An Independent Sequence Is a Process
A sequence of independent observations is already a stochastic process: its time index records the order of the observations, and each observation is a random variable on the same probability space. Thinking of the sequence this way makes its joint laws across time explicit. It also gives a direct route to a second kind of process: record at each time whether the observation falls above or below a chosen threshold.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. A real-valued discrete-time process is a family \((X_n)_{n\ge0}\) of random variables \(X_n:\Omega\to S\), where here the state space \(S\) is a subset of \(\mathbb{R}\) with its Borel sets. The process has time index \(n=0,1,2,\ldots\). The random variables are independent and identically distributed, or iid, if they have a common probability law \(\nu\) and, for every finite list of distinct times \(n_1,\ldots,n_k\) and Borel sets \(A_1,\ldots,A_k\subseteq S\),
The requirement that the times be distinct matters: independence applies to different random variables in the sequence. If a time is repeated, the corresponding event concerns the same random variable twice and must be combined as one event, not treated as an independent observation.
Finite-Dimensional Distributions of an Iid Process
The product formula is useful both for computing particular events and for identifying the entire joint distribution. It applies whether the state space is finite, countable, or a subset of the real line. In the discrete case, for example, the probability of a specified list of values is the product of their single-time probabilities. For real-valued observations, the same rule applies to events such as intervals or threshold ranges.
Proof. The definition of an iid sequence gives independence of the random variables at every finite collection of distinct times. By the definition of independence, the probability of the intersection of the events \(\{X_{n_\ell}\in A_\ell\}\) is the product of their individual probabilities. Identical distribution means \(\mathbb{P}(X_{n_\ell}\in A_\ell)=\nu(A_\ell)\) for each \(\ell\). Substitution gives the displayed formula. Since its right-hand side uses \(\nu\) and \(k\), but not the values of the distinct indices, the joint law is the same for every such selection of times. \(\square\)
For a real-valued iid process, a common special case uses cumulative distribution functions. If \(F(x)=\mathbb{P}(X_n\le x)\), then for distinct times \(n_1,\ldots,n_k\),
This is one way to read off joint probabilities from the single-coordinate distribution. The full product-law statement is stronger: it also covers sets that are not of the form \((-\infty,x]\).
Worked Example: Three Coordinates of an Iid Bernoulli Process
On a probability space \((\Omega,\mathcal{F},\mathbb{P})\), let \((X_n)_{n\ge0}\) be iid with state space \(S=\{0,1\}\) and \(\mathbb{P}(X_n=1)=\tfrac13\), \(\mathbb{P}(X_n=0)=\tfrac23\). Find \(\mathbb{P}(X_0=1,X_2=0,X_5=1)\).
The times \(0,2,5\) are distinct, so the three coordinates are independent. Their specified values have probabilities \(\tfrac13,\tfrac23,\tfrac13\), respectively. Therefore
Check: Summing over the two possible values of \(X_2\) gives \(\mathbb{P}(X_0=1,X_5=1)=\tfrac13(\tfrac23+\tfrac13)\tfrac13=\tfrac19\), which is also the product of the two endpoint probabilities. Summing the probabilities of all eight patterns at times \(0,2,5\) gives \((\tfrac13+\tfrac23)^3=1\), as required for a joint distribution.
Thresholding Each Coordinate
Now suppose the state space is real-valued and choose a fixed threshold \(a\in\mathbb{R}\). Define a new process by asking the same yes-or-no question of every coordinate. Using the strict threshold convention, set \(Y_n=\mathbf{1}_{\{X_n>a\}}\). Then the state space of \((Y_n)_{n\ge0}\) is \(\{0,1\}\): value \(1\) records that \(X_n>a\), and value \(0\) records that \(X_n\le a\). Its success probability is \(p=\mathbb{P}(X_n>a)=\nu((a,\infty))\), which is the same at every time.
Proof. The map \(g:S\to\{0,1\}\) defined by \(g(x)=\mathbf{1}_{\{x>a\}}\) is measurable, and \(Y_n=g(X_n)\). For a finite set of distinct times \(n_1,\ldots,n_k\) and a specified pattern \(b_1,\ldots,b_k\in\{0,1\}\), define \(C_\ell=(a,\infty)\cap S\) if \(b_\ell=1\), and \(C_\ell=(-\infty,a]\cap S\) if \(b_\ell=0\). Then \(\{Y_{n_\ell}=b_\ell\}=\{X_{n_\ell}\in C_\ell\}\). Independence and identical distribution of the \(X_n\) give \(\mathbb{P}(Y_{n_1}=b_1,\ldots,Y_{n_k}=b_k)=\prod_{\ell=1}^k\nu(C_\ell)\). Each factor is \(p\) when \(b_\ell=1\) and \(1-p\) when \(b_\ell=0\). Thus the pattern probability factors into the same one-time probabilities at every selected time. Since every event involving a finite collection of \(\{0,1\}\)-valued coordinates is a union of disjoint patterns, this factorization gives independence for the coordinates. The identical one-time probabilities give identical distribution. The argument remains valid when a factor is zero, so it also covers \(p=0\) and \(p=1\). \(\square\)
Worked Example: Thresholding Repeated Die Rolls
On a probability space \((\Omega,\mathcal{F},\mathbb{P})\), let \((X_n)_{n\ge0}\) be iid rolls of a fair six-sided die, with state space \(S=\{1,2,3,4,5,6\}\). Define \(Y_n=\mathbf{1}_{\{X_n>4\}}\). Compute the probability of the pattern \((Y_0,Y_1,Y_2,Y_3)=(0,1,0,1)\).
