Finite Information and Conditional Averages
When only partial information about an outcome is available, an expectation can be updated by averaging over the outcomes still compatible with that information. A finite measurable partition describes this situation precisely: each atom represents one possible information state, and the conditional expectation is constant within each atom. The average must use the probabilities of the underlying outcomes, not simply count them.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \(X:\Omega\to\mathbb{R}\) be an integrable random variable, meaning that \(\mathbb{E}[|X|]<\infty\). Let \(\mathcal{P}=\{P_1,\ldots,P_k\}\) be a finite measurable partition of \(\Omega\): its sets are measurable and pairwise disjoint, and their union is \(\Omega\). The sigma-algebra \(\sigma(\mathcal{P})\) consists of all unions of these atoms. It represents the information that tells us which atom contains the outcome, but not which point within that atom occurred.
The definition requires integrability so that the integrals over atoms are finite. On a positive-probability atom, the formula is an ordinary average with respect to the probability measure restricted to that atom and normalized to have total mass one. No average is defined by dividing by the probability of a null atom, since that would divide by zero. The choice on a null atom cannot affect any of the defining integrals.
Why the Atom-Average Formula Works
Proof. Define \(Y\) to have the indicated constant value on each atom. Since each atom is measurable and there are finitely many of them, \(Y\) is \(\sigma(\mathcal{P})\)-measurable. It is integrable: on each positive-probability atom, \(\mathbb{P}(P_i)|Y|=\left|\int_{P_i}X\,d\mathbb{P}\right|\le\int_{P_i}|X|\,d\mathbb{P}\), and on a null atom \(Y=0\). Summing these inequalities over the finitely many atoms gives \(\mathbb{E}[|Y|]\le\mathbb{E}[|X|]<\infty\).
For a positive-probability atom \(P_i\), constancy of \(Y\) gives \(\int_{P_i}Y\,d\mathbb{P}=\mathbb{P}(P_i)Y|_{P_i}=\int_{P_i}X\,d\mathbb{P}\), where \(Y|_{P_i}\) denotes its constant value there. If \(\mathbb{P}(P_i)=0\), then \(\int_{P_i}Y\,d\mathbb{P}=0\), and \(\int_{P_i}X\,d\mathbb{P}=0\) because \(\int_{P_i}|X|\,d\mathbb{P}=0\). Every \(A\in\sigma(\mathcal{P})\) is a union of some of the finitely many atoms. Add the atom identities over that union to obtain \(\int_A Y\,d\mathbb{P}=\int_A X\,d\mathbb{P}\). Thus \(Y\) satisfies both conditions in the definition. \(\square\)
This verification also explains how the construction uses the available information: the conditional expectation is constant on each atom, and within an atom its integral matches the integral of the original variable. Taking \(A=\Omega\) in the defining identity gives \(\mathbb{E}[Y]=\mathbb{E}[X]\).
Worked Example: Squared Outcome Given the Parity of a Die Roll
Let \(\Omega=\{1,2,3,4,5,6\}\) with each outcome having probability \(1/6\), and let \(X(j)=j^2\). Suppose the information revealed is whether the die roll is odd or even. The partition is \(P_{\mathrm{odd}}=\{1,3,5\}\) and \(P_{\mathrm{even}}=\{2,4,6\}\). Each atom has probability \(3/6=1/2\).
On the odd atom, the integral of \(X\) is \((1^2+3^2+5^2)/6=(1+9+25)/6=35/6\). Dividing by its probability gives the conditional average \((35/6)/(1/2)=35/3\). On the even atom, the integral is \((2^2+4^2+6^2)/6=(4+16+36)/6=56/6\), so the conditional average is \((56/6)/(1/2)=56/3\). Therefore the conditional expectation is \(35/3\) on odd rolls and \(56/3\) on even rolls.
Check: Its expectation is \((1/2)(35/3)+(1/2)(56/3)=91/6\). Directly, \(\mathbb{E}[X]=(1+9+25+4+16+36)/6=91/6\), so the two calculations agree.
Conditional Probabilities and Information States
Proof. The indicator \(\chi_B\) is integrable because \(0\le\chi_B\le1\) and \(\mathbb{P}(\Omega)=1\). Apply the theorem's atom-average formula. On a positive-probability atom, \(\int_{P_i}\chi_B\,d\mathbb{P}=\mathbb{P}(B\cap P_i)\), so dividing by \(\mathbb{P}(P_i)\) gives the stated value. On a null atom the construction assigns zero. The theorem verifies that these values satisfy the defining integral identity. \(\square\)
Thus, after observing that the outcome lies in a positive-probability atom \(P_i\), the value is the probability of \(B\) within that information state. This is a conditional probability on an atom. If the observed atom has probability zero, this formula does not apply: its denominator is zero, and the atom-average convention assigns a value only to specify a version of the random variable there.
