Tutorials › Probability Theory (Graduate) › Worked Problems in Conditional Expectation

Home  /  Tutorials  /  Conditional Expectation

Conditional Expectation · Tutorial 251 of 946

Worked Problems in Conditional Expectation

Practice combining partition calculations, independence, nested conditioning, pull-out identities, and conditional variance with the necessary checks made explicit.

Intermediate 10 min read

What You'll Learn

  • Compute conditional expectations directly on the cells of a finite partition.
  • Apply the tower property to nested sigma-algebras and check measurability.
  • Use independence to identify a conditional expectation with an ordinary expectation.
  • Apply the pull-out identity when the multiplier is measurable and bounded.
  • Calculate conditional variance and verify its decomposition across nested sigma-algebras.

From Conditional-Expectation Rules to Worked Problems

Conditional expectation problems often involve several ideas at once: a sigma-algebra generated by a finite partition, a larger sigma-algebra that contains it, or a random variable independent of the information being conditioned on. A reliable solution checks which information is available, whether each proposed answer is measurable with respect to that information, and whether the variables involved are integrable.

We will work through these checks in concrete settings. Throughout, conditional expectations are understood up to almost-sure equality. For an integrable random variable \(X\) and a sigma-algebra \(\mathcal G\), a version of \(\mathbb E[X\mid\mathcal G]\) is an integrable, \(\mathcal G\)-measurable random variable \(Y\) such that

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

This defining property is the main check behind the finite calculations below. We also use the tower property for nested sigma-algebras and the pull-out property for bounded measurable multipliers.

Conditional Expectation on a Finite Partition

Proposition: Let \(A_1,\ldots,A_n\) be a measurable partition of \(\Omega\), and let \(\mathcal G=\sigma(A_1,\ldots,A_n)\). If \(X\) is integrable, then on every cell with positive probability, \[ \mathbb E[X\mid\mathcal G] = \frac{\mathbb E[X\mathbf 1_{A_i}]}{\mathbb P(A_i)} \quad\text{on }A_i. \] On a cell of probability zero the value can be chosen arbitrarily.
Proof. Define \(Y\) to equal the displayed ratio on each cell \(A_i\) with positive probability, and set it to zero on any null cell. Since \(Y\) is constant on each cell, it is \(\mathcal G\)-measurable. It is integrable because \[ \int_{A_i}|Y|\,d\mathbb P = \left|\mathbb E[X\mathbf 1_{A_i}]\right| \leq \mathbb E[|X|\mathbf 1_{A_i}] \] on every positive-probability cell, and there are only finitely many cells. By construction, \(\int_{A_i}Y\,d\mathbb P=\int_{A_i}X\,d\mathbb P\); the same equality holds on null cells because both integrals are zero. Every set in \(\mathcal G\) is a union of cells, so adding these equalities proves the defining integral identity for every set in \(\mathcal G\). Thus \(Y\) is a version of \(\mathbb E[X\mid\mathcal G]\).

\(\square\)

Worked Example: Conditioning on Two Blocks of a Die

Let \(\Omega=\{1,2,3,4,5,6\}\) with the uniform probability measure, let \(X(\omega)=\omega\), and let \(\mathcal G\) be generated by \(A=\{1,2,3\}\) and \(B=\{4,5,6\}\). The variable \(X\) is integrable because it is bounded. A \(\mathcal G\)-measurable answer must be constant on each of these two cells.

On \(A\), the conditional mean is \((1+2+3)/3=2\); on \(B\), it is \((4+5+6)/3=5\). Therefore

$$ \mathbb E[X\mid\mathcal G]=2\mathbf 1_A+5\mathbf 1_B. $$

To check the defining identity, the integral over \(A\) of this answer is \(2(3/6)=1\), while \(\mathbb E[X\mathbf 1_A]=(1+2+3)/6=1\). On \(B\), the corresponding quantities are \(5(3/6)=5/2\) and \((4+5+6)/6=5/2\). The identity then holds on every union of \(A\) and \(B\).

Nested Sigma-Algebras: What Changes with More Information?

If \(\mathcal G\subseteq\mathcal H\), then \(\mathcal H\) represents at least as much information as \(\mathcal G\). The tower property says that conditioning first on \(\mathcal H\), then on \(\mathcal G\), gives the same result as conditioning directly on \(\mathcal G\): \(\mathbb E[\mathbb E[X\mid\mathcal H]\mid\mathcal G]=\mathbb E[X\mid\mathcal G]\) for integrable \(X\). In applying it, confirm both the inclusion of sigma-algebras and the integrability of \(X\).

