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

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Finite Probability Spaces and Random Variables — Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

A loaded die as a nonuniform finite probability space

Example

On $Ω = {1, 2, 3, 4, 5, 6}$ assign weights $w (1) = 0, w (2) = \frac{1}{10}, w (3) = w (4) = \frac{1}{5}, w (5) = w (6) = \frac{1}{4} .$ This is a loaded die with a possible zero-weight outcome. Let $A = {4, 5, 6}$ and $B = {2, 4, 6}$ .

Facts & Assumptions

Given: The weights and events in the Example.

[L1]

Finite probability spaces allow nonnegative outcome weights summing to $1$ (Finite probability spaces, outcome weights, events, and event probabilities).

[L2]

Complements and differences obey the finite probability laws (Normalization, nonnegativity, monotonicity, complements, and differences in a finite probability space).

[L3]

$P (A ∣ B) = P (A \cap B) / P (B)$ when $P (B) > 0$ (Conditional probability $P (A ∣ B)$ for $P (B) > 0$ ).

[L4]

The multiplication rule is $P (A \cap B) = P (A ∣ B) P (B)$ (The multiplication rule and finite chain rule for conditional probability).

Verification

technique · direct

1.1

The six weights sum to $1$ , so [L1] gives a finite probability space; ${1}$ is nonempty but has probability zero.

L1algebra

1.2

Direct summation gives $P (A) = 7 / 10$ , $P (B) = 11 / 20$ , and $P (A \cap B) = 9 / 20$ .

L1L2algebra

2.1

Since $P (B) > 0$ , [L3] gives $P (A ∣ B) = (9 / 20) / (11 / 20) = 9 / 11$ .

step 1.2L3algebra

3.1

Finally $(9 / 11) (11 / 20) = 9 / 20 = P (A \cap B)$ , verifying [L4] in this nonuniform space.

step 1.2step 2.1L4algebra∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Bayes' theorem for choosing one of two urns and observing a colour

Example

Choose urn $U_{1}$ with probability $1 / 3$ and urn $U_{2}$ with probability $2 / 3$ , independently choose a uniform auxiliary value $r \in {1, 2, 3, 4}$ , and declare the observation red when $U_{1}$ was chosen and $r \leq 3$ , or when $U_{2}$ was chosen and $r = 1$ . Thus the red likelihoods are $3 / 4$ and $1 / 4$ , respectively. The posterior probability of $U_{1}$ after observing red is $3 / 5$ .

Facts & Assumptions

Given: The finite two-stage experiment in the Example.

[L1]

Bayes' theorem over a positive-probability finite partition gives posterior probability as prior times likelihood divided by total likelihood (Bayes' theorem over a finite partition).

[L2]

A finite product experiment assigns product weights to its branches (The finite product of finite probability spaces).

[L3]

Conditional probability is the probability of an intersection divided by the positive probability of the conditioning event (Conditional probability $P (A ∣ B)$ for $P (B) > 0$ ).

Verification

technique · direct

1.1

In the product space, the three red branches over $U_{1}$ have total weight $(1 / 3) (3 / 4) = 1 / 4$ , and the one red branch over $U_{2}$ has weight $(2 / 3) (1 / 4) = 1 / 6$ . Hence $P (r e d) = 5 / 12 > 0$ .

L2algebra

2.1

Directly, $P (U_{1} ∣ r e d) = (1 / 4) / (5 / 12) = 3 / 5$ .

step 1.1L3algebra

3.1

Bayes' formula [L1] has the same numerator $1 / 4$ and denominator $1 / 4 + 1 / 6 = 5 / 12$ , so it also gives $3 / 5$ .

step 1.1step 2.1L1algebra∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Three pairwise-independent events that are not mutually independent

Example

On the uniform space $Ω = {00, 01, 10, 11}$ , let $A$ be the event that the first bit is $0$ , let $B$ be the event that the second bit is $0$ , and let $C$ be the event that the two bits agree. Then $A, B, C$ are pairwise independent but not mutually independent.

Facts & Assumptions

Given: The uniform four-point space and events $A, B, C$ in the Example.

