Finite joint distribution: validating a joint mass table and aggregating marginals and events from one joint law

On this page
  1. Goal and scope
  2. Givens and symbols
  3. Rules and sources
  4. Main derivation
  5. Substeps and lemmas
  6. Reverse audits
  7. Error controls and negative controls
  8. Result and limitations
  9. Verification and source binding

← Guide index · Original MKO

derivation-mko-finite-joint-distribution-01 · production-20261019-adopted-96

Goal and scope

Given the support S={(0,0),(0,1),(1,0),(1,1),(2,1)} with masses p(0,0)=1/6, p(0,1)=1/3, p(1,0)=1/4, p(1,1)=1/12, p(2,1)=1/6, verify that (S,p) is a two-dimensional finite joint distribution; from this one joint law compute the marginals of X and Y, the rectangle event P(X∈{1,2}, Y=1) and the single-pair mass P(X=2, Y=0); finally construct a different joint law with the same marginals, showing that two marginals do not in general determine the joint distribution.

Stays inside the MKO: a finite joint law whose support consists of distinct integer pairs with exact rational masses (assumption a1), treated as a mathematical model, not as data fitting or a continuous density (assumption a2). The guide validates one concrete table and proves the product-law lemma in general; it does not treat independence, covariance or conditional expectation (later nodes of the MKO), and does not change the canonical MKO or its English edition.

Givens and symbols

  • cartesian_product_def
  • frac_rules
  • finite_sum_rules
  • order_basics
  • numeric_evaluation
  • finite_sum_regroup
  • G1 — Table: S={(0,0),(0,1),(1,0),(1,1),(2,1)} with p(0,0)=1/6, p(0,1)=1/3, p(1,0)=1/4, p(1,1)=1/12, p(2,1)=1/6, all exact rationals.
  • G2 — X and Y are the two coordinates of the same pair: at (x,y), X=x and Y=y. Every event probability is aggregated from this one table.
  • G3 — A={1,2} and B={1}.
  • S — support: the listed distinct integer pairs
  • p(x,y) — exact mass of the pair (x,y); pairs outside S have mass 0
  • X,\ Y — coordinate projections: X=x and Y=y at the pair (x,y)
  • p_X,\ p_Y — marginal masses of X and of Y
  • A\times B — rectangle event: first coordinate in A and second coordinate in B
  • Q — the other joint law formed by multiplying the two marginals

Rules and sources

  • cartesian_product_def · prerequisite — Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d.
  • frac_rules · prerequisite — Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0.
  • finite_sum_rules · prerequisite — Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i.
  • order_basics · prerequisite — Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw.
  • numeric_evaluation · prerequisite — Exact evaluation of an explicit integer or rational expression; independently recomputable.
  • def_unfold · prerequisite — Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true.
  • case_analysis · prerequisite — Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility.
  • and_intro_elim · prerequisite — Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q.
  • prob_axioms · prerequisite — Probability axioms: P(A)≥0; P(Ω)=1; the probability of a (countable) union of pairwise disjoint events is the sum of their probabilities. Hence P(∅)=0 and P(Aᶜ)=1−P(A).
  • set_operations_def · prerequisite — Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A.
  • set_extensionality · prerequisite — Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements.
  • distributive · prerequisite — Distributivity: u(v+w)=uv+uw; with commutativity also (v+w)u=vu+wu.
  • mul_one · prerequisite — Multiplicative identity: 1·u=u.
  • function_def · prerequisite — Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every a.
  • exists_intro · prerequisite — ∃-introduction: if some a∈D satisfies P(a), then ∃x∈D, P(x).
  • quantifier_negation · prerequisite — Negating quantifiers: ¬(∀x, P(x)) is equivalent to ∃x, ¬P(x), and ¬(∃x, P(x)) is equivalent to ∀x, ¬P(x).
  • mko_joint_def · cited — A two-dimensional finite joint distribution lists distinct pairs (x,y) and nonnegative exact masses summing exactly to 1. Marginals, rectangle events and single-pair masses all aggregate from that same joint law. · mko-finite-joint-distribution
  • joint_event_sum · prerequisite — A finite joint law (S,p) gives event probabilities: for C⊆ℤ², P((X,Y)∈C)=Σ_{s∈S∩C} p(s); pairs not in S have mass 0. This is the additivity of the probability axioms over disjoint single-pair events.
  • marginal_def · prerequisite — Marginals: p_X(x)=P(X=x)=Σ_{y:(x,y)∈S} p(x,y) and p_Y(y)=P(Y=y)=Σ_{x:(x,y)∈S} p(x,y), the probabilities of the events {x}×ℤ and ℤ×{y}.
  • finite_sum_regroup · prerequisite — Regrouping a finite sum: if a finite set S is split into pairwise disjoint pieces S_1, S_2 and so on (each element in exactly one piece), then Σ_{s∈S} f(s) equals the total of the sums over the pieces.
  • L_product_law · lemma — Let p_X on a finite integer set S_X and p_Y on a finite integer set S_Y be nonnegative exact rational masses, each summing exactly to 1. Then Q(x,y)=p_X(x)p_Y(y) on S_X×S_Y is a finite joint law whose two marginals are exactly p_X and p_Y.

