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The Lumenary · First-year graduate

Measure-theoretic probability

Name: ______________________

Date: ______________________

You will learn to

Work with probability spaces, independence and the four modes of convergence, and know which implies which (with counterexamples)
Prove the Borel–Cantelli lemmas and the strong law of large numbers under a fourth-moment hypothesis, and state Etemadi's general version
Define conditional expectation through the Radon–Nikodym theorem, prove its main properties and recognise it as an orthogonal projection
Prove the optional stopping theorem and Doob's martingale convergence theorem via upcrossings, and apply them
Prove the central limit theorem from Lévy's continuity theorem using characteristic functions

1.
Let $X, X_1, X_2, \dots$ be random variables on one probability space. Which implications hold in general? Select all that apply.
- A. $X_n \to X$ almost surely $\Rightarrow$ $X_n \to X$ in probability.
- B. $X_n \to X$ in probability $\Rightarrow$ $X_n \to X$ almost surely.
- C. $X_n \to X$ in $L^2$ $\Rightarrow$ $X_n \to X$ in $L^1$ .
- D. $X_n \to X$ in probability $\Rightarrow$ $X_n \to X$ in distribution.
- E. $X_n \to X$ almost surely $\Rightarrow$ $X_n \to X$ in $L^1$ .
- F. $X_n \to X$ in distribution $\Rightarrow$ $X_n \to X$ in probability.
2.
Let $X_1, X_2$ be independent fair dice (values $1, \dots, 6$ ). The conditional expectation $\mathbb E[X_1 X_2 \mid X_1]$ is a random variable. What value does it take on the event $\{X_1 = 4\}$ ?
3.
Let $A_1, A_2, \dots$ be events with $P(A_n) = 1/n^2$ . Nothing is assumed about independence. Which statement is true?
- A.Almost surely, only finitely many of the $A_n$ occur.
- B.Nothing can be said without independence.
- C.Almost surely, infinitely many of the $A_n$ occur.
- D. $P(\limsup A_n)$ can be any number in $[0, 1]$ , depending on the dependence structure.
4.
The proof of the second Borel–Cantelli lemma runs: (1) the complement of $\limsup A_k$ is $\bigcup_n \bigcap_{k \ge n} A_k^c$ ; (2) $P(\bigcap_{k=n}^N A_k^c) = \prod_{k=n}^N (1 - P(A_k))$ ; (3) $1 - x \le e^{-x}$ ; (4) continuity from above. Which step uses the independence of the events?
- A.Step (1).
- B.Step (2).
- C.Step (3).
- D.Step (4).
5.
On $\Omega = [0, 1)$ with Lebesgue measure let $X(\omega) = \omega$ and let $\mathcal G$ be the σ-algebra generated by $B_1 = [0, \tfrac12)$ and $B_2 = [\tfrac12, 1)$ . What is $\mathbb E[X \mid \mathcal G]$ ?
- A.The number $\tfrac12$ .
- B. $\tfrac14 \mathbf 1_{B_1} + \tfrac34 \mathbf 1_{B_2}$ .
- C. $X$ itself.
- D. $\tfrac12 \mathbf 1_{B_1} + \mathbf 1_{B_2}$ .
6.
In Mathlib, Martingale M 𝓕 μ means: M is adapted and μ[M j | 𝓕 i] =ᵐ[μ] M i whenever i ≤ j. Which Lean statement correctly expresses a martingale has constant expectation (with [IsFiniteMeasure μ] and hM : Martingale M 𝓕 μ, hij : i ≤ j)?
- A.∫ ω, M j ω ∂μ = ∫ ω, M i ω ∂μ
- B.∀ ω, M j ω = M i ω
- C.μ[M i | 𝓕 j] =ᵐ[μ] M j
- D.μ[M j | 𝓕 i] = fun _ => ∫ ω, M j ω ∂μ
7.
Let $X_1, \dots, X_4$ be independent fair signs, $P(X_i = 1) = P(X_i = -1) = \tfrac12$ , and $S_4 = X_1 + X_2 + X_3 + X_4$ . Compute $\mathbb E[S_4^4]$ using the expansion from the proof of the strong law.
8.
A simple symmetric random walk starts at $3$ and stops at the first time $T$ it reaches $0$ or $10$ . Using optional stopping for the martingales $S_n$ and $S_n^2 - n$ , compute $\mathbb E T$ .
9.
Let $\xi_1, \xi_2, \dots$ be independent with $P(\xi_i = 0) = P(\xi_i = 2) = \tfrac12$ and $M_n = \xi_1 \cdots \xi_n$ . Then $(M_n)$ is a non-negative martingale with $\mathbb E M_n = 1$ , and $M_n \to 0$ almost surely. Why does this not contradict $\mathbb E M_\infty = \lim \mathbb E M_n$ ?
- A.Because the martingale is not uniformly integrable, so the a.s. convergence given by Doob's theorem need not hold in $L^1$ .
- B.Because $(M_n)$ is not $L^1$ -bounded, so Doob's theorem does not apply.
- C.Because $M_n$ does not actually converge almost surely.
- D.Because $(M_n)$ is not a martingale: $\mathbb E[M_{n+1} \mid \mathcal F_n] = 2 M_n$ .
10.
A finite probability space has outcomes $0, 1, \dots, n-1$ with probabilities probs, a random variable given by the list X, and a σ-algebra $\mathcal G$ generated by a partition blocks (a list of lists of outcomes). Write conditional_expectation(probs, X, blocks) returning the list whose $\omega$ -th entry is $\mathbb E[X \mid \mathcal G](\omega)$ . Assume every block has positive probability. Use Fraction so the answers are exact.
Computational — Practice

