Conditional Expectations and Independence
Y. Eddie Lu, Summer 2026
ECON 8002 course index · Lecture 4 of 9
Orientation
Conditional expectation: prediction from information
What is the best prediction of an uncertain payoff after a signal arrives?
How to use this page
Before opening a quiz, say what object is random, what information is available, and which theorem licenses the next line. Rebuild proofs from their defining integral identity.
How should I think about Lecture 4?
Information is a sub-\(\sigma\)-algebra \(\mathcal A\).
\(E[Y\mid\mathcal A]\) is the best \(\mathcal A\)-measurable forecast.
\(\longrightarrow\)
Conditional expectation: retain precisely the information one is allowed to use.
\(\longrightarrow\)
Independence makes some information irrelevant.
Random assignment turns observed mean differences into causal effects.
Course map
- Measure spaces
- Integration and Radon–Nikodym derivatives
- Distribution properties
- Conditioning and independence (current lecture)
- Convergence modes
- Transformations and Slutsky
- Laws of large numbers
- Weak limits and the delta method
- Central limit theorems and inference
Today supplies the information and independence language used in LLNs and causal identification.
Prerequisite retrieval
You need: a probability space \((\Omega,\mathcal F,P)\), a sub-\(\sigma\)-algebra \(\mathcal A\subseteq\mathcal F\), integrability \(E|Y|<\infty\), and the fact that a random variable is a measurable function on \(\Omega\).
Retrieval check.
If \(\mathcal A=\{\varnothing,\Omega\}\), what should \(E[Y\mid\mathcal A]\) be?
The constant \(E[Y]\): no nontrivial event is observable.
Conditional expectation from information
A concrete forecast
Let \(S\in\{\mathrm{good},\mathrm{bad}\}\) be a signal and \(Y\) a payoff. Before seeing \(S\), your forecast is \(E[Y]\). After it, the forecast has two possible values: \(E[Y\mid S]=m(S)\). It is random before the signal is realized.
Why not begin with \(E[Y\mid X=x]\)?
For discrete variables, conditional probabilities give a formula. For continuous \(X\), \(P(X=x)=0\) and a density may not exist. The foundational definition must condition on information, not divide by a probability-zero event.
\(E[Y\mid X=x]\) is a value of a chosen version of a function. The random variable is \(E[Y\mid X]\).
Definition: conditional expectation
Let \(Y\in L^1(P)\). A conditional expectation \(Z=E[Y\mid\mathcal A]\) is an \(\mathcal A\)-measurable random variable such that
\[\int_A Z\,dP=\int_A Y\,dP\qquad\text{for every }A\in\mathcal A.\]
It is unique up to equality almost surely.
Why existence and uniqueness hold
Apply the nonnegative Radon–Nikodym theorem separately to \(Y^+=\max\{Y,0\}\) and \(Y^-=\max\{-Y,0\}\). The finite measures
\[\nu^\pm(A)=\int_A Y^\pm\,dP\]
are absolutely continuous with respect to \(P|_{\mathcal A}\), so they have \(\mathcal A\)-measurable densities \(Z^\pm\). Then \(Z=Z^+-Z^-\) satisfies the defining integral identity, which constructs one version. If \(Z\) and \(\widetilde Z\) both satisfy the defining identity, then
\[\int_A(Z-\widetilde Z)\,dP=0\qquad\forall A\in\mathcal A,\]
so Corollary 2.10 gives \(Z=\widetilde Z\) almost surely.
Every phrase has a job
- \(\mathcal A\)-measurable: \(Z\) uses no information outside \(\mathcal A\).
- Equality on every \(A\in\mathcal A\): within every observable event, forecast and outcome have the same average.
- a.s. uniqueness: changing values on a null event changes nothing observable.
Retrieval check.
Does \(E[Y\mid\mathcal A]=Y\) always hold because their unconditional means agree?
No. It holds only if \(Y\) is \(\mathcal A\)-measurable. Matching one unconditional integral is far weaker than matching every \(A\in\mathcal A\).
Verify one conditional expectation
If \(\mathcal A=\sigma(B)\), every \(\mathcal A\)-measurable variable is constant on \(B\) and \(B^c\). Hence
\[E[Y\mid\sigma(B)]=E[Y\mid B]\mathbf1_B+E[Y\mid B^c]\mathbf1_{B^c},\]
when both conditional means are defined.
Conditional probability
For \(C\in\mathcal F\),
\[P(C\mid\mathcal A):=E[\mathbf1_C\mid\mathcal A].\]
It is an \([0,1]\)-valued random variable, not generally a scalar.
Conditioning on a random variable
Define
\[E[Y\mid X]:=E[Y\mid\sigma(X)].\]
The Doob–Dynkin lemma says that a \(\sigma(X)\)-measurable \(Z\) can be written \(Z=m(X)\) for a Borel function \(m\). Thus \(E[Y\mid X]=m(X)\).