The event \(Y_n=1\) means \(X_n\in\{5,6\}\), so its probability is \(2/6=1/3\). The event \(Y_n=0\) means \(X_n\in\{1,2,3,4\}\), so its probability is \(4/6=2/3\). Since the four times are distinct, independence gives
Check: There are \(2^4=16\) possible binary patterns at these four times. Summing their probabilities gives \((\tfrac23+\tfrac13)^4=1\). Also, summing over the two possible values of \(Y_3\) in the specified first-three-coordinate pattern gives \(\tfrac23\cdot\tfrac13\cdot\tfrac23(\tfrac23+\tfrac13)=\tfrac4{27}\), agreeing with the three-coordinate product.
Worked Example: A Thresholded Uniform Process
On \((\Omega,\mathcal{F},\mathbb{P})\), let \((X_n)_{n\ge0}\) be iid with the uniform distribution on state space \(S=(0,1)\). Set \(Y_n=\mathbf{1}_{\{X_n>3/4\}}\). Find the probability of the pattern \((Y_1,Y_4,Y_7)=(1,1,0)\).
For a uniform random variable on \((0,1)\), the interval \((3/4,1)\) has length \(1-3/4=1/4\), so \(p=\mathbb{P}(Y_n=1)=1/4\). The complementary event \(X_n\le3/4\) has probability \(3/4\). The three selected indices are distinct, and therefore
Check: Summing over both possible values at time \(7\) gives \(\tfrac14\cdot\tfrac14(\tfrac34+\tfrac14)=\tfrac1{16}\), the probability that \(Y_1=Y_4=1\). The threshold uses \(>\), so \(X_n=3/4\) belongs to the zero event; for this continuous distribution that single value has probability \(0\), but the strictness convention is still part of the definition.
What Thresholding Preserves—and a Common Mistake
Thresholding discards information about the precise value of each \(X_n\), but it preserves the independence across distinct times and the equality of the one-time distributions. The resulting process is therefore described completely by a single number \(p\): for any \(k\) distinct times and any binary pattern containing \(r\) ones and \(k-r\) zeros, its probability is \(p^r(1-p)^{k-r}\). This remains valid at \(p=0\) or \(p=1\), with the direct product of the relevant one-time probabilities being the safest interpretation.
Common pitfall: applying independence to a repeated time. In the Bernoulli example above, \(X_0\) has probability \(\tfrac13\) of being \(1\). A wrong calculation might treat the two statements \(X_0=1\) and \(X_0=1\) as independent and assign their intersection probability \(\tfrac13\cdot\tfrac13=\tfrac19\). They are the same event, however, so the correct calculation is \(\mathbb{P}(X_0=1,X_0=1)=\mathbb{P}(X_0=1)=\tfrac13\). The product rule applies to distinct coordinates, not duplicated descriptions of one coordinate.
These calculations are useful whenever a process records whether observations cross a boundary, satisfy a diagnostic cutoff, or meet a binary criterion. The key steps are to identify the one-time event, compute its probability under the common law, and then use independence only across distinct times. The joint pattern formula gives a complete finite-dimensional description of the resulting indicator process.
Check Your Understanding
Use the product rule only after identifying the distinct time indices and the one-time probabilities.
- An iid process on \(\{0,1\}\) has \(\mathbb{P}(X_n=1)=\tfrac14\). What is \(\mathbb{P}(X_2=1,X_6=0,X_9=0)\)?
- If \(Y_n=\mathbf{1}_{\{X_n>a\}}\), write the probabilities of \(Y_n=0\) and \(Y_n=1\) in terms of the common law \(\nu\).
- For iid fair die rolls, let \(Y_n=\mathbf{1}_{\{X_n\ge5\}}\). Find the probability of the pattern \((Y_0,Y_1,Y_2)=(1,0,1)\).
- Why does the formula for a specified binary pattern still establish independence when an event allows either value at one of the selected times?
- For any random variable \(X_0\), compare \(\mathbb{P}(X_0\in A,X_0\in B)\) with the product \(\mathbb{P}(X_0\in A)\mathbb{P}(X_0\in B)\). When is it valid to use the product rule?