Main derivation

S01 · definition

Before

\text{table }(S,p)

After

\text{(i) distinct pairs, (ii) } p(s)\ge 0 \text{ exact, (iii) } \sum_{s\in S}p(s)=1

Conditions: none

Rule: mko_joint_def · Uses: G1, G2

Leaf basis: mko_joint_def + def_unfold

By the MKO, a table is a two-dimensional finite joint distribution only if it meets three requirements at once: (i) the listed pairs are distinct; (ii) every mass is a nonnegative exact number; (iii) the masses sum to exactly 1, not approximately 1. Each requirement is checked below, and no marginal or event is computed before all three pass.

S02 · implies

Before

(0,0),\ (0,1),\ (1,0),\ (1,1),\ (2,1)

After

\text{five pairwise distinct pairs}

Conditions: none

Rule: cartesian_product_def · Uses: S01, G1, cartesian_product_def

Leaf basis: cartesian_product_def + case_analysis

Two ordered pairs are equal only if both coordinates agree. Group by first coordinate: two pairs with different first coordinates differ; with first coordinate 0, (0,0) and (0,1) differ in the second coordinate; with first coordinate 1, (1,0) and (1,1) differ in the second coordinate; first coordinate 2 has only (2,1). So all C(5,2)=10 comparisons give 'not equal', and requirement (i) holds.

S03 · implies

Before

p=\tfrac16,\ \tfrac13,\ \tfrac14,\ \tfrac1{12},\ \tfrac16

After

p(s)>0\ \text{for every } s\in S

Conditions: none

Rule: order_basics · Uses: S01, G1

Leaf basis: order_basics + frac_rules

Each mass is a quotient of two positive integers, so it is a positive exact rational and in particular nonnegative. Requirement (ii) holds, and exact rationals also meet MKO assumption a1.

S04 · equivalent

Before

\sum_{s\in S}p(s)=\tfrac16+\tfrac13+\tfrac14+\tfrac1{12}+\tfrac16

After

\sum_{s\in S}p(s)=1

Conditions: none

Rule: frac_rules · Uses: S01, G1

Leaf basis: frac_rules + numeric_evaluation

Compute the total with exact fraction arithmetic (details in S04.1 and S04.2): the result is exactly 1, so requirement (iii) holds. Rounded decimals may not be used instead.

S05 · implies

Before

\text{(i), (ii), (iii) checked}

After

(S,p)\ \text{is a finite joint law}

Conditions: none

Rule: and_intro_elim · Uses: S01, S02, S03, S04

Leaf basis: mko_joint_def + and_intro_elim

The three requirements hold by S02, S03 and S04; combining them, by the definition in S01, (S,p) is a two-dimensional finite joint distribution. From now on any event probability may be aggregated from this one table.

S06 · definition

Before

C\subseteq\mathbb{Z}^2

After

P((X,Y)\in C)=\sum_{s\in S\cap C}p(s)

Conditions: (S,p)\ \text{valid}

Rule: joint_event_sum · Uses: S05, G2, joint_event_sum

Leaf basis: joint_event_sum + prob_axioms

A valid joint law determines every event probability: add the masses of the support pairs inside the event; pairs outside S contribute 0. This is additivity over disjoint single-pair events. For example C=ℤ² gives the total 1 of S04, in line with P(Ω)=1.

