From Integrals on Events to a Density
Conditioning on a sub-sigma-algebra means retaining only the information recorded by that sigma-algebra. An integrable random variable \(X\) may depend on more information than this, so we seek a \(\mathcal{G}\)-measurable random variable whose integrals agree with those of \(X\) on every event in \(\mathcal{G}\). The Radon–Nikodym theorem gives a systematic construction: first encode those event integrals as a signed measure, then take its density with respect to probability restricted to \(\mathcal{G}\).
Throughout, \((\Omega,\mathcal{F},\mathbb{P})\) is a probability space, and \(\mathcal{G}\subseteq\mathcal{F}\) is a sub-sigma-algebra. No completeness assumption is required. The measure space \((\Omega,\mathcal{G},\mathbb{P}|_{\mathcal{G}})\), where \(\mathbb{P}|_{\mathcal{G}}(A)=\mathbb{P}(A)\) for \(A\in\mathcal{G}\), is finite because its total measure is \(1\). Let \(X:\Omega\to\mathbb{R}\) be an \(\mathcal{F}\)-measurable, integrable random variable; integrability means \(\mathbb{E}|X|<\infty\).
The definition asks for agreement on all events visible to \(\mathcal{G}\), not pointwise agreement with \(X\). To construct such a \(Y\), define a set function on \(\mathcal{G}\) by $$ \nu(A)=\int_A X\,d\mathbb{P},\qquad A\in\mathcal{G}. $$ Because \(X\) may have either sign, \(\nu\) is a signed measure rather than necessarily a probability measure. The construction rests on showing that it is finite and absolutely continuous with respect to \(\mathbb{P}|_{\mathcal{G}}\).
The Induced Signed Measure
Proof. For each \(A\in\mathcal{G}\), the integral \(\int_A X\,d\mathbb{P}\) is finite since \(\int_A|X|\,d\mathbb{P}\le\mathbb{E}|X|<\infty\). Also, if \(\mathbb{P}(A)=0\), then \(\int_A|X|\,d\mathbb{P}=0\), so \(\nu(A)=0\). This proves absolute continuity.
To check countable additivity, let \(A_1,A_2,\ldots\) be pairwise disjoint sets in \(\mathcal{G}\), and put \(A=\bigcup_{k=1}^{\infty}A_k\). The partial sums \(X\sum_{k=1}^N\chi_{A_k}\) converge pointwise to \(X\chi_A\), and their absolute values are bounded by \(|X|\). Since \(|X|\) is integrable, the Dominated Convergence Theorem applies and gives $$ \nu(A)=\lim_{N\to\infty}\int X\sum_{k=1}^N\chi_{A_k}\,d\mathbb{P} =\lim_{N\to\infty}\sum_{k=1}^N\nu(A_k). $$ Thus \(\nu\) is countably additive. Finally, for any finite measurable partition \(A_1,\ldots,A_r\) of \(\Omega\) by sets in \(\mathcal{G}\), $$ \sum_{j=1}^r|\nu(A_j)| \le\sum_{j=1}^r\int_{A_j}|X|\,d\mathbb{P} =\mathbb{E}|X|. $$ Taking the supremum over such partitions shows that the total variation of \(\nu\) is finite and at most \(\mathbb{E}|X|\). \(\square\)
Construction by the Radon–Nikodym Theorem
Proof. Use the induced signed measure \(\nu\) from the proposition on the finite measure space \((\Omega,\mathcal{G},\mathbb{P}|_{\mathcal{G}})\). By the signed Radon–Nikodym theorem, a finite signed measure absolutely continuous with respect to a finite measure has an integrable measurable density. Thus there is a \(\mathcal{G}\)-measurable integrable function \(Y\) such that, for every \(A\in\mathcal{G}\), $$ \nu(A)=\int_A Y\,d(\mathbb{P}|_{\mathcal{G}}). $$ The restricted measure agrees with \(\mathbb{P}\) on \(\mathcal{G}\), while \(\nu(A)=\int_A X\,d\mathbb{P}\) by definition. Therefore \(\int_A Y\,d\mathbb{P}=\int_A X\,d\mathbb{P}\) for every \(A\in\mathcal{G}\), as required. The density is finite almost everywhere; if necessary, set it to zero on the \(\mathcal{G}\)-measurable null set where it is not finite, obtaining a real-valued version. \(\square\)
The Radon–Nikodym theorem used here is the standard theorem for finite signed measures. The construction does not require \(X\) to be nonnegative: the signed-measure formulation handles positive and negative values together. Taking \(A=\Omega\) in the defining identity also gives \(\mathbb{E}[Y]=\mathbb{E}[X]\).
Worked Example: Conditioning a Four-Outcome Variable on Parity
Let \(\Omega=\{1,2,3,4\}\), let \(\mathcal{F}\) be its power set, and give each outcome probability \(1/4\). Let \(X(i)=i\), and let \(\mathcal{G}=\{\varnothing,\{1,3\},\{2,4\},\Omega\}\), the sigma-algebra generated by parity. Compute the Radon–Nikodym density of \(\nu(A)=\mathbb{E}[X\chi_A]\).
For the odd event \(O=\{1,3\}\), $$ \nu(O)=\frac14(1)+\frac14(3)=1,\qquad \mathbb{P}(O)=\frac24=\frac12. $$ Thus the density on \(O\) is \(\nu(O)/\mathbb{P}(O)=1/(1/2)=2\). For the even event \(E=\{2,4\}\), $$ \nu(E)=\frac14(2)+\frac14(4)=\frac32,\qquad \mathbb{P}(E)=\frac24=\frac12, $$ so the density on \(E\) is \((3/2)/(1/2)=3\). Hence define \(Y=2\) on \(O\) and \(Y=3\) on \(E\). These four sets exhaust \(\mathcal{G}\): the identity is zero on \(\varnothing\), holds on \(O\) and \(E\) by construction, and on \(\Omega\) both sides equal \(1+3/2=5/2\).
