Tutorials › Probability Theory (Graduate) › Testing Conditional Expectations Beyond Events

Home  /  Tutorials  /  Conditional Expectation

Conditional Expectation · Tutorial 231 of 946

Testing Conditional Expectations Beyond Events

See why testing on events is enough to obtain the conditional-expectation identity for all bounded measurable tests, and how to use that extension in examples.

Intermediate 10 min read

What You'll Learn

  • State the event-based definition of conditional expectation.
  • Extend the defining identity from indicators to simple test functions.
  • Approximate bounded measurable tests uniformly by simple functions.
  • Identify where integrability guarantees the limiting expectations exist.
  • Apply the test-function identity in symmetry and independence examples.
  • Recognize why the boundedness condition matters for this argument.

From Events to Test Functions

Conditional expectation is defined by testing against events in the conditioning sigma-algebra. This event-by-event definition is precise, but it can be cumbersome: in applications, the quantity being tested is often a bounded measurable random variable rather than an indicator.

The key fact is that the event identity already contains all the bounded tests. Indicators give the identity for simple functions by finite linearity, and bounded measurable functions can be uniformly approximated by simple functions. Integrability makes the resulting passage to the limit valid.

The Defining Identity

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 real-valued random variable. A conditional expectation of \(X\) given \(\mathcal G\) is an integrable, \(\mathcal G\)-measurable random variable \(Y\) such that

$$ \mathbb E[X\mathbf 1_A]=\mathbb E[Y\mathbf 1_A] \qquad\text{for every }A\in\mathcal G. $$

Here \(\mathbf 1_A\) denotes the indicator of \(A\). The event identity is meaningful because both \(X\) and \(Y\) are integrable. We will show that it implies an identity for every bounded, real-valued, \(\mathcal G\)-measurable test random variable \(Z\).

Definition: A simple random variable is a measurable random variable taking only finitely many values. A simple \(\mathcal G\)-measurable random variable can be written as \(S=\sum_{j=1}^m c_j\mathbf 1_{A_j}\), where \(m\) is finite, the \(A_j\) are disjoint sets in \(\mathcal G\), and the \(c_j\) are real numbers.

If \(S\) is such a simple test, the event identity immediately gives

$$ \mathbb E[XS] =\sum_{j=1}^m c_j\mathbb E[X\mathbf 1_{A_j}] =\sum_{j=1}^m c_j\mathbb E[Y\mathbf 1_{A_j}] =\mathbb E[YS]. $$

All terms are finite: \(S\) is bounded, and \(X,Y\) are integrable. The important step is extending this calculation beyond simple tests.

The Bounded-Test Theorem

Theorem: Suppose \(X\) is integrable and \(Y\) is an integrable, \(\mathcal G\)-measurable random variable. Then \(Y\) satisfies the event identity for \(X\) if and only if
$$ \mathbb E[XZ]=\mathbb E[YZ] \qquad \text{for every bounded, real-valued, }\mathcal G\text{-measurable }Z. $$
Proof. First suppose the event identity holds. If \(S\) is a simple \(\mathcal G\)-measurable random variable, express it as \(S=\sum_{j=1}^m c_j\mathbf 1_{A_j}\) with disjoint \(A_j\in\mathcal G\). Applying the event identity to each \(A_j\) and summing gives \(\mathbb E[XS]=\mathbb E[YS]\). Now let \(Z\) be bounded and \(\mathcal G\)-measurable. Choose a finite constant \(M\) such that \(|Z|\le M\). For each \(n\in\mathbb N\), define
$$ S_n=2^{-n}\left\lfloor 2^nZ\right\rfloor. $$
Because \(Z\) is \(\mathcal G\)-measurable, \(S_n\) is \(\mathcal G\)-measurable. Since \(Z\) is bounded, \(S_n\) takes only finitely many values, so it is simple. The definition of the floor function gives \(|Z-S_n|<2^{-n}\) pointwise. Therefore
$$ \left|\mathbb E[XZ]-\mathbb E[XS_n]\right| \le \mathbb E\!\left[|X|\,|Z-S_n|\right] \le 2^{-n}\mathbb E[|X|], $$
and, using integrability of \(Y\),
$$ \left|\mathbb E[YZ]-\mathbb E[YS_n]\right| \le 2^{-n}\mathbb E[|Y|]. $$
The simple-test identity gives \(\mathbb E[XS_n]=\mathbb E[YS_n]\) for every \(n\). Consequently,
$$ \left|\mathbb E[XZ]-\mathbb E[YZ]\right| \le 2^{-n}\bigl(\mathbb E[|X|]+\mathbb E[|Y|]\bigr). $$
The right-hand side tends to zero as \(n\) tends to infinity, so \(\mathbb E[XZ]=\mathbb E[YZ]\). Conversely, suppose the identity holds for every bounded \(\mathcal G\)-measurable \(Z\). For any \(A\in\mathcal G\), choose \(Z=\mathbf 1_A\), which is bounded and \(\mathcal G\)-measurable. Then \(\mathbb E[X\mathbf 1_A]=\mathbb E[Y\mathbf 1_A]\). This is exactly the event identity. \(\square\)