S07 · implies

Before

p_X(x)=\sum_{y:(x,y)\in S}p(x,y)

After

p_X(0)=\tfrac12,\ p_X(1)=\tfrac13,\ p_X(2)=\tfrac16

Conditions: none

Rule: marginal_def · Uses: S06, marginal_def, G1

Leaf basis: marginal_def + joint_event_sum + frac_rules

The marginal of X is the probability of the event {X=x}={x}×ℤ; by S06 add the masses of the support pairs in that event. First list the pairs for each x (S07.1), then add (S07.2).

S08 · equivalent

Before

\sum_{x}p_X(x)=\sum_{x}\sum_{y:(x,y)\in S}p(x,y)

After

\sum_{x}p_X(x)=\sum_{s\in S}p(s)=1

Conditions: \text{the fibres }S\cap(\{x\}\times\mathbb{Z})\text{ partition }S

Rule: finite_sum_regroup · Uses: S04, S07.1, S07.2, finite_sum_regroup

Leaf basis: finite_sum_regroup + numeric_evaluation

The three sets of S07.1 partition S, so by regrouping the finite sum, adding the three marginal values gives the total mass of the support, which is 1. This holds for every valid finite joint law, not only this table; the numeric check here is 1/2+1/3+1/6=1.

S09 · implies

Before

p_Y(y)=\sum_{x:(x,y)\in S}p(x,y)

After

p_Y(0)=\tfrac{5}{12},\ p_Y(1)=\tfrac{7}{12}

Conditions: none

Rule: marginal_def · Uses: S06, marginal_def, G1

Leaf basis: marginal_def + joint_event_sum + frac_rules

Group by second coordinate: the pairs with Y=0 are (0,0) and (1,0), with total 2/12+3/12=5/12; the pairs with Y=1 are (0,1), (1,1) and (2,1), with total 4/12+1/12+2/12=7/12. They add to 1, consistent with regrouping as in S08.

S10 · implies

Before

P(X\in A,\ Y\in B)=P((X,Y)\in A\times B)

After

P(X\in A,\ Y\in B)=\tfrac14

Conditions: none

Rule: joint_event_sum · Uses: S06, G3

Leaf basis: joint_event_sum + cartesian_product_def

A rectangle event requires both coordinates to fall in their sets at the same time, so it is the event that (X,Y) belongs to A×B. By S06, first find S∩(A×B) (S10.1), then add the masses (S10.2).

S11 · implies

Before

P(X\in A)\,P(Y\in B)=\left(\tfrac13+\tfrac16\right)\tfrac{7}{12}

After

\tfrac{7}{24}\neq\tfrac{6}{24}=P(X\in A,\ Y\in B)

Conditions: none

Rule: numeric_evaluation · Uses: S07, S09, S10

Leaf basis: numeric_evaluation + frac_rules + prob_axioms

P(X∈A)=p_X(1)+p_X(2)=1/3+1/6=1/2 (the events {X=1} and {X=2} are disjoint) and P(Y∈B)=p_Y(1)=7/12, with product 7/24; but S10 gives the rectangle probability 1/4=6/24. So for this joint law the rectangle probability is not the product of the marginal probabilities and must be aggregated from the joint table itself.

S12 · implies

Before

P(X=2,\ Y=0)

After

0\neq\tfrac{5}{72}=p_X(2)\,p_Y(0)

Conditions: none

Rule: joint_event_sum · Uses: S06, S07, S09, G1

Leaf basis: joint_event_sum + finite_sum_rules + frac_rules

The only pair of S with first coordinate 2 is (2,1), so (2,0)∉S; the event {(2,0)} contains no support pair and by S06 its probability is the empty sum 0. On the other hand p_X(2)p_Y(0)=(1/6)(5/12)=5/72>0. A single-pair mass must be read from the joint law, not obtained by multiplying marginals.

