega:\omega(t)\in A\}\) is a generating measurable set. The same argument shows that every finite-coordinate map \(\pi_F\) is measurable: the inverse image of a measurable rectangle \(A_1\times\cdots\times A_k\) is \(\bigcap_{j=1}^k X_{t_j}^{-1}(A_j)\). These rectangles generate \(\mathcal{A}^{\otimes k}\).
Reading Finite-Dimensional Laws
Proof. The coordinate maps are measurable by the generating-set calculation above, so \(\pi_F\) is measurable. The defining property of the extended measure is \(P(\pi_F^{-1}(B))=\mu_F(B)\) for every \(B\in\mathcal{A}^{\otimes k}\). By the definition of pushforward measure, \(P\circ\pi_F^{-1}=\mu_F\), which is precisely the asserted finite-dimensional law. This conclusion identifies the law of each finite coordinate vector; it does not say that a finite-dimensional vector determines the entire path. \(\square\)
This proposition is often the practical use of the extension: a prescribed law for a finite list of times becomes the probability of the corresponding coordinate event under \(P\). In particular, for \(A_j\in\mathcal{A}\),
The finite-dimensional measure must be evaluated on the joint set, not inferred by multiplying marginal probabilities unless independence has separately been established.
Worked Example: A Four-Path Measure and an Eventually-One Event
Take \(T=\mathbb{N}\), \(S=\{0,1\}\), and \(\Omega=S^{\mathbb{N}}\) with its product sigma-algebra. Let \(P\) assign mass \(1/4\) to each of the four paths \(a=(0,0,0,\ldots)\), \(b=(1,1,1,\ldots)\), \(c=(0,1,0,1,\ldots)\), and \(d=(1,0,1,0,\ldots)\), and mass zero to the rest of \(\Omega\). This is a probability measure: the four distinct singleton paths have total mass \(4(1/4)=1\). For each \(n\), \(X_n(\omega)=\omega(n)\) is measurable.
For every \(N\ge1\), define \(E_N=\bigcap_{n=N}^{\infty}\{X_n=1\}\). These are measurable, since each coordinate event is measurable and sigma-algebras are closed under countable intersections. The sets increase: if \(\omega\in E_N\), then \(\omega(n)=1\) for all \(n\ge N\), hence also for all \(n\ge N+1\), so \(\omega\in E_{N+1}\). Among the four paths with positive mass, only \(b\) belongs to any \(E_N\): \(a\) is always zero, and each of \(c,d\) has infinitely many zero coordinates. Therefore \(P(E_N)=1/4\) for every \(N\).
The union \(E=\bigcup_{N=1}^{\infty}E_N\) is the event that the path is eventually always one. Continuity from below for measures gives \(P(E)=\lim_{N\to\infty}P(E_N)=1/4\). Check: directly among the four positive-mass paths, \(b\) is the only one eventually always one, so the probability is \(1\cdot(1/4)=1/4\), in agreement.
It is important that \(E_N\) is not the singleton \(\{b\}\) as a subset of \(\Omega\). For example, the path \((0,1,1,1,\ldots)\) is not \(b\), but belongs to \(E_2\). It has probability zero under this particular \(P\), so it does not alter the probability calculation. The event and the support of a measure are different objects.
For completeness, continuity from below used here is the following standard measure property: if \(A_1\subseteq A_2\subseteq\cdots\) are measurable in a measure space \((X,\mathcal{M},\mu)\), then \(\mu(\bigcup_n A_n)=\lim_n\mu(A_n)\), with infinity allowed. It follows by writing \(D_1=A_1\) and \(D_n=A_n\setminus A_{n-1}\) for \(n\ge2\). The \(D_n\) are disjoint, \(A_N=\bigcup_{n=1}^N D_n\), and \(\bigcup_n A_n=\bigcup_nD_n\). Countable additivity gives \(\mu(A_N)=\sum_{n=1}^N\mu(D_n)\) and \(\mu(\bigcup_nA_n)=\sum_{n=1}^\infty\mu(D_n)\), so the finite partial sums increase to the latter value.
