2021.Q1: Conditional Expectation via Symmetry and Radon–Nikodym
2021 Probability Prelim Exam (PDF)
Short Introduction
This problem is testing:
- Your ability to read a σ-algebra geometrically, not just symbolically.
- Your understanding of conditional expectation as a Radon–Nikodym derivative, not as a heuristic “average”.
- Your ability to verify a conditional expectation using the defining integral property, rather than pointwise arguments.
Main tools needed:
- σ-algebra generated by a symmetry (pairing $\omega$ with $\omega+1$)
- Measurability via preimages
- Radon–Nikodym theorem
- Definition and uniqueness of conditional expectation
- Careful use of integrals $\int_B dP = P(B)$
Pattern recognition cue:
- Whenever a σ-algebra “forgets” information by identifying points, expect averaging over atoms.
Problem Statement (verbatim)
Problem 1. Let $(\Omega,\mathcal F, P)$ be a probability space where $\Omega=[-1,1]\subset\mathbb R$,
$\mathcal F$ represents the Borel sets contained in $\Omega$ and
$P=\lambda/2$, where $\lambda$ is the Lebesgue measure.
In what follows $W,X,Y$ are random variables which are defined on $(\Omega,\mathcal F,P)$.
Let
\(\mathcal G=\{B\in\mathcal F: B=A\cup(A+1),\ A\subset[-1,0]\},\) where $A+1={\omega+1:\omega\in A}$.Example: $A=(-\tfrac23,-\tfrac13)$, $A+1=(\tfrac13,\tfrac23)$, and
$B=(-\tfrac23,-\tfrac13)\cup(\tfrac13,\tfrac23)$.You can assume without proof that $\mathcal G\subset\mathcal F$ is a σ-algebra.
(a) Let $X$ be a random variable that satisfies
$X(\omega)=X(\omega+1)$, $\omega\in[-1,0]$.
(Example $X(\omega)=\sin(2\pi\omega)$).
Prove that ${X\le x}\in\mathcal G$, $x\in\mathbb R$, namely $X$ is $\mathcal G$-measurable.(b) Let $Y$ be a random variable. Prove that \(\tilde Y(\omega)\equiv \begin{cases} \dfrac{Y(\omega)+Y(\omega+1)}{2}, & -1\le\omega\le 0,\\)6pt] \dfrac{Y(\omega)+Y(\omega-1)}{2}, & 0<\omega\le 1 \end{cases} $$ is $\mathcal G$-measurable.
(c) Let \(W(\omega)= \begin{cases} 1, & -1\le\omega\le-\tfrac12,\\ 0, & -\tfrac12<\omega\le 1. \end{cases}\) What is $E(W\mid\mathcal G)$? Verify that your answer is correct.
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Solutions by Part
Part (a) — Measurability under the symmetry
Claim.
If $X(\omega)=X(\omega+1)$ for $\omega\in[-1,0]$, then $X$ is $\mathcal G$-measurable.
Proof.
Fix $x\in\mathbb R$ and define
\(A=\{\omega\in[-1,0]: X(\omega)\le x\}.\)
Then
\(\{X\le x\}
=\{\omega\in[-1,0]:X(\omega)\le x\}
\cup
\{\omega\in[0,1]:X(\omega)\le x\}.\)
Write $\omega=\lambda+1$ with $\lambda\in[-1,0]$. Since $X(\lambda)=X(\lambda+1)$,
\(\{\omega\in[0,1]:X(\omega)\le x\}
=
\{\lambda+1:\lambda\in[-1,0],X(\lambda)\le x\}
=A+1.\)
Hence ${X\le x}=A\cup(A+1)\in\mathcal G$. ∎
Conclusion.
${X\le x}\in\mathcal G$ for all $x$, so $X$ is $\mathcal G$-measurable.
Key Takeaways.
- Measurability is about preimages of Borel sets, not values.
- Symmetry conditions often imply measurability via σ-algebra generators.
- Write sets explicitly and rewrite using the symmetry.
Part (b) — Averaging produces a $\mathcal G$-measurable variable
Claim.
$\tilde Y$ defined by pairwise averaging is $\mathcal G$-measurable.
Proof.