The proof uses two separate facts. Simple tests reduce the claim to a finite sum of event identities. Uniform approximation then controls the error by an integrable quantity, namely \(|X|\) or \(|Y|\). No exchange of an uncontrolled limit with expectation is needed.

Continuity in the Test Function

The same estimates give a useful quantitative statement. For a bounded test \(Z\), each of the maps \(Z\mapsto\mathbb E[XZ]\) and \(Z\mapsto\mathbb E[YZ]\) changes continuously when the test changes in uniform size.

Proposition: If \(X\) is integrable and \(Z,W\) are bounded random variables, then
$$ \left|\mathbb E[XZ]-\mathbb E[XW]\right| \le \|Z-W\|_\infty\,\mathbb E[|X|], $$

where \(\|Z-W\|_\infty\) may be taken as any bound for \(|Z-W|\) outside a null set.

Proof. Since \(X\) is integrable and \(Z,W\) are bounded, \(XZ\) and \(XW\) are integrable. Linearity of expectation and the triangle inequality give
$$ \left|\mathbb E[XZ]-\mathbb E[XW]\right| =\left|\mathbb E[X(Z-W)]\right| \le \mathbb E[|X|\,|Z-W|] \le \|Z-W\|_\infty\,\mathbb E[|X|]. $$

The proof applies with \(Y\) in place of \(X\) as well. This continuity estimate explains why a uniform approximation is especially convenient: the approximation error is multiplied only by a finite first absolute moment.

Worked Example: Symmetry and Conditioning on Absolute Value

Worked Example: Symmetry and Conditioning on Absolute Value

Let \(\Omega=[-1,1]\) with the uniform probability distribution, let \(X(x)=x\), and let \(\mathcal G=\sigma(|X|)\). We claim that \(Y=0\) is a version of \(\mathbb E[X\mid\mathcal G]\).

Every set \(A\in\mathcal G\) is symmetric about zero: membership in \(A\) depends only on \(|x|\), so \(x\in A\) if and only if \(-x\in A\). The density is \(1/2\), and a change of variable \(x\mapsto -x\) gives

$$ \mathbb E[X\mathbf 1_A] =\frac12\int_A x\,dx =\frac12\int_A (-x)\,dx =-\mathbb E[X\mathbf 1_A]. $$

Thus \(\mathbb E[X\mathbf 1_A]=0=\mathbb E[Y\mathbf 1_A]\) for every \(A\in\mathcal G\). The candidate \(Y=0\) is integrable and \(\mathcal G\)-measurable, so it is a conditional expectation. The theorem now gives \(\mathbb E[XZ]=0\) for every bounded \(\mathcal G\)-measurable test \(Z\). For instance, \(Z=1+|X|\) is such a test, and direct symmetry also yields

$$ \mathbb E[X(1+|X|)] =\frac12\int_{-1}^1 x(1+|x|)\,dx=0. $$

Worked Example: A Test Depending on One Coordinate

Worked Example: A Test Depending on One Coordinate

Let \(U\) and \(V\) be the coordinate functions on the unit square \([0,1]^2\) with uniform probability measure. Set \(X=U+V\), and let \(\mathcal G=\sigma(U)\). We check that \(Y=U+\tfrac12\) is a conditional expectation.

