From a Random Variable to a Measure on Its Values
A random variable is a measurable map from a probability space into a space of possible values. Its law answers a specific question: for a Borel set of values, what is the probability that the random variable lands in that set? The answer is obtained by taking the preimage in the original sample space and measuring that event. This construction turns the probabilities of events into a probability measure on the value space.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space: \(\mathcal{F}\) is a sigma-algebra on \(\Omega\), and \(\mathbb{P}\) is a measure with \(\mathbb{P}(\Omega)=1\). It is finite, and hence sigma-finite; completeness is not assumed. Write \(\mathcal{B}_{\mathbb{R}}\) for the Borel sigma-algebra on \(\mathbb{R}\). A real-valued random variable \(X\) is an \(\mathcal{F}/\mathcal{B}_{\mathbb{R}}\)-measurable function \(X:\Omega\to\mathbb{R}\). Measurability means that \(X^{-1}(B)=\{\omega\in\Omega:X(\omega)\in B\}\) belongs to \(\mathcal{F}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\).
The definition uses Borel sets in the value space and measurable preimages in the sample space. It does not require that a set of values itself be an event in \(\mathcal{F}\). In particular, \(\mathbb{P}_X\) is a measure on values, while \(\mathbb{P}\) is a measure on outcomes.
Why the Push-Forward Is a Probability Measure
Proof. For every Borel set \(B\), measurability of \(X\) gives \(X^{-1}(B)\in\mathcal{F}\), so \(\mathbb{P}_X(B)\) is defined. The preimage of the empty set is empty, and therefore \(\mathbb{P}_X(\varnothing)=\mathbb{P}(\varnothing)=0\). The preimage of \(\mathbb{R}\) is all of \(\Omega\), so \(\mathbb{P}_X(\mathbb{R})=\mathbb{P}(\Omega)=1\). Now let \((B_j)_{j\ge1}\) be pairwise disjoint Borel sets. Their preimages are pairwise disjoint: if an outcome belonged to both \(X^{-1}(B_i)\) and \(X^{-1}(B_j)\), its value under \(X\) would lie in \(B_i\cap B_j=\varnothing\). Also, preimages preserve unions, so \(X^{-1}(\bigcup_{j\ge1}B_j)=\bigcup_{j\ge1}X^{-1}(B_j)\). Countable additivity of \(\mathbb{P}\) now gives \(\mathbb{P}_X(\bigcup_{j\ge1}B_j)=\mathbb{P}(\bigcup_{j\ge1}X^{-1}(B_j))=\sum_{j\ge1}\mathbb{P}(X^{-1}(B_j))=\sum_{j\ge1}\mathbb{P}_X(B_j)\). Thus \(\mathbb{P}_X\) is countably additive, nonnegative, and has total mass one. \(\square\)
This proof uses only measurability, the probability axioms, and elementary properties of preimages. Although the source probability space is not assumed complete, the law is still a probability measure on the Borel sigma-algebra; it is not asserted to be complete.
Worked Example: A Finite-Valued Random Variable
Let \(\Omega=\{a,b,c,d\}\), let \(\mathcal{F}\) be its power set, and assign probabilities \(\mathbb{P}(\{a\})=1/10\), \(\mathbb{P}(\{b\})=2/10\), \(\mathbb{P}(\{c\})=3/10\), and \(\mathbb{P}(\{d\})=4/10\). This is a finite, complete probability space: the four point probabilities sum to \((1+2+3+4)/10=1\), and every subset is measurable. Define \(X(a)=0\), \(X(b)=1\), \(X(c)=1\), and \(X(d)=3\). Since every subset of \(\Omega\) is measurable, \(X\) is a real-valued random variable.
For any Borel set \(B\), its preimage consists exactly of the outcomes whose assigned values belong to \(B\). Thus the law places mass \(1/10\) at \(0\), mass \(2/10+3/10=5/10\) at \(1\), and mass \(4/10\) at \(3\). For example, with \(B=\{1,3\}\), the preimage is \(\{b,c,d\}\), and
Check: the complementary value set \(\mathbb{R}\setminus\{1,3\}\) has preimage \(\{a\}\), whose probability is \(1/10\). The two law probabilities add to \(9/10+1/10=1\), as required.
The Law of a Discrete Random Variable
Proof. The events \(\{X=x_i\}\) are measurable because \(\{x_i\}\) is Borel and \(X\) is measurable. They are pairwise disjoint because the values \(x_i\) are distinct. Their union has probability one by the assumption that \(X\) takes values in the listed set almost surely. Countable additivity therefore gives \(\sum_{i\in I}p_i=1\). For a Borel set \(B\), the event \(\{X\in B\}\), apart from a possible null subset of outcomes where \(X\) is outside the listed set, is the disjoint union of \(\{X=x_i\}\) over those indices with \(x_i\in B\). The omitted subset is null, so countable additivity gives the stated formula. If the list is finite, the same argument is finite additivity, which follows from countable additivity. \(\square\)
Worked Example: A Countable Discrete Law
Take \(\Omega=\mathbb{N}=\{1,2,\ldots\}\), with \(\mathcal{F}\) its power set and \(\mathbb{P}(\{n\})=2^{-n}\). This is a probability space with counting support: the geometric sum is \(\sum_{n=1}^{\infty}2^{-n}=(1/2)/(1-1/2)=1\). It is finite and sigma-finite, and it is complete because every nonempty subset has positive probability, so the only null set is empty. Let \(X(n)=n\). For the Borel set \(B=\{1,3,5,\ldots\}\), the preimage is the odd positive integers, so
Here the series is geometric with first term \(2^{-1}=1/2\) and ratio \(2^{-2}=1/4\), whose absolute value is less than one. Check: the complementary even values have probability \(\sum_{k=1}^{\infty}2^{-2k}=(1/4)/(1-1/4)=1/3\). The two probabilities sum to \(2/3+1/3=1\).
