From Outcomes to Numbers
A probability model describes possible outcomes and assigns probabilities to events. A random variable adds a numerical description: it assigns a real number to each outcome. The number might represent a count, a payment, a waiting time, or another quantity of interest. The essential point is that a random variable is a function defined on the sample space, not a random choice made independently of the outcomes.
To calculate a probability such as “the random variable is at most 2,” we must identify which original outcomes produce values at most 2. That set of outcomes is called a preimage. It is the bridge between a statement about numerical values and an event to which the probability model can assign a probability.
For a set \(B\) of numerical values, \(X^{-1}(B)\) collects every outcome whose assigned number lies in \(B\). The notation \(X^{-1}(B)\) means a preimage; it does not mean that we calculate a reciprocal of \(X\). In particular, the event that \(X\) is at most a real number \(a\) is
The set \((-\infty,a]\) describes numbers, while \(X^{-1}((-\infty,a])\) describes outcomes. Once the preimage is identified, its probability is well-defined by measurability:
This notation is shorthand for the probability of the event on the right. It does not require \(X\) to take different values on different outcomes: several outcomes can produce the same number, and all of them belong to the preimage when that number is in the specified set.
Why the Preimage Gives the Right Event
The definition follows directly from what it means for an outcome to satisfy a condition on \(X\). An outcome \(\omega\) belongs to the event \(X\le a\) exactly when \(X(\omega)\le a\). The latter condition says precisely that the numerical value \(X(\omega)\) belongs to \((-\infty,a]\). Thus the two descriptions select the same outcomes.
Because these sets of outcomes are equal, the probability model assigns them the same probability. More generally, for any Borel set \(B\), the event “\(X\) takes a value in \(B\)” is \(X^{-1}(B)\), so \(P(X\in B)\) means \(P(X^{-1}(B))\). This is a direct consequence of the definition of a function and the probability assigned to events.
For example, “the payment is at most 2,000” becomes \(X\le 2000\).
Apply the function \(X\) to each outcome and determine whether its value satisfies the condition.
Collect exactly the outcomes whose assigned values satisfy the condition.
Calculate the probability of that event, respecting which subsets are measurable.
Worked Applications
Worked Example: Counting Failed Components
A hypothetical system has two components, A and B. Their operating results are independent, with probabilities \(0.9\) and \(0.8\) of functioning, respectively. Let \(F\) mean failed and \(W\) mean working. The sample space is \(\Omega=\{(W,W),(W,F),(F,W),(F,F)\}\), and all subsets are measurable. Define \(X\) as the number of failed components. Find \(P(X\le 1)\).
Method: define the function, take the preimage, then calculate its probability. The function assigns values as follows: \(X(W,W)=0\), \(X(W,F)=1\), \(X(F,W)=1\), and \(X(F,F)=2\).
The event \(X\le1\) therefore has preimage \(\{(W,W),(W,F),(F,W)\}\). Its complement is the outcome that both components fail. Independence gives
Hence \(P(X\le1)=1-0.02=0.98\). As a check, the three included outcome probabilities are \(0.9(0.8)=0.72\), \(0.9(0.2)=0.18\), and \(0.1(0.8)=0.08\); their sum is \(0.72+0.18+0.08=0.98\).
Common mistake avoided: The condition \(X\le1\) does not mean that exactly one component works. It includes both the outcome with zero failures and the two outcomes with one failure.
Worked Example: A Deductible Payment as a Function
For a hypothetical policy, the loss \(L\) is 0 dollars with probability \(0.40\), 4,000 dollars with probability \(0.35\), or 9,000 dollars with probability \(0.25\). The policy has a 3,000-dollar deductible, no limit, and pays \(\max(L-3000,0)\). Let \(X\) denote the payment, in dollars. Find the probability that the payment is at most 2,000 dollars.
Method: apply the payment rule to each loss outcome before forming the event. The possible loss outcomes are \(\Omega=\{0,4000,9000\}\), with all subsets measurable. The function values are \(X(0)=\max(0-3000,0)=0\), \(X(4000)=\max(4000-3000,0)=1000\), and \(X(9000)=\max(9000-3000,0)=6000\).
Now test the condition \(X\le2000\) on each outcome. The first two payments, 0 and 1,000, satisfy it; 6,000 does not. Thus the preimage is \(\{0,4000\}\), where the entries name loss outcomes, not payment amounts. Therefore,
The remaining outcome has probability \(0.25\), so \(0.75+0.25=1.00\), a check that the outcomes have been accounted for. Common mistake avoided: applying the condition to the gross loss would give a different event. The question concerns the payment after the deductible, so first evaluate the function \(X\).