Retrieval check.
Which is random before \(X\) is observed: \(E[Y\mid X]\) or \(E[Y\mid X=x]\)?
\(E[Y\mid X]=m(X)\) is random. For fixed \(x\), \(E[Y\mid X=x]=m(x)\) is a number.
Conditional densities recover the familiar formula
Suppose \((Y,X)\) has joint density \(f(y,x)\) with respect to \(\lambda\otimes\nu\), where \(\lambda,\nu\) are \(\sigma\)-finite reference measures. If \(E|g(Y,X)|<\infty\), then, on \(\{f_X(X)>0\}\),
\[f_{Y\mid X}(y\mid x)=\frac{f(y,x)}{f_X(x)},\qquad E[g(Y,X)\mid X]=\int g(y,X)f_{Y\mid X}(y\mid X)\,d\lambda(y)\quad\mathrm{a.s.}\]
Lebesgue measure gives the familiar continuous formula; counting measure gives the discrete one.
Example: bivariate normal forecasting
If \((Y,X)\) is jointly normal and \(\sigma_X>0\), then
\[E[Y\mid X=x]=\mu_Y+\rho\frac{\sigma_Y}{\sigma_X}(x-\mu_X).\]
The conditional mean is linear here because of normality. It need not be linear in general.
Rules, limits, and optimal prediction
Core algebra
For integrable \(Y,W\), constants \(a,b\), and \(Y\le W\) a.s.:
\[E[aY+bW\mid\mathcal A]=aE[Y\mid\mathcal A]+bE[W\mid\mathcal A],\qquad E[Y\mid\mathcal A]\le E[W\mid\mathcal A]\ \mathrm{a.s.}\]
If \(Y\) is \(\mathcal A\)-measurable, then \(E[Y\mid\mathcal A]=Y\) a.s.
The law of iterated expectations
\[E[E[Y\mid\mathcal A]]=E[Y].\]
Forecast first using information, then average over information: the original mean returns.
Tower property
If \(\mathcal A_0\subseteq\mathcal A\), then
\[E[E[Y\mid\mathcal A]\mid\mathcal A_0]=E[Y\mid\mathcal A_0] =E[E[Y\mid\mathcal A_0]\mid\mathcal A].\]
The first equality discards finer information. The second holds because \(E[Y\mid\mathcal A_0]\) is already \(\mathcal A\)-measurable.
Retrieval check.
May one replace \(E[E[Y\mid X]\mid Z]\) by \(E[Y\mid Z]\) for arbitrary \(X,Z\)?
No. One needs \(\sigma(Z)\subseteq\sigma(X)\), or another argument establishing the equality.
Take out what is known
If \(H\) is \(\mathcal A\)-measurable and the products are integrable, then
\[E[HY\mid\mathcal A]=H E[Y\mid\mathcal A].\]
The condition is information, not merely algebraic appearance.
Conditional variance and limits
If \(Y\in L^2\), conditional Jensen gives
\[E[Y\mid\mathcal A]^2\le E[Y^2\mid\mathcal A],\]
so \(\operatorname{Var}(Y\mid\mathcal A)\ge0\). If \(Y_n\ge0\), conditional Fatou gives
\[E[\liminf_nY_n\mid\mathcal A]\le\liminf_nE[Y_n\mid\mathcal A]\quad\mathrm{a.s.}\]
If \(Y_n\to Y\) a.s. and \(|Y_n|\le W\) for an integrable \(W\), conditional DCT gives \(E[Y_n\mid\mathcal A]\to E[Y\mid\mathcal A]\) a.s.
Best predictor theorem
Let \(Y\in L^2(P)\). Among all \(\sigma(X)\)-measurable \(G\in L^2(P)\),
\[m(X)=E[Y\mid X]\]
is the a.s.-unique minimizer of \(E[(Y-G)^2]\).
This is the population regression function. OLS is the best linear approximation to it.
Proof blueprint: projection identity
For any square-integrable \(g(X)\), write \(m(X)=E[Y\mid X]\):
\[E[(Y-g)^2]=E[(Y-m)^2]+E[(m-g)^2]+2E[(Y-m)(m-g)].\]
The final term is zero by conditioning on \(X\) and pulling out \(m(X)-g(X)\).
Retrieval check.
Why is \(E[(Y-m(X))(m(X)-g(X))\mid X]=0\)?
The second factor is \(\sigma(X)\)-measurable and \(E[Y-m(X)\mid X]=0\).
A loss-function boundary
Squared loss selects the conditional mean. Absolute loss selects a conditional median. There is no loss-free claim that “the conditional mean is the best forecast.”
Independence and causal identification
Independence: events first
Events \(A_1,\ldots,A_k\) are mutually independent if
\[P\left(\bigcap_{j=1}^k A_j\right)=\prod_{j=1}^kP(A_j)\]
for every finite selection of distinct events. Pairwise independence alone is weaker.