Answer key — Measure-theoretic probability

1.
True: (a), (c), (d). False: (b) (independent $A_n$ with $P(A_n) = 1/n$ ), (e) ( $n\mathbf 1_{(0,1/n)}$ ) and (f) ( $X_n = -X$ with $X$ symmetric). The general picture: a.s. $\Rightarrow$ in probability $\Rightarrow$ in distribution, and $L^p \Rightarrow L^q$ ( $q \le p$ ) $\Rightarrow$ in probability; in probability gives a.s. convergence only along a subsequence.
2.
Taking out what is known and then using independence: $\mathbb E[X_1 X_2 \mid X_1] = X_1 \, \mathbb E[X_2 \mid X_1] = X_1 \, \mathbb E X_2 = \tfrac72 X_1$ . On $\{X_1 = 4\}$ this equals $\tfrac72 \cdot 4 = 14$ . The answer is a random variable, $\tfrac72 X_1$ ; its expectation is $\tfrac72 \cdot \tfrac72 = \mathbb E[X_1 X_2]$ , as property 3 requires.
3.
For every $n$ , $P(\limsup_k A_k) \le P(\bigcup_{k \ge n} A_k) \le \sum_{k \ge n} 1/k^2 \to 0$ . So $P(\limsup A_n) = 0$ : almost surely only finitely many $A_n$ occur. This is the first Borel–Cantelli lemma, valid for arbitrary events.
4.
Step (2): $P(A_n^c \cap \dots \cap A_N^c) = \prod (1 - P(A_k))$ is the product rule for independent events. If all $A_k$ equal one event $A$ with $P(A) = \tfrac12$ , the left side is $\tfrac12$ while the right side is $2^{-(N - n + 1)}$ , and the conclusion $P(\limsup A_k) = 1$ fails.
5.
$Y = c_1 \mathbf 1_{B_1} + c_2 \mathbf 1_{B_2}$ with $c_1 \cdot \tfrac12 = \int_0^{1/2} \omega \, d\omega = \tfrac18$ and $c_2 \cdot \tfrac12 = \int_{1/2}^1 \omega \, d\omega = \tfrac38$ . So $c_1 = \tfrac14$ , $c_2 = \tfrac34$ . It is a random variable; its expectation is $\tfrac12 \cdot \tfrac14 + \tfrac12 \cdot \tfrac34 = \tfrac12 = \mathbb E X$ . Geometrically, $Y$ is the orthogonal projection of $X$ onto the two-dimensional space of functions constant on each block.
6.
(a). The proof: ∫ M j = ∫ μ[M j | 𝓕 i] by integral_condExp (integrating a conditional expectation does not change the integral), and μ[M j | 𝓕 i] =ᵐ[μ] M i by the martingale property, so integral_congr_ae gives ∫ M i. Option (b) confuses constant expectation with constant paths; (c) reverses the indices; (d) treats a conditional expectation as a number.
7.
Here $\mathbb E X_i^4 = \mathbb E X_i^2 = 1$ , so $\mathbb E S_4^4 = 4 \cdot 1 + 3 \cdot 4 \cdot 3 \cdot 1 = 40$ . Check directly: $S_4 \in \{\pm 4, \pm 2, 0\}$ with probabilities $\tfrac{2}{16}, \tfrac{8}{16}, \tfrac{6}{16}$ , so $\mathbb E S_4^4 = \tfrac{2}{16} \cdot 256 + \tfrac{8}{16} \cdot 16 = 32 + 8 = 40$ . The growth $\mathbb E S_n^4 \le C n^2$ is what makes $\sum P(|S_n/n| > \varepsilon)$ converge.
8.
Optional stopping (justified by $\mathbb E T < \infty$ and $|S_{T \wedge n}| \le 10$ ) gives $\mathbb E S_T = 3$ , so $10 \, P(S_T = 10) = 3$ and $P(S_T = 10) = \tfrac{3}{10}$ . Then $\mathbb E S_T^2 = 100 \cdot \tfrac{3}{10} = 30$ and $\mathbb E[S_T^2 - T] = 3^2 = 9$ , so $\mathbb E T = 30 - 9 = 21 = a(N - a)$ .
9.
(a). $M_n$ equals $2^n$ with probability $2^{-n}$ and $0$ otherwise. It is $L^1$ -bounded, so Doob gives $M_n \to M_\infty$ a.s., and $M_\infty = 0$ . But $\{M_n\}$ is not uniformly integrable ( $\mathbb E[M_n \mathbf 1_{\{M_n > K\}}] = 1$ once $2^n > K$ ), so $\mathbb E M_\infty = 0 \ne 1$ . For a uniformly integrable martingale one would have $M_n = \mathbb E[M_\infty \mid \mathcal F_n]$ .
10.
```
from fractions import Fraction

def conditional_expectation(probs, X, blocks):
    result = [None] * len(X)
    for block in blocks:
        p_block = sum(Fraction(probs[w]) for w in block)
        value = sum(Fraction(probs[w]) * X[w] for w in block) / p_block
        for w in block:
            result[w] = value
    return result
```
On a block $B$ of the partition, $\mathbb E[X \mid \mathcal G] = \mathbb E[X \mathbf 1_B]/P(B)$ : the function is constant on blocks ( $\mathcal G$ -measurable) and has the same integral as $X$ over each block, hence over every set in $\mathcal G$ . The last test checks property 3, $\mathbb E[\mathbb E[X \mid \mathcal G]] = \mathbb E X$ .