S13 · theorem_application

Before

p_X\ \text{on}\ \{0,1,2\},\ p_Y\ \text{on}\ \{0,1\}

After

Q(x,y)=p_X(x)\,p_Y(y)\ \text{is a finite joint law with marginals}\ p_X,\ p_Y

Conditions: p_X\ge 0,\ \sum_x p_X(x)=1; p_Y\ge 0,\ \sum_y p_Y(y)=1

Rule: L_product_law · Uses: S07, S08, S09

Leaf basis: L_product_law + numeric_evaluation

Apply Lemma L_product_law with S_X={0,1,2} and S_Y={0,1}; its hypotheses come from S07, S08 and S09. The six masses of Q are Q(0,0)=5/24, Q(0,1)=7/24, Q(1,0)=5/36, Q(1,1)=7/36, Q(2,0)=5/72 and Q(2,1)=7/72. The lemma itself is proved in general in S13.1 to S13.3.

S14 · implies

Before

J(2,0)=0,\ Q(2,0)=\tfrac{5}{72}

After

J\neq Q\ \text{while}\ J,\ Q\ \text{have the same marginals}

Conditions: none

Rule: function_def · Uses: S07, S09, S12, S13

Leaf basis: function_def + exists_intro + quantifier_negation

Regard both joint laws as mass functions on ℤ² (zero off the support). The original table J has mass 0 at (2,0) (S12), while Q has mass 5/72 there; the two functions differ at a point, so J≠Q. Yet both have the marginals p_X and p_Y (J by S07 and S09, Q by S13). One counterexample refutes the universal claim that two marginals determine the joint distribution.

Substeps and lemmas

S04.1 · equivalent

Before

\tfrac16+\tfrac13+\tfrac14+\tfrac1{12}+\tfrac16

After

\tfrac{2}{12}+\tfrac{4}{12}+\tfrac{3}{12}+\tfrac{1}{12}+\tfrac{2}{12}

Conditions: k\neq 0

Rule: frac_rules · Uses: G1

Leaf basis: frac_rules

Use a/b=(ak)/(bk) for k≠0 to give every fraction the denominator 12: k=2 for 1/6, k=4 for 1/3, k=3 for 1/4 and k=1 for 1/12. No term changes its value.

S04.2 · equivalent

Before

\tfrac{2}{12}+\tfrac{4}{12}+\tfrac{3}{12}+\tfrac{1}{12}+\tfrac{2}{12}

After

\tfrac{12}{12}=1

Conditions: 12\neq 0

Rule: frac_rules · Uses: S04.1

Leaf basis: frac_rules + numeric_evaluation

Add over the common denominator, a/c+b/c=(a+b)/c: the numerators give 2+4+3+1+2=12, so the total is 12/12=1.

S07.1 · equivalent

Before

S\cap(\{x\}\times\mathbb{Z}),\ x=0,1,2

After

\{(0,0),(0,1)\},\ \{(1,0),(1,1)\},\ \{(2,1)\}

Conditions: none

Rule: set_operations_def · Uses: G1, set_operations_def

Leaf basis: set_operations_def + set_extensionality

Check the first coordinate of each of the five pairs of S and put it into the matching set. Each pair lies in exactly one set; the three sets are pairwise disjoint and their union is S.

S07.2 · equivalent

Before

p(0,0)+p(0,1),\ p(1,0)+p(1,1),\ p(2,1)

After

\tfrac12,\ \tfrac13,\ \tfrac16

Conditions: none

Rule: frac_rules · Uses: S07.1, G1

Leaf basis: frac_rules + numeric_evaluation

1/6+1/3=1/6+2/6=3/6=1/2; 1/4+1/12=3/12+1/12=4/12=1/3; p(2,1)=1/6.

S10.1 · equivalent

Before

S\cap(A\times B)

After

\{(1,1),\ (2,1)\}

Conditions: none

Rule: cartesian_product_def · Uses: G1, G3, cartesian_product_def

Leaf basis: cartesian_product_def + set_operations_def + set_extensionality

Check each pair: (0,0) and (0,1) have x=0∉A; (1,0) has y=0∉B; (1,1) has 1∈A and 1∈B; (2,1) has 2∈A and 1∈B. So the intersection is exactly {(1,1),(2,1)}.