[L1]

In a uniform finite space, probability is cardinality divided by the outcome count (The uniform probability space on a nonempty finite set).

[L2]

Pairwise independence requires each pairwise intersection to have product probability, while mutual independence also requires the triple identity (Independent events, pairwise independence, and mutual independence of a finite family).

Verification

technique · direct

1.1

Each of $A, B, C$ has two outcomes and probability $1 / 2$ . Moreover $A \cap B = A \cap C = B \cap C = {00}$ , so every pairwise intersection has probability $1 / 4$ .

2.1

Therefore every pair satisfies $P (E \cap F) = 1 / 4 = (1 / 2) (1 / 2)$ .

step 1.1L2algebra

2.2

The triple intersection is ${00}$ and has probability $1 / 4$ , whereas the product of the three probabilities is $1 / 8$ .

step 1.1L1algebra

3.1

Thus the events are pairwise independent but not mutually independent.

step 2.1step 2.2L2∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Events with the same probability need not be independent

Statement refuted

If two events have the same probability, then they are independent.

Facts & Assumptions

Given: The uniform probability space on $Ω = {1, 2, 3}$ .

[L1]

Uniform event probability is event cardinality divided by $∣ Ω ∣$ (The uniform probability space on a nonempty finite set).

[L2]

Independence requires $P (A \cap B) = P (A) P (B)$ (Independent events, pairwise independence, and mutual independence of a finite family).

Counterexample

technique · constructive

1.1

Let $A = {1, 2}$ and $B = {2, 3}$ . Then $P (A) = P (B) = 2 / 3$ .

L1construct

2.1

Their intersection is ${2}$ , so $P (A \cap B) = 1 / 3 \neq 4 / 9 = P (A) P (B)$ .

step 1.1L1algebra

3.1

Thus equal event probabilities do not imply independence.

step 2.1L2discharge-construct∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

Uncorrelated finite random variables need not be independent

Statement refuted

If $Cov (X, Y) = 0$ , then the finite random variables $X$ and $Y$ are independent.

Facts & Assumptions

Given: The uniform space $Ω = {- 1, 0, 1}$ , the identity variable $X (ω) = ω$ , and $Y = X^{2}$ .

[L1]

Uniform probabilities are cardinality ratios (The uniform probability space on a nonempty finite set).

[L2]

Independence requires every joint attained-value probability to factor (Pairwise and mutual independence of finite-valued random variables).

[L3]

Covariance is defined by centred products and equals $E [X Y] - E [X] E [Y]$ (Variance, standard deviation, and covariance on a finite probability space, $Var (X) = E [X^{2}] - E [X]^{2}$ and $Cov (X, Y) = E [X Y] - E [X] E [Y]$ ).

Counterexample

technique · constructive

1.1

On the stated three-point space, symmetry gives $E [X] = 0$ and $E [X Y] = E [X^{3}] = 0$ , while $E [Y] = 2 / 3$ .

givenL1algebraconstruct

1.2

But $P (X = 0, Y = 0) = 1 / 3$ , whereas $P (X = 0) P (Y = 0) = (1 / 3) (1 / 3) = 1 / 9$ .

L1algebra

2.1

Hence $Cov (X, Y) = 0$ .

step 1.1L3algebra

3.1

Thus $X, Y$ are uncorrelated but not independent.

step 2.1step 1.2L2discharge-construct∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

For dependent variables, $E [X Y]$ need not equal $E [X] E [Y]$

Statement refuted

Every pair of finite random variables satisfies $E [X Y] = E [X] E [Y]$ .

Facts & Assumptions

Given: A uniform random sign $X$ on ${- 1, 1}$ and the variable $Y = X$ .

[L1]

A uniform two-point space gives both outcomes probability $1 / 2$ (The uniform probability space on a nonempty finite set).

[L2]

Independence requires joint attained-value probabilities to factor (Pairwise and mutual independence of finite-valued random variables).

[L3]

Expectation is the weighted finite sum of values (Expectation of a real random variable on a finite probability space).