Interval-Valued Random Variables
A law need not be concentrated on finitely or countably many values. Consider the probability space \(([0,1],\mathcal{L},m)\), where \(\mathcal{L}\) is the Lebesgue sigma-algebra restricted to \([0,1]\) and \(m\) is Lebesgue measure. This space is complete and has total measure \(m([0,1])=1\), so it is finite and sigma-finite. The coordinate map \(U(\omega)=\omega\) is a real-valued random variable: for each Borel \(B\subseteq\mathbb{R}\), its preimage in \([0,1]\) is \(B\cap[0,1]\), which is Lebesgue measurable.
Proof. For a Borel set \(B\), the definition of \(U\) gives \(U^{-1}(B)=\{\omega\in[0,1]:\omega\in B\}=B\cap[0,1]\). Therefore \(\mathbb{P}_U(B)=m(U^{-1}(B))=m(B\cap[0,1])\). Taking \(B=\mathbb{R}\) yields \(m([0,1])=1\), consistent with the probability-measure proposition. \(\square\)
Worked Example: Measuring a Borel Set Under the Uniform Law
Use the preceding coordinate random variable and take \(B=[1/4,3/5)\cup\{1\}\), a Borel set. Its preimage is exactly \([1/4,3/5)\cup\{1\}\) inside \([0,1]\). The singleton has Lebesgue measure zero, and the interval has length \(3/5-1/4=12/20-5/20=7/20\). Hence
Check: within \([0,1]\), the complement of \(B\) consists of \([0,1/4)\), \([3/5,1)\), and the point \(1\) is excluded from that second interval but belongs to \(B\). Their measures are \(1/4=5/20\), \(1-3/5=2/5=8/20\), and \(0\), respectively. The complement therefore has measure \(13/20\), and \(7/20+13/20=1\).
Two Useful Consequences and a Common Pitfall
The law depends on the probabilities of the value events, not on the names of the outcomes in \(\Omega\). In the finite example, changing the labels \(a,b,c,d\) while preserving their probabilities and the values assigned to them leaves \(\mathbb{P}_X\) unchanged. More generally, an almost-sure change in a random variable leaves its law unchanged.
Proof. Since \(X\) and \(Y\) are measurable, the set \(N=\{\omega:X(\omega)\ne Y(\omega)\}\) is measurable: it is the inverse image of \(\mathbb{R}\setminus\{0\}\) under the measurable map \(X-Y\). The almost-everywhere hypothesis means \(\mathbb{P}(N)=0\). For any Borel set \(B\), the indicators of the events \(\{X\in B\}\) and \(\{Y\in B\}\) agree on \(N^c\), because \(X(\omega)=Y(\omega)\) there. Thus their symmetric difference is contained in \(N\). The two events consequently have the same probability: each differs from their intersection only by a subset of the null set \(N\). Hence \(\mathbb{P}_X(B)=\mathbb{P}(X\in B)=\mathbb{P}(Y\in B)=\mathbb{P}_Y(B)\) for every Borel \(B\), proving equality of the measures. \(\square\)
Common pitfall: confusing a set of values with its preimage. In the finite example, the Borel set \(B=\{1\}\) contains one number, but the event \(X\in B\) contains two outcomes, \(b\) and \(c\). Its probability is therefore \(\mathbb{P}_X(\{1\})=\mathbb{P}(\{b,c\})=2/10+3/10=5/10\), not \(1\), and not the probability of a set of outcomes called \(\{1\}\). The correct operation is always to form \(X^{-1}(B)\) first and then apply \(\mathbb{P}\).
The push-forward construction is useful because it separates the measurable values of a random variable from the particular probability space used to produce them. Once \(\mathbb{P}_X\) is known, probabilities of Borel sets of values are computed as \(\mathbb{P}_X(B)\). For discrete variables this becomes a sum of point probabilities; for the coordinate variable on an interval it becomes the Lebesgue measure of the corresponding preimage. In both cases, the measure-theoretic definition is the same.
Check Your Understanding
In each question, identify the Borel set of values and take its preimage before calculating its probability.
- On a finite probability space, \(X\) takes values \(0,2,2,5\) at four equally likely outcomes. Find \(\mathbb{P}_X(\{2,5\})\).
- For the countable law \(\mathbb{P}(\{n\})=2^{-n}\) and \(X(n)=n\), find the law probability of the set of positive integers divisible by \(3\).
- For \(U(\omega)=\omega\) on \(([0,1],\mathcal{L},m)\), compute \(\mathbb{P}(U\in[1/3,5/6])\). Explain why including both endpoints does not change the value.
- Give the preimage \(X^{-1}(B)\) in the finite-valued worked example when \(B=\{0,3\}\), and compute \(\mathbb{P}_X(B)\).
- Explain why two real-valued random variables that are equal almost everywhere have the same probability for every Borel set of values.