For any \(A\in\mathcal G\), there is a Borel set \(B\subseteq[0,1]\) such that \(A=\{U\in B\}\). Fubini's theorem gives

$$ \mathbb E[X\mathbf 1_A] =\int_B\int_0^1(u+v)\,dv\,du =\int_B\left(u+\frac12\right)\,du =\mathbb E[Y\mathbf 1_A]. $$

The candidate \(Y\) is \(\mathcal G\)-measurable and integrable, so the event identity verifies the claim. To see the bounded-test version explicitly, choose \(Z=U^2\). Then

$$ \mathbb E[ZX] =\int_0^1\int_0^1u^2(u+v)\,dv\,du =\frac14+\frac16 =\frac5{12}, $$

while

$$ \mathbb E[ZY] =\int_0^1u^2\left(u+\frac12\right)\,du =\frac14+\frac16 =\frac5{12}. $$

The theorem gives the same equality for every bounded measurable function of \(U\), not only for this particular choice \(U^2\).

Worked Example: An Independent Integrable Factor

Worked Example: An Independent Integrable Factor

Let \(U\) be uniform on \([0,1]\), let \(V\) be an independent exponential random variable with density \(e^{-v}\) for \(v\ge0\), and set \(X=UV\). Let \(\mathcal G=\sigma(U)\). Since \(V\) is nonnegative and has mean

$$ \mathbb E[V]=\int_0^\infty ve^{-v}\,dv=1, $$

the random variable \(X\) is integrable. We claim that \(Y=U\) is a conditional expectation of \(X\) given \(\mathcal G\).

For \(A=\{U\in B\}\in\mathcal G\), independence gives

$$ \mathbb E[X\mathbf 1_A] =\mathbb E[U\mathbf 1_{\{U\in B\}}]\,\mathbb E[V] =\mathbb E[U\mathbf 1_A] =\mathbb E[Y\mathbf 1_A]. $$

Now take the bounded \(\mathcal G\)-measurable test \(Z=1/(1+U)\). Since \(u/(1+u)=1-1/(1+u)\), direct integration gives

$$ \mathbb E[ZX] =\mathbb E\!\left[\frac{U}{1+U}\right]\mathbb E[V] =\int_0^1\frac{u}{1+u}\,du =1-\ln 2. $$

For the candidate \(Y\), the same calculation yields

$$ \mathbb E[ZY] =\mathbb E\!\left[\frac{U}{1+U}\right] =\int_0^1\frac{u}{1+u}\,du =1-\ln 2. $$

This illustrates how the test identity can be used with a non-indicator function that directly matches the quantity in an application.

Why Boundedness Matters Here

Boundedness ensures that \(XZ\) and \(YZ\) are integrable whenever \(X\) and \(Y\) are integrable. It also provides the error estimates that make the simple-function approximation argument work. For example, if \(|Z|\le C\), then \(|XZ|\le C|X|\), whose expectation is finite.

For an unbounded \(\mathcal G\)-measurable test, the products \(XZ\) and \(YZ\) need not be integrable. Even when they are, uniform approximation by bounded simple functions is no longer available in the same form. Thus the theorem makes a specific claim about bounded tests; extending it to an unbounded test requires additional integrability conditions and a separate justification.

The practical lesson is to choose the test class deliberately. When the test is bounded and measurable with respect to the conditioning sigma-algebra, the event definition supplies the desired identity automatically. The proof is not an extra assumption about conditional expectation; it is a consequence of the defining event identities and integrability.

Check Your Understanding

Use the event identity and the approximation argument to answer these questions.

  1. Why does the defining identity hold for every simple \(\mathcal G\)-measurable test once it holds for indicators of events in \(\mathcal G\)?
  2. How does the quantized approximation \(S_n=2^{-n}\lfloor 2^nZ\rfloor\) control the error in \(\mathbb E[XZ]\)?
  3. Which hypotheses ensure that both \(\mathbb E[XZ]\) and \(\mathbb E[YZ]\) are finite for a bounded test \(Z\)?
  4. How does choosing \(Z=\mathbf 1_A\) prove the converse direction of the bounded-test theorem?
  5. Why does the proof not automatically establish the same identity for every unbounded \(\mathcal G\)-measurable test?

Previous

Next

Back to Tutorials
Clinical Biostats