Counterexample

technique · constructive

1.1

For the constructed pair $Y = X$ , directly $E [X] = E [Y] = 0$ , while $X Y = X^{2} = 1$ everywhere and hence $E [X Y] = 1$ .

givenL1L3algebraconstruct

1.2

Also $P (X = 1, Y = 1) = 1 / 2 \neq 1 / 4 = P (X = 1) P (Y = 1)$ , so the variables are dependent.

L1L2algebra

2.1

Thus $E [X Y] = 1 \neq 0 = E [X] E [Y]$ , refuting the statement.

step 1.1step 1.2discharge-construct∎

False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

False: linearity of expectation requires independence

Statement

The identity $E [X + Y] = E [X] + E [Y]$ is valid only when $X$ and $Y$ are independent.

Facts & Assumptions

Given: Finite real random variables on one finite probability space.

[L1]

Expectation is linear for every finite family of random variables, without any independence hypothesis (Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L2]

A uniform two-point space assigns each outcome probability $1 / 2$ (The uniform probability space on a nonempty finite set).

[L3]

Independence requires every joint attained-value probability to factor (Pairwise and mutual independence of finite-valued random variables).

Refutation

technique · direct

1.1

On the uniform random-sign space, take $X (ω) = ω$ and $Y = X$ . Then $P (X = 1, Y = 1) = 1 / 2 \neq 1 / 4 = P (X = 1) P (Y = 1)$ , so [L3] shows that $X$ and $Y$ are dependent.

L2L3construct

2.1

Nevertheless [L1] gives $E [X + Y] = E [X] + E [Y]$ . Therefore independence is not necessary for linearity, so the statement is false.

step 1.1L1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

The union bound can be strict for overlapping events

Example

On the uniform four-point space, let $A = {1, 2}$ , $B = {2, 3}$ , and $C = {2, 4}$ . Then the three events cover the space, but the sum of their probabilities is $3 / 2$ .

Facts & Assumptions

Given: The uniform space and events in the Example.

[L1]

Uniform event probabilities are cardinality ratios (The uniform probability space on a nonempty finite set).

[L2]

Two-event inclusion-exclusion subtracts the probability of the overlap (Two-event inclusion-exclusion: $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ ).

[L3]

The finite union bound compares union probability with the sum of marginal probabilities (The finite union bound).

Verification

technique · direct

1.1

Each event has probability $1 / 2$ , while $A \cup B \cup C = Ω$ has probability $1$ .

2.1

Thus [L3] reads $1 \leq 3 / 2$ , and the inequality is strict.

step 1.1L3algebra

3.1

The common outcome $2$ is counted in all three marginal probabilities; [L2] displays this overlap correction already for each pair.

step 1.1L2∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

A symmetric two-point distribution attains equality in Chebyshev's inequality

Example

For every $ε > 0$ , let $X$ be uniform on ${- ε, + ε}$ . At the weak threshold $ε$ , equality holds in Chebyshev's inequality.

Facts & Assumptions

Given: A real $ε > 0$ and the random variable in the Example.

[L1]

The two points in a uniform finite space each have probability $1 / 2$ (The uniform probability space on a nonempty finite set).

[L2]

A finite real random variable and its expectation are defined by finite weighted sums (Real random variables on finite probability spaces and their finite distributions, Expectation of a real random variable on a finite probability space).

[L3]

Variance is the expectation of the squared centred variable (Variance, standard deviation, and covariance on a finite probability space).

[L4]

Chebyshev states $P (∣ X - E [X] ∣ \geq t) \leq Var (X) / t^{2}$ for $t > 0$ (Chebyshev's inequality on a finite probability space).

Verification

technique · direct

1.1

Symmetry gives $E [X] = 0$ , and $X^{2} = ε^{2}$ everywhere, so $Var (X) = ε^{2}$ .

