A stochastic process is a family, not one noisy curve
Objective: What is the difference between a sample path and the random variable X_t?
A weather trace, queue length, share price, or particle position changes with time, but before observation its future is not one fixed curve. A stochastic process packages every possible time trace together with probabilities. Looking across time on one outcome gives a sample path; fixing a time and looking across all outcomes gives an ordinary random variable.
Write X_t(ω) with two inputs: t selects the time and ω selects the realized outcome. Finite-dimensional distributions describe the joint law at finitely many chosen times; they are stronger than separate one-time histograms because they retain dependence across time. Name the probability space, the information currently available, the time index, and every integrability or regularity assumption before manipulating a formula. A stochastic statement is about a family of random outcomes, so one simulated path can illustrate the statement but cannot prove its distribution, expectation, convergence, or error.
The first boundary to test is: Matching every one-time distribution does not determine temporal dependence: independent coin tosses and one coin copied forever can each be Bernoulli(1/2) at every time. This is not a footnote. It identifies the earliest assumption whose loss can change the answer, make a conditional quantity undefined, or turn a finite-time calculation into a false long-run claim.
The joint law of (X_t1,…,X_tn) determines the probability of every event involving only those selected times.
An event involving only t1,…,tn is the inverse image of a measurable set B under the random vector (X_t1,…,X_tn).
The joint law assigns that event probability P((X_t1,…,X_tn)∈B).
Therefore every finite-time probability question is answered by the corresponding finite-dimensional distribution.
Let X_n be independent fair coin indicators. Find P(X_1=1,X_2=0) and compare it with a process Y_n=Y_1 for all n where Y_1 is fair.
- Independence gives P(X_1=1,X_2=0)=P(X_1=1)P(X_2=0)=1/4.
- For Y, the event Y_1=1 forces Y_2=1 because the same value is copied.
- Hence P(Y_1=1,Y_2=0)=0 although X_1,X_2,Y_1,Y_2 all have the same one-time Bernoulli law.
Result: The independent process gives 1/4; the copied process gives 0, proving that marginal laws do not determine a process.