S10.2 · equivalent

Before

p(1,1)+p(2,1)

After

\tfrac{1}{12}+\tfrac{2}{12}=\tfrac14

Conditions: none

Rule: frac_rules · Uses: S10.1, G1

Leaf basis: frac_rules + numeric_evaluation

1/12+1/6=1/12+2/12=3/12=1/4.

S13.1 · implies

Before

Q(x,y)=p_X(x)\,p_Y(y),\ (x,y)\in S_X\times S_Y

After

\text{distinct pairs},\ Q(x,y)\ge 0\ \text{exact rational}

Conditions: p_X(x)\ge 0; p_Y(y)\ge 0

Rule: cartesian_product_def · Uses: cartesian_product_def, order_basics

Leaf basis: cartesian_product_def + order_basics + frac_rules

The elements of S_X×S_Y are listed by the choice of first and second coordinate, and each choice gives a different ordered pair, so the pairs are distinct. The product of two nonnegative exact rationals is again a nonnegative exact rational. This gives requirements (i) and (ii) of a joint law.

S13.2 · equivalent

Before

\sum_{x\in S_X}\sum_{y\in S_Y}p_X(x)\,p_Y(y)

After

\sum_{x\in S_X}p_X(x)\cdot\sum_{y\in S_Y}p_Y(y)=1\cdot 1=1

Conditions: \sum_x p_X(x)=1; \sum_y p_Y(y)=1

Rule: distributive · Uses: S13.1, finite_sum_rules, distributive, finite_sum_regroup

Leaf basis: distributive + finite_sum_rules + finite_sum_regroup + mul_one

Regroup the sum over S_X×S_Y by first coordinate into a double sum; in the inner sum Σ_y p_X(x)p_Y(y) the factor p_X(x) is constant and comes out, giving p_X(x)Σ_y p_Y(y)=p_X(x)·1; the outer sum then gives Σ_x p_X(x)=1. Requirement (iii) holds.

S13.3 · equivalent

Before

\sum_{y\in S_Y}Q(x,y),\ \sum_{x\in S_X}Q(x,y)

After

p_X(x)\sum_{y}p_Y(y)=p_X(x),\ p_Y(y)\sum_{x}p_X(x)=p_Y(y)

Conditions: \sum_x p_X(x)=1; \sum_y p_Y(y)=1

Rule: finite_sum_rules · Uses: S13.1, finite_sum_rules, marginal_def

Leaf basis: finite_sum_rules + mul_one + marginal_def

The X-marginal of Q fixes x and sums over y: the constant p_X(x) comes out and the remaining Σ_y p_Y(y) is 1, so it equals p_X(x); the Y-marginal equals p_Y(y) in the same way. S13.1 to S13.3 together prove Lemma L_product_law.