Check: \(\mathbb{E}[Y]=(2+2+3+3)/4=10/4=5/2\), and \(\mathbb{E}[X]=(1+2+3+4)/4=10/4=5/2\), agreeing with the identity for \(\Omega\).
Worked Example: Conditioning the Coordinate on Three Intervals
Let \(\Omega=[0,1]\), \(\mathcal{F}=\mathcal{L}\), and \(\mathbb{P}\) be Lebesgue measure on this interval. Take \(X(x)=x\) and let \(\mathcal{G}\) be generated by the partition \(A_1=[0,1/3)\), \(A_2=[1/3,2/3)\), \(A_3=[2/3,1]\). Each interval has probability \(1/3\). Find the density of the induced measure on these atoms.
Direct integration gives $$ \nu(A_1)=\int_0^{1/3}x\,dx=\frac{(1/3)^2}{2}=\frac1{18}, $$ $$ \nu(A_2)=\int_{1/3}^{2/3}x\,dx=\frac{(2/3)^2-(1/3)^2}{2}=\frac{4/9-1/9}{2}=\frac16, $$ and $$ \nu(A_3)=\int_{2/3}^{1}x\,dx=\frac{1-(2/3)^2}{2}=\frac{1-4/9}{2}=\frac5{18}. $$ Dividing each value by the atom's probability \(1/3\) gives densities \(1/6\), \(1/2\), and \(5/6\), respectively. Thus \(Y\) takes these three values on \(A_1,A_2,A_3\). Every set in \(\mathcal{G}\) is a union of these atoms, so additivity gives the required identity on every set in \(\mathcal{G}\).
Check: The integral of \(Y\) is \((1/6)(1/3)+(1/2)(1/3)+(5/6)(1/3)=1/18+3/18+5/18=1/2\). Independently, \(\mathbb{E}[X]=\int_0^1x\,dx=1/2\). The two computations agree.
Worked Example: An Integrable Unbounded Variable
Let \(\Omega=(0,1)\) with Lebesgue probability, \(X(x)=-\ln x\), and \(\mathcal{G}=\{\varnothing,B,B^c,\Omega\}\), where \(B=(0,1/2)\). The function \(X\) is nonnegative and integrable since \(\int_0^1-\ln x\,dx=1\). Determine the density on the two nontrivial atoms.
Using the antiderivative \(x(1-\ln x)\), whose limit at \(0\) is \(0\), we obtain $$ \nu(B)=\int_0^{1/2}-\ln x\,dx =\frac12(1+\ln2). $$ Since \(\mathbb{P}(B)=1/2\), the density on \(B\) is \(1+\ln2\). The integral over the complement is the total integral minus the integral over \(B\): $$ \nu(B^c)=1-\frac12(1+\ln2)=\frac12(1-\ln2). $$ As \(\mathbb{P}(B^c)=1/2\), the density on \(B^c\) is \(1-\ln2\). Therefore \(Y=(1+\ln2)\chi_B+(1-\ln2)\chi_{B^c}\). The four sets of \(\mathcal{G}\) are \(\varnothing,B,B^c,\Omega\); the identities hold on the two atoms by calculation and on \(\varnothing,\Omega\) by zero integrals and addition.
Check: \(\mathbb{E}[Y]=\frac12(1+\ln2)+\frac12(1-\ln2)=1\), which equals \(\int_0^1-\ln x\,dx=1\). The logarithmic terms cancel exactly.
Why the Sub-Sigma-Algebra Matters
The density is taken relative to \(\mathbb{P}\) restricted to \(\mathcal{G}\), not relative to all of \(\mathcal{F}\). This restriction is what forces the resulting density to be \(\mathcal{G}\)-measurable: it can express only the information available in \(\mathcal{G}\). When \(\mathcal{G}\) is a finite partition sigma-algebra, the density on each positive-probability atom is the average of \(X\) on that atom. The Radon–Nikodym construction extends the same event-integral idea beyond finite partitions.
A common error is to replace the conditional expectation by the constant \(\mathbb{E}[X]\) regardless of \(\mathcal{G}\). In the parity example, this constant would be \(5/2\). On the odd event, however, its integral would be \((5/2)(1/2)=5/4\), whereas \(\mathbb{E}[X\chi_O]=1\). Since \(5/4\ne1\), the defining identity fails. The unconditional expectation matches the integral on \(\Omega\), but conditioning requires matching on every event in \(\mathcal{G}\).
Another point to keep distinct is that \(\nu\) need not be a probability measure: for signed \(X\), \(\nu(A)\) can be negative. Its role is to record the event integrals of \(X\); absolute continuity and finite variation make it eligible for the signed Radon–Nikodym theorem. The resulting density then supplies precisely the required \(\mathcal{G}\)-measurable representative.
Check Your Understanding
Use the construction and its defining event identities to answer the following.
- For integrable \(X\), why does \(\nu(A)=\int_A X\,d\mathbb{P}\) vanish whenever \(A\in\mathcal{G}\) has probability zero?
- In the four-outcome example, compute \(\nu(\{2,4\})\) and divide by \(\mathbb{P}(\{2,4\})\) to recover the density on the even atom.
- For the three-interval example, verify the defining identity on \(A_1\cup A_3\) by computing both integrals.
- Why is the Dominated Convergence Theorem applicable when proving countable additivity of \(\nu\)? State the dominating integrable function.
- In the unbounded-variable example, compute the integral of the proposed density over \(B^c\), and compare it with \(\int_{B^c}-\ln x\,dx\).