For $\omega\in[-1,0]$,
\(\tilde Y(\omega)=\frac{Y(\omega)+Y(\omega+1)}{2}.\)
Then $\omega+1\in[0,1]$ and
\(\tilde Y(\omega+1)
=\frac{Y(\omega+1)+Y(\omega+1-1)}{2}
=\frac{Y(\omega+1)+Y(\omega)}{2}
=\tilde Y(\omega).\)
Thus $\tilde Y(\omega)=\tilde Y(\omega+1)$ on $[-1,0]$. By part (a), this implies $\tilde Y$ is $\mathcal G$-measurable. ∎
Conclusion.
Pairwise averaging along the symmetry produces a $\mathcal G$-measurable random variable.
Key Takeaways.
- σ-algebras defined by identifications force constancy on atoms.
- Averaging is not heuristic, it is forced by measurability.
- This construction is the prototype for conditional expectation here.
Part (c) — Computing $E(W\mid\mathcal G)$
Claim.
\(E(W\mid\mathcal G)
=\frac12\,\mathbf 1_{[-1,-1/2]\cup[0,1/2]}.\)
Proof.
Step 1: Non-measurability.
Since $W(\omega)\neq W(\omega+1)$ on a set of positive measure, $W$ is not $\mathcal G$-measurable.
Step 2: Define the induced measure.
Define a finite signed measure $\nu$ on $(\Omega,\mathcal G)$ by
\(\nu(B)=\int_B W\,dP,\qquad B\in\mathcal G.\)
Then $\nu\ll P|_{\mathcal G}$.
Step 3: Apply Radon–Nikodym.
By the Radon–Nikodym theorem, there exists a unique (a.s.) $\mathcal G$-measurable function $\bar W$ such that
\(\int_B \bar W\,dP=\int_B W\,dP\quad\forall B\in\mathcal G.\)
By definition, $\bar W=E(W\mid\mathcal G)$.
Step 4: Identify $\bar W$ using part (b).
By part (b), the function
\(\tilde W(\omega)=
\begin{cases}
\dfrac{W(\omega)+W(\omega+1)}{2}, & \omega\in[-1,0],
\dfrac{W(\omega)+W(\omega-1)}{2}, & \omega\in[0,1]
\end{cases}\)
is $\mathcal G$-measurable.
A direct computation shows:
- $\tilde W(\omega)=\tfrac12$ for $\omega\in[-1,-\tfrac12]\cup[0,\tfrac12]$,
- $\tilde W(\omega)=0$ otherwise.
Step 5: Verification.
For any $B=A\cup(A+1)\in\mathcal G$,
\(\int_B \tilde W\,dP
=\int_B W\,dP.\)
Hence, by uniqueness of the Radon–Nikodym derivative,
$\tilde W=\bar W$ a.s.
Conclusion.
\(E(W\mid\mathcal G)
=\frac12\,\mathbf 1_{[-1,-1/2]\cup[0,1/2]}.\)
Key Takeaways.
- Conditional expectation is defined by integral identities, not pointwise agreement.
- $\int_B dP=P(B)$ is a critical identity.
- The conditional expectation of an indicator is a conditional probability, hence can take values in $[0,1]$.
- Uniqueness is up to a.s. equality of functions, not measures.
Master Key Takeaways
- Conditional expectation replaces unknown outcomes by conditional probabilities.
- σ-algebra atoms dictate where averaging occurs.
- Radon–Nikodym provides existence and uniqueness of $E(\cdot\mid\mathcal G)$.
- Never test conditional expectation on points or arbitrary intervals.
Cheat Sheet Entries to Extract
- If $\nu(B)=\int_B X\,dP$ on $\mathcal G$, then $E(X\mid\mathcal G)=\frac{d\nu}{dP}$.
- If $Z$ is constant on $B\in\mathcal G$, then $\int_B Z\,dP=Z\cdot P(B)$.
- For indicators: $E(\mathbf 1_A\mid\mathcal G)=P(A\mid\mathcal G)$.
Notes on My Original Work
- Strong geometric intuition about “partial information” was correct.
- Initial confusion came from mixing pointwise reasoning with integral identities.
- Once sets $B\in\mathcal G$ were treated correctly, the solution became straightforward.
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