Finite Spaces and Measurable Events
If every singleton \(\{\omega\}\) is an event, as happens when \(\mathcal{F}\) is the power set of a finite sample space, probabilities of distinct outcomes can be added to find the probability of a preimage. In that setting,
That formula requires the singleton events in the sum to be measurable. A finite sample space alone does not guarantee this. In a general finite probability space, the measurable events are unions of atoms, where an atom is a nonempty measurable event containing no smaller nonempty measurable event. A preimage is measurable, so it is a union of such atoms; its probability can be found by adding the probabilities of the atoms it contains. We cannot necessarily assign or add probabilities to individual outcomes inside an atom.
Worked Example: A Coarse Probability Model
Let \(\Omega=\{a,b,c,d\}\), but suppose the only measurable events are unions of the two sets \(A=\{a,b\}\) and \(C=\{c,d\}\). In particular, the measurable events are \(\varnothing,A,C,\Omega\). Let \(P(A)=0.35\) and \(P(C)=0.65\). Define \(X(a)=X(b)=1\) and \(X(c)=X(d)=4\). Find \(P(X\le2)\), and explain why individual-outcome probabilities should not be used.
Method: identify the preimage as a measurable union of atoms. The outcomes \(a\) and \(b\) both receive value 1, which is at most 2. The outcomes \(c\) and \(d\) receive value 4, which is greater than 2. Hence
The result is well-defined because \(A\) is measurable. The individual sets \(\{a\}\) and \(\{b\}\) are not measurable in this model, so \(P(\{a\})\) and \(P(\{b\})\) are not defined by \(P\). It would therefore be invalid to write \(P(\{a\})+P(\{b\})\), even though their union \(A\) has a specified probability. The atoms \(A\) and \(C\) partition the sample space, and their probabilities check: \(0.35+0.65=1\).
Common mistake avoided: treating each listed outcome as if it automatically had its own probability. Measurability, not merely being named in \(\Omega\), determines which sets have probabilities.
Common Exam Traps and Why the Idea Matters
A numerical rule is useful only when its connection to outcomes is handled correctly. In actuarial work, a random variable might represent a claim count or a policy payment; an event about that number must still be translated back to the outcomes of the model before its probability is evaluated. This habit also helps prevent errors when different outcomes lead to the same recorded value.
- Confusing values with outcomes. The statement \(X\le a\) is about the numbers assigned by \(X\); the event used in the sample space consists of outcomes \(\omega\) satisfying \(X(\omega)\le a\).
- Assuming one value identifies one outcome. If several outcomes map to the same number, the preimage includes all of them.
- Using a nonmeasurable set. A probability is assigned to events in \(\mathcal{F}\), not automatically to every subset of \(\Omega\). The random-variable measurability condition ensures that preimages of Borel sets are events.
- Adding singleton probabilities without checking. This is valid when the relevant singleton sets are measurable, such as on a finite space with the power-set sigma-algebra. It is not valid for every finite probability space.
Check Your Understanding
For each problem, define the relevant preimage before calculating its probability.
- A hypothetical inspection selects one of three machine states: normal, warning, or shutdown, with probabilities \(0.70\), \(0.20\), and \(0.10\). A score \(X\) is 0, 1, or 3, respectively. Find \(P(X\le1)\). (Answer: 0.90)
- A hypothetical two-part device has independent parts with failure probabilities \(0.05\) and \(0.10\). Let \(X\) be the number of failed parts. Find \(P(X=0)\) by expressing the event in terms of outcomes. (Answer: 0.8550)
- A hypothetical claim has loss 0, 2,000, or 8,000 dollars with probabilities \(0.50\), \(0.30\), and \(0.20\). A policy pays \(\max(L-1{,}000,0)\). Let \(X\) be the payment. Find \(P(X<1{,}000)\). (Answer: 0.50)
- On \(\Omega=\{r,s,t,u\}\), the measurable events are unions of \(A=\{r,s\}\) and \(C=\{t,u\}\), with \(P(A)=0.40\). Define \(X(r)=X(s)=2\) and \(X(t)=X(u)=5\). Find \(P(X>3)\), and state whether \(P(\{t\})\) is defined. (Answer: 0.60; \(P(\{t\})\) is not defined.)
- A finite model has all singleton outcomes measurable. Four outcomes have probabilities \(0.10,0.25,0.30,0.35\), and a function \(X\) assigns values \(0,2,2,5\) to them in that order. Find the preimage and probability of \(X\le2\). (Answer: the first three outcomes; 0.65)