Independence of random elements
Random elements \(X_i\) are independent when their generated \(\sigma\)-algebras \(\sigma(X_i)\) are independent. This defines independence for vectors and processes, not only scalar variables.
Retrieval check.
If \(U,V\) are independent \(\operatorname{Bernoulli}(1/2)\) variables, are \(U,V,U\oplus V\) mutually independent?
No. They are pairwise independent, but \(U\oplus V\) is determined by \(U,V\).
C.d.f. factorization and transformations
For random variables \(X_1,\ldots,X_k\), independence is equivalent to
\[F_{X_1,\ldots,X_k}(x_1,\ldots,x_k)=\prod_{j=1}^kF_{X_j}(x_j).\]
If \(X\perp Z\), then \(g(X)\perp h(Z)\) for Borel \(g,h\).
From independent classes to independent \(\sigma\)-algebras
Lemma 4.4: let \(\mathcal C_i\) be independent collections of events, each closed under finite intersections. Then the generated \(\sigma\)-algebras \(\sigma(\mathcal C_i)\) are independent.
It is enough to verify factorization on tractable generating classes. A monotone-class argument extends the equality to every event in the generated \(\sigma\)-algebras.
Independent information is irrelevant
If \((Y,X_1)\perp X_2\), then
\[E[Y\mid X_1,X_2]=E[Y\mid X_1]\quad\mathrm{a.s.}\]
The strong assumption includes independence of \(X_2\) from both \(Y\) and \(X_1\) jointly.
Retrieval check.
Does \(Y\perp X_2\) alone imply \(E[Y\mid X_1,X_2]=E[Y\mid X_1]\)?
No. \(X_2\) can add information about \(Y\) jointly with \(X_1\) unless the stronger joint-independence condition holds.
Random assignment identifies an ATE
Let \(Y=DY(1)+(1-D)Y(0)\), where \(E|Y(d)|<\infty\) for \(d\in\{0,1\}\) and \(0<P(D=1)<1\). If the joint random-assignment condition \((Y(0),Y(1))\perp D\) holds, then
\[E[Y\mid D=1]-E[Y\mid D=0]=E[Y(1)-Y(0)].\]
The equality is an identification result, not a statement that treatment is harmless.
Synthesis and retrieval
Assumption audit
| Claim | Assumption doing the work |
|---|---|
| conditional expectation exists | \(E|Y|<\infty\) |
| pull-out rule | \(H\) is \(\mathcal A\)-measurable and product integrable |
| tower property | nested \(\sigma\)-algebras |
| density formula | joint density, \(\sigma\)-finite references, integrability, \(f_X(X)>0\) a.s. on the stated set |
| observed difference \(=\) ATE | integrable potential outcomes, overlap, and \((Y(0),Y(1))\perp D\) |
Common errors
- Do not divide by \(P(X=x)\) in general.
- Do not treat \(E[Y\mid X]\) as a scalar.
- Do not remove conditioning bars without nesting.
- Zero covariance is not independence.
- Pairwise independence is not mutual independence.
A short independence check
If \(Y\perp X\), then \(E[Y\mid X]=E[Y]\) a.s. But the converse for one particular \(Y\) does not establish independence of \(Y\) and \(X\).
Retrieval check.
Does \(E[Y\mid X]=E[Y]\) prove \(Y\perp X\)?
Not in general. It states mean independence; full independence requires the analogous condition for all bounded measurable functions of \(Y\).
Proof and calculation blueprint
- State the information set \(\mathcal A\).
- Verify measurability of the candidate forecast.
- Verify its integral identity on every \(A\in\mathcal A\).
- For a best-prediction claim, expand the square and kill the cross term by the tower property.
- For causal interpretation, state the potential-outcome independence assumption before simplifying.
Exit ticket 1
Let \(P(B)=1/2\), \(E[Y\mid B]=4\), and \(E[Y\mid B^c]=0\). Write \(E[Y\mid\sigma(B)]\) and \(E[Y]\).
\(E[Y\mid\sigma(B)]=4\mathbf1_B\) and \(E[Y]=2\).
Exit ticket 2
State the two defining properties of \(E[Y\mid\mathcal A]\) without looking.
It is \(\mathcal A\)-measurable and satisfies \(\int_AE[Y\mid\mathcal A]\,dP=\int_A Y\,dP\) for every \(A\in\mathcal A\).
Exit ticket 3
Give one condition sufficient for \(E[Y\mid X,Z]=E[Y\mid X]\).
One sufficient condition is \((Y,X)\perp Z\), with \(Y\) integrable.
Mastery checklist
- I can define \(E[Y\mid\mathcal A]\) and explain a.s. uniqueness.
- I can distinguish \(E[Y\mid X]\) from \(E[Y\mid X=x]\).
- I can use total expectation, the tower property, and the pull-out rule.
- I can prove the conditional-mean projection identity.
- I can state mutual independence and the random-assignment condition.