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Conditional Expectation · Tutorial 238 of 946

Positivity and Order Preservation

Learn why conditional expectation preserves nonnegativity and how to compare conditional expectations of ordered integrable random variables.

Intermediate 9 min read

What You'll Learn

  • State the defining integral property of conditional expectation.
  • Prove that the conditional expectation of a nonnegative integrable random variable is nonnegative almost surely.
  • Deduce order preservation for two integrable random variables.
  • Compare conditional expectations using finite partitions.
  • Distinguish strict order on some outcomes from strict order almost surely.

Why Order Matters Under Conditioning

Conditional expectation replaces a random variable with a quantity that reflects the information in a chosen sigma-algebra. Because it averages over information that is not retained, it is natural to ask whether this operation respects inequalities. If one integrable random variable is everywhere no larger than another, can their conditional expectations reverse order?

They cannot: conditional expectation preserves order, up to almost-sure equality. The key step is to show that a nonnegative random variable has a nonnegative conditional expectation. The proof uses the defining integral property and tests what would happen on the event where the conditional expectation is negative.

Definition and Almost-Sure Conventions

Let \((\Omega,\mathcal F,\mathbb P)\) be a probability space, let \(\mathcal G\subseteq\mathcal F\) be a sigma-algebra, and let \(X\) be an integrable random variable. A conditional expectation of \(X\) given \(\mathcal G\), written \(\mathbb E[X\mid\mathcal G]\), is an integrable, \(\mathcal G\)-measurable random variable \(Y\) such that

$$ \int_A Y\,d\mathbb P=\int_A X\,d\mathbb P \qquad\text{for every }A\in\mathcal G. $$

The existence and almost-sure uniqueness of this random variable are part of the earlier material on conditional expectation. In this tutorial, inequalities between conditional expectations mean inequalities almost surely. This qualification matters because a conditional expectation is determined only up to changes on a probability-zero set.

Definition: An integrable random variable \(X\) is nonnegative almost surely if \(\mathbb P(X\ge 0)=1\). For random variables \(X\) and \(Z\), the notation \(X\le Z\) almost surely means \(\mathbb P(X\le Z)=1\).

Worked Example: Averaging Nonnegative Values on a Partition

Worked Example: Averaging Nonnegative Values on a Partition

Let \(\Omega=\{1,2,3,4\}\), with probability \(1/4\) assigned to each outcome. Let \(\mathcal G\) be generated by the partition \(\{\{1,2\},\{3,4\}\}\), and define \(X(1)=0\), \(X(2)=2\), \(X(3)=1\), and \(X(4)=5\). On each partition cell, the conditional expectation is the average of \(X\) on that cell.

$$ \mathbb E[X\mid\mathcal G] = \begin{cases} (0+2)/2=1,&\omega\in\{1,2\},\\ (1+5)/2=3,&\omega\in\{3,4\}. \end{cases} $$

Both values are nonnegative, as expected. To check the defining property directly, on \(\{1,2\}\) the integral of the conditional expectation is \((1/4)+(1/4)=1/2\), while the integral of \(X\) is \(0/4+2/4=1/2\). On \(\{3,4\}\), these integrals are \(3/4+3/4=3/2\) and \(1/4+5/4=3/2\), respectively. The equality also holds on \(\varnothing\) and on \(\Omega\), by adding the cell equalities.

Conditional Expectation Preserves Nonnegativity

The finite calculation illustrates the general fact. The proof does not require a finite partition: the defining integral identity applies to every event in \(\mathcal G\), including the event on which a proposed conditional expectation is negative.

Theorem: If \(X\) is integrable and \(X\ge 0\) almost surely, then \(\mathbb E[X\mid\mathcal G]\ge 0\) almost surely.
Proof. Let \(Y\) be a version of \(\mathbb E[X\mid\mathcal G]\). Since \(Y\) is \(\mathcal G\)-measurable, the event \(A=\{Y<0\}\) belongs to \(\mathcal G\). Suppose, for contradiction, that \(\mathbb P(A)>0\). The events \(A_n=\{Y\le -1/n\}\), for \(n\in\mathbb N\), have union \(A\). Therefore at least one \(A_n\) has positive probability; otherwise their countable union would have probability zero. On \(A\), \(Y\le 0\), and on \(A_n\), \(Y\le -1/n\). Consequently,
$$ \int_A Y\,d\mathbb P \le -\frac{1}{n}\mathbb P(A_n)<0. $$
But \(A\in\mathcal G\), so the defining property gives \(\int_A Y\,d\mathbb P=\int_A X\,d\mathbb P\). Since \(X\ge0\) almost surely, the latter integral is nonnegative. This contradicts the strict inequality above. Hence \(\mathbb P(A)=0\), proving \(Y\ge0\) almost surely. \(\square\)

The argument depends on testing the conditional expectation on a \(\mathcal G\)-measurable event. If \(Y\) were negative on a set of positive probability, one of the sets where it is bounded away from zero on the negative side would have positive probability. Its integral would then be strictly negative, contrary to the integral identity and the nonnegativity of \(X\).

Order Preservation

Linearity of conditional expectation gives the order result from the nonnegativity theorem. In particular, for integrable \(X\) and \(Z\), linearity says that \(\mathbb E[Z-X\mid\mathcal G]=\mathbb E[Z\mid\mathcal G]-\mathbb E[X\mid\mathcal G]\) almost surely.

