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✓ 6 results · all verified · 5 also independently AI-judged

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Independence Borel Cantelli and Zero One Laws - Examples

1 · Prerequisites

2 · Summary

These examples isolate the main behaviors proved on the A page. Independence is not the same as disjointness, measurable functions of disjoint independent blocks stay independent, and Borel-Cantelli turns a summable tail estimate into an eventual almost-sure bound.

The counterexamples keep the scope exact. A divergent sum of probabilities does not force infinitely many occurrences without independence, and even when two events start independent, overlapping unions can destroy that independence.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Independent events need not be disjoint

Example

On the uniform space $Ω = {00, 01, 10, 11}$ , let $A : = {00, 01}, B : = {00, 10} .$ Then $A$ and $B$ are independent, but they are not disjoint because $A \cap B = {00}$ .

Facts & Assumptions

Given: The uniform four-point space and the events $A, B$ displayed above.

[L1]

In a uniform finite space, event probability is cardinality divided by the total number of outcomes. (The uniform probability space on a nonempty finite set)

[L2]

Two events are independent exactly when $P (A \cap B) = P (A) P (B)$ . (Independent events, pairwise independence, and mutual independence of a finite family)

Verification

technique · direct

1.1

By [L1], the events $A$ and $B$ each have probability $2 / 4 = 1 / 2$ , while $A \cap B = {00}$ has probability $1 / 4$ .

2.1

Step 1.1 gives $P (A \cap B) = \frac{1}{4} = \frac{1}{2} \cdot \frac{1}{2} = P (A) P (B),$ so [L2] shows that $A$ and $B$ are independent. Since $00 \in A \cap B$ , they are not disjoint.

step 1.1L2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Functions of disjoint independent coordinate blocks remain independent

Example

Let $X_{0}, X_{1}, X_{2}, X_{3}$ be independent random elements. Put $Y : = (X_{0}, X_{1}), Z : = (X_{2}, X_{3}) .$ If $f$ and $g$ are measurable maps on the targets of $Y$ and $Z$ , then $f (Y)$ and $g (Z)$ are independent.

This is the standard way to pass from coordinate independence to independence of functions built from disjoint coordinate blocks.

Facts & Assumptions

Given: Independent random elements $X_{0}, X_{1}, X_{2}, X_{3}$ and measurable maps $f$ and $g$ as in the Example.

[L1]

Disjoint groups of an independent sigma-algebra family remain independent. (Disjoint groups of an independent sigma-algebra family remain independent)

[L2]

Measurable coordinatewise functions preserve independence. (Measurable coordinatewise functions preserve independence)

[L3]

Independence of random elements is defined through independence of their generated sigma-algebras. (Independent random elements)

Verification

technique · direct

1.1

The sigma-algebras $σ (X_{0}), σ (X_{1}), σ (X_{2}), σ (X_{3})$ are independent by [L3]. Grouping the first two and last two coordinates, [L1] shows that the block sigma-algebras $σ (X_{0}, X_{1})$ and $σ (X_{2}, X_{3})$ are independent. Therefore the block random elements $Y$ and $Z$ are independent.

L1L3

2.1

Applying [L2] to the independent pair $Y, Z$ and the measurable maps $f, g$ gives independence of $f (Y)$ and $g (Z)$ .

step 1.1L2∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

A summable tail bound becomes an almost-sure eventual bound

Example

Let $(X_{n})_{n \in N}$ be real random variables and let $(a_{n})_{n \in N}$ be positive real numbers. If $\sum_{n = 0}^{\infty} P (∣ X_{n} ∣ > a_{n}) < + \infty,$ then with probability $1$ only finitely many indices $n$ satisfy $∣ X_{n} ∣ > a_{n}$ . Equivalently, almost surely there is $N$ such that $∣ X_{n} ∣ \leq a_{n}$ for every $n \geq N$ .

Facts & Assumptions

Given: Real random variables $X_{n}$ and positive thresholds $a_{n}$ with $\sum_{n = 0}^{\infty} P (∣ X_{n} ∣ > a_{n}) < + \infty$ .