L1L2L3algebra

2.1

The event $∣ X - E [X] ∣ \geq ε$ is all of the outcome space and has probability $1$ .

step 1.1L1

3.1

The right side of [L4] at $t = ε$ is $ε^{2} / ε^{2} = 1$ , so equality holds. Positivity of $ε$ licenses the division.

step 1.1step 2.1L4algebra∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

A two-valued random variable attains equality in Markov's inequality

Example

Let $a > 0$ and $p \in [0, 1]$ . On a two-outcome finite probability space, give an event $E$ probability $p$ and define $X = a$ on $E$ and $X = 0$ off $E$ . Then equality holds in Markov's inequality at threshold $a$ .

Facts & Assumptions

Given: Parameters $a > 0$ , $p \in [0, 1]$ , and the construction in the Example.

[L1]

Zero outcome weights are permitted in a finite probability space (Finite probability spaces, outcome weights, events, and event probabilities).

[L2]

Real random variables and expectation are finite functions and weighted sums (Real random variables on finite probability spaces and their finite distributions, Expectation of a real random variable on a finite probability space).

[L3]

Markov gives $P (X \geq a) \leq E [X] / a$ for nonnegative $X$ and $a > 0$ (Markov's inequality on a finite probability space).

Verification

technique · direct

1.1

Give the two outcomes weights $p$ and $1 - p$ ; these are nonnegative and sum to $1$ , including at $p = 0, 1$ .

2.1

The variable is nonnegative, $E [X] = a p$ , and ${X \geq a} = E$ has probability $p$ .

step 1.1L2algebra

3.1

Hence $P (X \geq a) = p = a p / a = E [X] / a$ , so [L3] is sharp.

step 2.1L3algebra∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13Open item page →

Markov's conclusion can fail without nonnegativity

Statement refuted

The conclusion of Markov's inequality remains valid when the hypothesis $X \geq 0$ is removed.

Facts & Assumptions

Given: The uniform two-point probability space and a random variable $X$ taking values $1$ and $- 3$ .

[L1]

Each point in a uniform two-point space has probability $1 / 2$ (The uniform probability space on a nonempty finite set).

[L2]

A real random variable and its expectation are a function and its weighted finite sum (Real random variables on finite probability spaces and their finite distributions, Expectation of a real random variable on a finite probability space).

[L3]

Markov's theorem assumes that $X$ is nonnegative (Markov's inequality on a finite probability space).

Counterexample

technique · constructive

1.1

For the constructed two-valued variable, direct calculation gives $E [X] = (1 - 3) / 2 = - 1$ .

givenL1L2algebraconstruct

2.1

At threshold $a = 1$ , $P (X \geq 1) = 1 / 2$ , while $E [X] / a = - 1$ .

step 1.1L1algebra

3.1

The purported conclusion would be $1 / 2 \leq - 1$ , which is false. This refutes removal of the nonnegativity hypothesis, not [L3].

step 2.1L3discharge-construct∎

Sources

Standard references

Recommended treatments; not extraction sources.

Finite Probability Spaces and Random Variables — Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

A loaded die as a nonuniform finite probability space

Example

Facts & Assumptions

Verification

Bayes' theorem for choosing one of two urns and observing a colour

Example

Facts & Assumptions

Verification

Three pairwise-independent events that are not mutually independent

Example

Facts & Assumptions

Verification

Events with the same probability need not be independent

Statement refuted

Facts & Assumptions

Counterexample

Uncorrelated finite random variables need not be independent

Statement refuted

Facts & Assumptions

Counterexample

For dependent variables, E[XY]E[XY] need not equal E[X]E[Y]E[X]E[Y]

Statement refuted

Facts & Assumptions

Counterexample

False: linearity of expectation requires independence

Statement

Facts & Assumptions

Refutation

The union bound can be strict for overlapping events

Example

Facts & Assumptions

Verification

A symmetric two-point distribution attains equality in Chebyshev's inequality

Example

Facts & Assumptions

Verification

A two-valued random variable attains equality in Markov's inequality

Example

Facts & Assumptions

Verification

Markov's conclusion can fail without nonnegativity

Statement refuted

Facts & Assumptions

Counterexample

Sources

Standard references

For dependent variables, $E [X Y]$ need not equal $E [X] E [Y]$