Worked Example: Refining a Die Partition

Again let \(X(\omega)=\omega\) on a fair die. Let \(\mathcal G=\{\varnothing,\Omega\}\), the trivial sigma-algebra, and let \(\mathcal H\) be generated by the odd and even outcomes. Both are sub-sigma-algebras of the die's underlying sigma-algebra, and \(\mathcal G\subseteq\mathcal H\). Since \(X\) is bounded, it is integrable.

On the odd cell \(\{1,3,5\}\), the average is \(3\), while on the even cell \(\{2,4,6\}\), it is \(4\). Thus

$$ \mathbb E[X\mid\mathcal H] = 3\mathbf 1_{\{1,3,5\}}+4\mathbf 1_{\{2,4,6\}}. $$

This answer is \(\mathcal H\)-measurable. Its expectation is \((3+4)/2=7/2\), and \(\mathbb E[X]=21/6=7/2\). Conditioning this \(\mathcal H\)-measurable variable on the trivial sigma-algebra therefore gives the constant \(7/2\), which is also \(\mathbb E[X\mid\mathcal G]\). This verifies the tower property in this example.

Independence Removes the Conditioning Information

If \(X\) is integrable and independent of a sigma-algebra \(\mathcal G\), then its conditional expectation given \(\mathcal G\) is the constant \(\mathbb E[X]\). Independence is the key hypothesis: without it, the conditional mean can vary with the event being observed.

Proof. The constant \(c=\mathbb E[X]\) is integrable and \(\mathcal G\)-measurable. For every \(A\in\mathcal G\), independence of \(\sigma(X)\) and \(\mathcal G\) gives \[ \mathbb E[X\mathbf 1_A] = \mathbb E[X]\mathbb P(A) = \int_A c\,d\mathbb P. \] The first equality follows first for indicator functions of events in \(\sigma(X)\), then for integrable \(X\) by simple-function approximation. Thus \(c\) satisfies the defining integral identity, proving the claim.

\(\square\)

Worked Example: An Independent Bernoulli Variable

Suppose \(X\) and \(Y\) are independent, \(X\) takes values \(0\) and \(1\) with \(\mathbb P(X=1)=1/3\), and \(Y\) takes values \(a\) and \(b\), with probabilities \(1/4\) and \(3/4\). Take \(\mathcal G=\sigma(Y)\). The variable \(X\) is bounded and hence integrable, and independence means \(\sigma(X)\) and \(\mathcal G\) are independent.

The expectation is \(\mathbb E[X]=1/3\), so the result above gives \(\mathbb E[X\mid\mathcal G]=1/3\). Directly, \(\mathbb P(Y=a)=1/4\) and \(\mathbb P(X=1,Y=a)=1/12\), giving conditional mean \((1/12)/(1/4)=1/3\) on \(\{Y=a\}\). For \(b\), the joint probability is \(1/4\), so the conditional mean is \((1/4)/(3/4)=1/3\). The proposed constant is \(\mathcal G\)-measurable and agrees with the conditional mean on both cells.

Pulling Out a Measurable Factor

The pull-out property allows a \(\mathcal G\)-measurable bounded factor \(Z\) to be taken outside conditional expectation: \(\mathbb E[ZX\mid\mathcal G]=Z\mathbb E[X\mid\mathcal G]\), provided \(X\) is integrable. Boundedness ensures \(ZX\) is integrable. The right side is integrable as well, since conditional expectation is integrable and \(Z\) is bounded.

Worked Example: Pulling Out a Partition-Measurable Multiplier

On \(\Omega=\{1,2,3,4\}\) with uniform probabilities, set \(X(1)=1\), \(X(2)=3\), \(X(3)=2\), and \(X(4)=6\). Let \(\mathcal G\) be generated by \(A=\{1,2\}\) and \(B=\{3,4\}\), and let \(Z=2\) on \(A\) and \(Z=-1\) on \(B\). The variables \(X\) and \(Z\) are bounded, so \(X\) and \(ZX\) are integrable; \(Z\) is \(\mathcal G\)-measurable.

The partition formula gives \(\mathbb E[X\mid\mathcal G]=2\) on \(A\), since \((1+3)/2=2\), and \(4\) on \(B\), since \((2+6)/2=4\). On \(A\), the mean of \(ZX\) is \((2+6)/2=4\); on \(B\), it is \((-2-6)/2=-4\). Hence

$$ \mathbb E[ZX\mid\mathcal G] = 4\mathbf 1_A-4\mathbf 1_B = Z\mathbb E[X\mid\mathcal G]. $$

The equality is checked on each cell, which is enough because every event in \(\mathcal G\) is a union of those cells.

Conditional Variance Across Nested Information

For a square-integrable \(X\), define \(\operatorname{Var}(X\mid\mathcal G)=\mathbb E[(X-\mathbb E[X\mid\mathcal G])^2\mid\mathcal G]\). Square-integrability ensures this conditional variance is well defined. For nested sigma-algebras, the conditional variance decomposition separates the variation remaining within the finer information from the variation in the conditional mean.