Theorem: If \(X\) and \(Z\) are integrable and \(X\le Z\) almost surely, then
$$ \mathbb E[X\mid\mathcal G]\le \mathbb E[Z\mid\mathcal G] \qquad\text{almost surely}. $$
Proof. Define \(D=Z-X\). The hypothesis \(X\le Z\) almost surely means \(D\ge0\) almost surely, and \(D\) is integrable because \(X\) and \(Z\) are integrable. By positivity and linearity,
$$ 0\le \mathbb E[D\mid\mathcal G] =\mathbb E[Z-X\mid\mathcal G] =\mathbb E[Z\mid\mathcal G]-\mathbb E[X\mid\mathcal G] \qquad\text{almost surely}. $$
Rearranging this inequality proves the claim. \(\square\)

Worked Example: Comparing Two Conditional Expectations

Worked Example: Comparing Two Conditional Expectations

Again take four equally likely outcomes and let \(\mathcal G\) be generated by \(\{1,2\}\) and \(\{3,4\}\). Define \(X=(1,4,0,2)\) and \(Z=(2,5,3,2)\), where the entries list the values at outcomes \(1,2,3,4\), respectively. At each outcome, \(X\le Z\): the four comparisons are \(1\le2\), \(4\le5\), \(0\le3\), and \(2\le2\).

Averaging over the two cells gives

$$ \mathbb E[X\mid\mathcal G] = \begin{cases} (1+4)/2=5/2,&\omega\in\{1,2\},\\ (0+2)/2=1,&\omega\in\{3,4\}, \end{cases} $$ $$ \mathbb E[Z\mid\mathcal G] = \begin{cases} (2+5)/2=7/2,&\omega\in\{1,2\},\\ (3+2)/2=5/2,&\omega\in\{3,4\}. \end{cases} $$

On the first cell, \(5/2\le7/2\); on the second, \(1\le5/2\). Thus the conditional expectations preserve the original order on every cell of this partition.

Worked Example: Strict Inequality Need Not Hold Everywhere

Worked Example: Strict Inequality Need Not Hold Everywhere

Let \(\Omega=\{1,2,3,4\}\) again, with equal probabilities and the same conditioning sigma-algebra. Set \(X=0\) at every outcome and set \(Z=(1,0,0,0)\). Then \(X\le Z\) at all outcomes, and \(X<Z\) at outcome \(1\), an event of probability \(1/4\). But the conditional expectations are

$$ \mathbb E[X\mid\mathcal G] = \begin{cases} 0,&\omega\in\{1,2\},\\ 0,&\omega\in\{3,4\}, \end{cases} \qquad \mathbb E[Z\mid\mathcal G] = \begin{cases} (1+0)/2=1/2,&\omega\in\{1,2\},\\ (0+0)/2=0,&\omega\in\{3,4\}. \end{cases} $$

The conditional expectations are strictly ordered on \(\{1,2\}\), but equal on \(\{3,4\}\). Thus \(X\le Z\) almost surely, with strict inequality on a set of positive probability, does not by itself imply strict inequality between the conditional expectations almost surely. The order-preservation theorem asserts a weak inequality; it does not assert that the inequality is strict everywhere.

How to Use the Result Carefully

The practical procedure is short: subtract the lower random variable from the upper one, verify that the difference is nonnegative almost surely, and apply positivity to that difference. Linearity then translates the resulting nonnegative conditional expectation into the desired comparison.

1
Check integrability.
Both variables must be integrable so their conditional expectations and their difference are defined in the usual integrable sense.
2
Form the difference.
If \(X\le Z\) almost surely, then \(Z-X\ge0\) almost surely.
3
Apply positivity and linearity.
The conditional expectation of \(Z-X\) is nonnegative almost surely and equals the difference of the two conditional expectations.

A common pitfall is to read these statements pointwise for every outcome. Conditional expectations are only defined up to almost-sure equality, and the theorems accordingly establish inequalities almost surely. Another is to infer strict order from a weak-order theorem: strictness requires additional information, and strict inequality on only part of the sample space need not produce strict conditional inequality everywhere.

Takeaway: For integrable random variables, \(X\ge0\) almost surely implies \(\mathbb E[X\mid\mathcal G]\ge0\) almost surely. Applying this to \(Z-X\) shows that \(X\le Z\) almost surely implies \(\mathbb E[X\mid\mathcal G]\le\mathbb E[Z\mid\mathcal G]\) almost surely.

Check Your Understanding

Use the defining integral property and the order-preservation result to answer these questions.

  1. In the positivity proof, why is the event \(\{\mathbb E[X\mid\mathcal G]<0\}\) available as a test event?
  2. Why does positive probability of \(\{Y<0\}\) imply positive probability for \(\{Y\le-1/n\}\) for at least one \(n\in\mathbb N\)?
  3. If \(X\le Z\) almost surely, which random variable should be tested with the positivity theorem to compare their conditional expectations?
  4. For the four-outcome example with \(X=(0,0,0,0)\) and \(Z=(1,0,0,0)\), on which partition cell are the conditional expectations equal?
  5. What does “almost surely” account for when stating an inequality between conditional expectations?

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