Constructing a Path-Space Law from a Process
Proof. For any \(t\in T\) and \(A\in\mathcal{A}\), \(\Phi^{-1}(\{\eta\in S^T:\eta(t)\in A\})=\{\omega:X_t(\omega)\in A\}\in\mathcal{F}\), by measurability of \(X_t\). Finite intersections of these coordinate sets generate \(\mathcal{A}^{\otimes T}\). The collection of subsets of \(S^T\) whose inverse images under \(\Phi\) lie in \(\mathcal{F}\) is a sigma-algebra, so it contains the sigma-algebra generated by those sets. Thus \(\Phi\) is measurable. The pushforward \(P\) is a probability measure because \(P(S^T)=Q(\Omega_0)=1\). If \(\pi_F\) is a finite-coordinate map, then \(\pi_F\circ\Phi=(X_{t_1},\ldots,X_{t_k})\). Hence for each measurable \(B\subseteq S^k\), \(P(\pi_F^{-1}(B))=Q(\{\omega:(X_{t_1}(\omega),\ldots,X_{t_k}(\omega))\in B\})\), proving equality of the finite-dimensional laws. \(\square\)
Worked Example: The Law of a Constant-Path Process
Let \(\Omega_0=[0,1]\), let \(\mathcal{F}\) be the Lebesgue sigma-algebra, and let \(Q=m\) restricted to \([0,1]\), so \(Q(\Omega_0)=1\). Take \(S=[0,1]\) with its Borel sigma-algebra and \(T=\mathbb{N}\). Define \(X_n(x)=x\) for every \(n\), and map \(x\) to the constant path \(\Phi(x)=(x,x,\ldots)\). The coordinate functions are measurable, so the proposition makes \(P=Q\circ\Phi^{-1}\) a path-space probability measure.
For any two distinct indices \(i,j\), the event \(\{X_i\in[1/4,3/4],\,X_j\in[1/2,1]\}\) has probability equal to the Lebesgue measure of the intersection \([1/4,3/4]\cap[1/2,1]=[1/2,3/4]\). Its length is \(3/4-1/2=1/4\), so the probability is \(1/4\). This computation uses the diagonal nature of the paths: both coordinate restrictions apply to the same \(x\).
Check: the path-space event is \(\{\eta:\eta(i)\in[1/4,3/4],\,\eta(j)\in[1/2,1]\}\). Its preimage under \(\Phi\) is exactly \([1/2,3/4]\), whose \(Q\)-measure is \(1/4\). The direct preimage computation agrees with the finite-dimensional calculation.
Worked Example: A Random Alternating Path
Let \(\Omega_0=\{u,v\}\), let \(\mathcal{F}\) contain all subsets, and put \(Q(\{u\})=Q(\{v\})=1/2\). Define paths \(X(u)=(0,1,0,1,\ldots)\) and \(X(v)=(1,0,1,0,\ldots)\), and let \(P\) be the resulting path-space law. For the first three coordinates, the vector is \((0,1,0)\) with probability \(1/2\) and \((1,0,1)\) with probability \(1/2\). Therefore \(P(X_1=1,X_2=0,X_3=1)=1/2\), while \(P(X_1=X_2)=0\).
Each coordinate has marginal probability \(1/2\) of being one: at odd indices the path \(v\) has value one, and at even indices the path \(u\) has value one. Multiplying these two marginal probabilities would give \(1/4\), not the actual probability \(P(X_1=1,X_2=0)=1/2\). Thus the coordinates are dependent. Check: exactly one of the two equally likely paths begins \(1,0\), giving probability \(1/2\); exactly one also has \(X_1=1\), giving marginal probability \(1/2\).
Using Path-Space Laws Carefully
Finite-dimensional laws answer questions that involve finitely many coordinates. For a general event in \(\mathcal{A}^{\otimes T}\), its probability is assigned by the path-space measure, but it need not be a single finite-dimensional calculation. When \(T=\mathbb{N}\), events such as “eventually always one” can be written using countable unions and intersections of coordinate events, as in the first example. The probability of an increasing union can then be obtained by continuity from below.
A common error is to replace a measurable event by its intersection with the support and then assert that the two sets are equal. In a measure concentrated on finitely many paths, they may have the same probability without being the same subset of path space. The correct procedure is to define the event in the full product space, verify measurability there, and calculate its measure using the support only as a way to sum the positive masses. Another error is to multiply marginal probabilities without a proved independence assumption. Coordinate processes need not have independent coordinates; the alternating and constant-path examples show this directly.
Path-space laws make the distinction between finite observations and complete paths precise. The extension theorem supplies a measure whose coordinate vectors have the prescribed distributions; the coordinate maps then form a process on that space. Conversely, any measurable process induces a path-space measure by pushforward. In either direction, statements about probability must be tied to the measure and the event actually defined, not merely to the paths one happens to use when calculating.
Check Your Understanding
For each question, identify the measurable space and the event before computing its probability.
- For a probability measure \(P\) on \((S^T,\mathcal{A}^{\otimes T})\), write the finite-dimensional law of \((X_{t_1},X_{t_2})\) as a pushforward measure.
- In the four-path example, why does the path \((0,1,1,1,\ldots)\) belong to \(E_2\) even though it has probability zero?
- Show that if \(E_N=\bigcap_{n=N}^{\infty}\{X_n=1\}\), then \(E_N\subseteq E_{N+1}\).
- For the constant-path process, compute \(P(X_1\in[0,1/3],X_2\in[1/4,1/2])\).
- In the alternating-path example, compute \(P(X_2=1,X_3=0)\) and compare it with \(P(X_2=1)P(X_3=0)\). What does the comparison show?