[L1]

If the probabilities of events are summable, then the corresponding events occur infinitely often with probability $0$ . (First Borel-Cantelli lemma for events)

Verification

technique · direct

1.1

Apply [L1] to the events $A_{n} : = {∣ X_{n} ∣ > a_{n}}$ . Then $P (∣ X_{n} ∣ > a_{n} i . o .) = 0.$

2.1

The complement of the event in step 1.1 is exactly the event that there is an index $N$ such that $∣ X_{n} ∣ \leq a_{n}$ for all $n \geq N$ . Hence that eventual bound holds almost surely.

step 1.1algebra∎

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-09-05Open item page →

Convergence of an independent random series is a zero-one event

Example

If $(X_{n})_{n \in N}$ is an independent sequence of real random variables, then the event ${\sum_{n = 0}^{\infty} X_{n} converges}$ has probability either $0$ or $1$ .

The zero-one law does not decide which value occurs; it says only that no intermediate probability is possible.

Facts & Assumptions

Given: An independent sequence of real random variables $(X_{n})_{n \in N}$ .

[L1]

Almost-sure convergence of an independent series is a tail event of probability $0$ or $1$ . (Almost-sure convergence of an independent series is a zero-one event)

Verification

technique · direct

1.1

The displayed event is exactly the convergence event covered by [L1].

2.1

Therefore its probability belongs to ${0, 1}$ .

step 1.1algebra∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

A divergent probability sum does not force infinitely many occurrences without independence

Statement refuted

If $(A_{n})_{n \in N}$ is a sequence of events with $\sum_{n = 0}^{\infty} P (A_{n}) = + \infty,$ then $P (A_{n} i . o .) = 1.$

Facts & Assumptions

Given: An event $A$ with $0 < P (A) < 1$ , and define $A_{n} : = A$ for every $n \in N$ .

[L1]

The event $A_{n} i . o .$ is the event that infinitely many of the $A_{n}$ occur. (Limsup and the infinitely often event)

[L2]

Probability measures respect complements and monotone set identities. (Basic identities for a probability measure)

Counterexample

technique · constructive

1.1

Since every $A_{n}$ equals $A$ , one has $\sum_{n = 0}^{\infty} P (A_{n}) = \sum_{n = 0}^{\infty} P (A) = + \infty .$

givenalgebraconstruct

1.2

For every outcome $ω$ , either $ω \in A$ and then $ω \in A_{n}$ for all $n$ , or $ω \notin A$ and then $ω \notin A_{n}$ for all $n$ . Hence the infinitely-often event is exactly $A$ itself: $A_{n} i . o . = A .$

2.1

Therefore $P (A_{n} i . o .) = P (A),$ and the chosen hypothesis $0 < P (A) < 1$ makes this probability strictly between $0$ and $1$ . So the displayed implication is false without an independence hypothesis.

step 1.1step 1.2L2discharge-construct∎

CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05Open item page →

Unions of overlapping independent events need not remain independent

Statement refuted

Whenever $A$ and $B$ are independent events, and $C$ overlaps both of them, the unions $A \cup C$ and $B \cup C$ are also independent.

Facts & Assumptions

Given: The uniform four-point space $Ω = {00, 01, 10, 11}$ and the events $A : = {00, 01}, B : = {00, 10}, C : = {00, 11} .$

[L1]

In a uniform finite space, event probability is cardinality divided by the total number of outcomes. (The uniform probability space on a nonempty finite set)

[L2]

Two events are independent exactly when $P (E \cap F) = P (E) P (F)$ . (Independent events, pairwise independence, and mutual independence of a finite family)

Counterexample

technique · constructive

1.1

By [L1], the events $A$ and $B$ have probability $1 / 2$ , and $A \cap B = {00}$ has probability $1 / 4$ . Hence $A$ and $B$ are independent by [L2]. Also $C$ overlaps both of them because $00 \in A \cap C$ and $00 \in B \cap C$ .

L1L2construct

2.1

The unions are $A \cup C = {00, 01, 11}, B \cup C = {00, 10, 11},$ so each has probability $3 / 4$ , while $(A \cup C) \cap (B \cup C) = {00, 11}$ has probability $1 / 2$ .

step 1.1L1

3.1

Since $P ((A \cup C) \cap (B \cup C)) = \frac{1}{2} \neq \frac{9}{16} = \frac{3}{4} \cdot \frac{3}{4},$ [L2] shows that $A \cup C$ and $B \cup C$ are not independent. This refutes the statement.

step 2.1L2discharge-construct∎