Theorem: If \(X\) is square-integrable and \(\mathcal G\subseteq\mathcal H\), then \[ \operatorname{Var}(X\mid\mathcal G) = \mathbb E[\operatorname{Var}(X\mid\mathcal H)\mid\mathcal G] + \operatorname{Var}(\mathbb E[X\mid\mathcal H]\mid\mathcal G) \quad\text{almost surely}. \]
Proof. Write \(M=\mathbb E[X\mid\mathcal H]\) and \(N=\mathbb E[X\mid\mathcal G]\). Conditional expectation is an \(L^2\) contraction, so \(M\) is square-integrable; the tower property gives \(\mathbb E[M\mid\mathcal G]=N\). The identity \[ X-N=(X-M)+(M-N) \] yields, after squaring, a cross term \(2(X-M)(M-N)\). The factor \(M-N\) is \(\mathcal H\)-measurable because \(\mathcal G\subseteq\mathcal H\). Moreover, it is square-integrable, and \(X-M\) is square-integrable. The pull-out property, extended to these products by truncation, gives \[ \mathbb E[(X-M)(M-N)\mid\mathcal H] = (M-N)\mathbb E[X-M\mid\mathcal H] = 0. \] Here \(\mathbb E[X-M\mid\mathcal H]=0\); the product is integrable by the Cauchy–Schwarz inequality. Taking conditional expectations given \(\mathcal G\), the squared identity therefore becomes \[ \mathbb E[(X-N)^2\mid\mathcal G] = \mathbb E[(X-M)^2\mid\mathcal G] + \mathbb E[(M-N)^2\mid\mathcal G]. \] The tower property identifies the first term on the right with \(\mathbb E[\operatorname{Var}(X\mid\mathcal H)\mid\mathcal G]\). The second is \(\operatorname{Var}(M\mid\mathcal G)\), since \(N=\mathbb E[M\mid\mathcal G]\). The left side is \(\operatorname{Var}(X\mid\mathcal G)\), proving the decomposition.

\(\square\)

Worked Example: Checking the Variance Decomposition

Let \(\Omega=\{1,2,3,4\}\) be uniform, let \(\mathcal G\) be trivial, and let \(\mathcal H\) be generated by \(A=\{1,2\}\) and \(B=\{3,4\}\). Set \(X(1)=0\), \(X(2)=2\), \(X(3)=2\), and \(X(4)=6\). These variables are bounded, so \(X\) is square-integrable, and \(\mathcal G\subseteq\mathcal H\).

On \(A\), the conditional mean is \(1\) and the conditional variance is \(((0-1)^2+(2-1)^2)/2=1\). On \(B\), the mean is \(4\) and the variance is \(((2-4)^2+(6-4)^2)/2=4\). Thus \(\mathbb E[X\mid\mathcal H]\) takes values \(1,4\), each with probability \(1/2\), and \(\operatorname{Var}(X\mid\mathcal G)=\operatorname{Var}(X)=((0-5/2)^2+(2-5/2)^2+(2-5/2)^2+(6-5/2)^2)/4=19/4\).

The first term on the right side of the decomposition is the mean of the two cell variances, \((1+4)/2=5/2\). The second is the variance of the two conditional means, \(((1-5/2)^2+(4-5/2)^2)/2=9/4\). Their sum is \(5/2+9/4=19/4\), matching the unconditional variance.

Checks That Prevent Common Errors

A numerical answer is not yet a conditional expectation until it passes the relevant checks. For a finite partition, the answer must be constant on each conditioning cell, and its cell integral must match that of the original variable. For nested conditioning, verify the inclusion \(\mathcal G\subseteq\mathcal H\) before using the tower property. For independence, identify the independent sigma-algebra, not merely a zero covariance. For a pull-out step, check measurability and integrability of the multiplier and product. For conditional variance, start with square-integrability.

Check Your Understanding

For each problem, identify the conditioning information and state the measurability and integrability checks before calculating.

  1. For a uniform die roll, find \(\mathbb E[X\mid\mathcal G]\) when \(\mathcal G\) is generated by \(\{1,2\}\) and \(\{3,4,5,6\}\).
  2. If \(\mathcal G\subseteq\mathcal H\) and \(X\) is integrable, what does the tower property say about conditioning \(\mathbb E[X\mid\mathcal H]\) on \(\mathcal G\)?
  3. Why does independence of \(X\) from \(\mathcal G\) imply that \(\mathbb E[X\mid\mathcal G]\) is constant?
  4. In the pull-out property, why is boundedness of a \(\mathcal G\)-measurable multiplier a useful sufficient condition?
  5. In the variance decomposition, which two sources of variation appear on the right-hand side?

Previous

Next

Back to Tutorials
Clinical Biostats