Worked Example: Expected Payoff After a Category Is Revealed
Let the finite probability space have outcomes \(\Omega=\{a,b,c,d\}\), with probabilities \(\mathbb{P}(\{a\})=1/2\), \(\mathbb{P}(\{b\})=1/4\), \(\mathbb{P}(\{c\})=1/8\), and \(\mathbb{P}(\{d\})=1/8\). These add to \(1/2+1/4+1/8+1/8=1\). A payoff \(X\) has values \(0,8,4,12\) at \(a,b,c,d\), respectively. The revealed category distinguishes \(P_1=\{a,b\}\) from \(P_2=\{c,d\}\).
The first category has probability \(1/2+1/4=3/4\). Its payoff integral is \(0(1/2)+8(1/4)=0+2=2\), so its conditional average is \(2/(3/4)=8/3\). The second category has probability \(1/8+1/8=1/4\). Its payoff integral is \(4(1/8)+12(1/8)=1/2+3/2=2\), so its conditional average is \(2/(1/4)=8\).
Check: Weighting these conditional averages by the category probabilities gives \((3/4)(8/3)+(1/4)(8)=2+2=4\). Directly, \(\mathbb{E}[X]=0(1/2)+8(1/4)+4(1/8)+12(1/8)=0+2+1/2+3/2=4\). For the event \(B=\{b,d\}\), the conditional probability of \(B\) in the first category is \((1/4)/(3/4)=1/3\); in the second it is \((1/8)/(1/4)=1/2\).
Worked Example: A Null Atom in a Finite Probability Space
Take \(\Omega=\{u,v,w\}\), with probabilities \(1/2,1/2,0\), respectively, and partition \(\mathcal{P}=\{\{u\},\{v,w\}\}\). Let \(X(u)=6\), \(X(v)=2\), and \(X(w)=100\). The first atom has probability \(1/2\), so the conditional average there is \((6\cdot(1/2))/(1/2)=6\). The second atom has probability \(1/2+0=1/2\); its payoff integral is \(2(1/2)+100(0)=1+0=1\), so its conditional average is \(1/(1/2)=2\).
Check: The constructed conditional expectation equals \(6\) at \(u\) and \(2\) at both \(v\) and \(w\). Its expectation is \(6(1/2)+2(1/2)=3+1=4\). The original expectation is \(6(1/2)+2(1/2)+100(0)=3+1+0=4\). In particular, the value \(100\) at the zero-probability outcome contributes nothing. If instead an entire partition atom had probability zero, the chosen conditional value on that atom would not be determined by an average; our convention would set it to zero.
A Common Averaging Error
Conditional averages are probability-weighted, not necessarily unweighted averages over the points in an atom. In the category example, an incorrect calculation on \(\{a,b\}\) might take the ordinary mean of the two payoff values and report \((0+8)/2=4\). That gives equal weight to \(a\) and \(b\), although their probabilities are \(1/2\) and \(1/4\).
The correction is to divide the payoff integral on the atom by the atom's probability. The numerator is \(0(1/2)+8(1/4)=2\), and the denominator is \(1/2+1/4=3/4\), so the correct conditional average is \(2/(3/4)=8/3\), not \(4\). Equivalently, within that category the two outcomes have normalized probabilities \((1/2)/(3/4)=2/3\) and \((1/4)/(3/4)=1/3\); the weighted mean is \(0(2/3)+8(1/3)=8/3\). Equal weighting is justified only when the outcomes in the atom have equal probabilities.
The finite-partition construction is useful whenever information has finitely many states: the atoms can represent observed categories, signals, or a finite summary of data. Compute the probability of each state, compute the payoff integral over it, and divide by that state probability when it is positive. The result is the expected value compatible with the information revealed.
Check Your Understanding
Use the atom-average formula and verify each answer by checking the relevant probabilities or integrals.
- A fair die has partition \(\{\{1,2,3\},\{4,5,6\}\}\). For \(X(j)=j\), what is the conditional expectation on each atom?
- For an event \(B\) and a positive-probability atom \(P_i\), derive the conditional probability of \(B\) on \(P_i\) from the formula for \(\chi_B\).
- In the category example, calculate the conditional probability of \(\{a,c\}\) on each category atom.
- Why does the conditional expectation construction not divide by the probability of a null atom?
- Give a finite partition example in which an unweighted average on an atom differs from the conditional expectation, and identify the outcome probabilities responsible for the difference.