Reverse audits

  • S14 · sufficient_goal_reduction · S12, S13, function_def, exists_intro — Goal G: marginals do not in general determine the joint law. Sufficient condition H: two different joint laws with the same marginals exist; H⇒G because one counterexample refutes a universal claim. H is constructed forwards in S12 and S13; G is never used as a premise.
  • S13.3 · inverse_equivalence · finite_sum_rules, mul_one — Σ c·a_i=c·Σ a_i and multiplying by 1 are rewrites of equal values, so the step is two-way. This sub-step does not use L_product_law.
  • S13.2 · inverse_equivalence · distributive, finite_sum_rules, finite_sum_regroup — Pulling out a constant factor and regrouping a finite sum are rewrites of equal values and can be reversed; the conditions used are that each marginal totals 1. This sub-step does not use L_product_law.
  • S13.1 · premise_audit · cartesian_product_def, order_basics — Nonnegative times nonnegative is nonnegative; no converse is claimed. This sub-step does not use L_product_law.
  • S13 · premise_audit · S07, S08, S09, L_product_law — The lemma needs each marginal to be nonnegative with total 1, which S07 to S09 confirm. The step only constructs Q and does not claim that Q equals the original joint law.
  • S12 · premise_audit · S06, S07, S09, G1 — It needs the support list to be complete (assumption a1): the mass of (2,0) is 0 precisely because (2,0) is not listed.
  • S11 · premise_audit · S07, S09, S10, numeric_evaluation — It only compares two exact numbers; it says they differ for this table, not for every joint law (for the product law Q they agree).
  • S10.2 · inverse_equivalence · frac_rules, numeric_evaluation — An exact rewrite of equal fractions, two-way.
  • S10.1 · inverse_equivalence · cartesian_product_def, set_extensionality — All five pairs of S were decided, so both sides are the same set and the step holds both ways.
  • S10 · premise_audit · S06, G3, cartesian_product_def — It needs 'at the same time': (x,y)∈A×B exactly when x∈A and y∈B. Using 'or' instead gives a different event with a different value.
  • S09 · premise_audit · S06, marginal_def — It requires the list for each y to be complete: S has exactly two pairs with second coordinate 0 and exactly three with 1, five in all, matching the size of S.
  • S08 · inverse_equivalence · finite_sum_regroup, S07.1 — Regrouping changes only how the finite sum is grouped, not its value; both sides are the same number, so the step is two-way. The condition is the partition: each pair is counted exactly once.
  • S07.2 · inverse_equivalence · frac_rules, numeric_evaluation — Exact fraction arithmetic rewrites equal values, so the two sides are the same three rationals.
  • S07.1 · inverse_equivalence · set_operations_def, set_extensionality — All five pairs were decided one by one, so both sides are the same sets element by element, and the step holds both ways.
  • S07 · premise_audit · S06, marginal_def — It requires the listed pairs to be the complete intersection of S with {x}×ℤ; missing one pair gives a wrong marginal. No converse is claimed: the individual masses cannot be recovered from the marginal.
  • S06 · premise_audit · S05, joint_event_sum, prob_axioms — It needs S05: if the masses did not sum to 1 or some mass were negative, the numbers obtained would not be probabilities. The step applies a definition and derives nothing else.
  • S05 · sufficient_goal_reduction · S01, S02, S03, S04, and_intro_elim — Goal G: (S,p) is a finite joint law. Sufficient condition H: requirements (i), (ii) and (iii) all hold; H⇒G by the definition. H is proved forwards in S02 to S04; G is never used as a premise.
  • S04.2 · inverse_equivalence · frac_rules, numeric_evaluation — Integer addition and combining over a common denominator are rewrites of equal values and can be undone step by step.
  • S04.1 · inverse_equivalence · frac_rules — Backward, cancel the same nonzero factor k to return to the original fractions; the two sides agree term by term.
  • S04 · inverse_equivalence · frac_rules, numeric_evaluation — Both sides name the same rational number, and S04.2 can be read backwards to the form in S04.1, so the step holds both ways.
  • S03 · premise_audit · G1, order_basics — It needs only the sign of each of the five masses; positive implies nonnegative, and no converse is claimed.
  • S02 · premise_audit · G1, cartesian_product_def — The grouping covers all 10 comparisons with none missed; one differing coordinate suffices for two pairs to differ, and no converse is needed.
  • S01 · premise_audit · G1, mko_joint_def — This step only unfolds the definition and lists the three obligations; it derives no new fact.

Error controls and negative controls

Two marginal tables are not a joint law

The common mistake named by the MKO is to take two marginal tables as the joint distribution. Here J and Q have exactly the same marginals (1/2, 1/3, 1/6 and 5/12, 7/12), yet their masses at (2,0) are 0 and 5/72 (S14). Marginals are aggregates of the joint table, and aggregation loses information, so they cannot be inverted.

The total must be exactly 1

Replacing the masses by 0.166, 0.333, 0.25, 0.083 and 0.166 still gives exact rational numbers when the finite decimals are read literally. However, their total is 0.998 rather than 1, so the table fails requirement (iii). If intended merely as truncated approximations, they must not be treated as the original exact masses. S04 uses the original fractions and obtains exactly 12/12=1.

A rectangle event needs both coordinates at once

P(X∈A, Y∈B) is the event that both coordinate conditions hold at once, here 1/4. Computing P(X∈A)+P(Y∈B)−P(X∈A, Y∈B)=1/2+7/12−1/4=5/6 gives the 'or' event; computing P(X∈A)P(Y∈B)=7/24 treats the marginals as the joint law. Neither is the rectangle probability (S10, S11).

  • Multiplying the marginals gives the joint law: P(X=x,Y=y)=p_X(x)p_Y(y) holds for this table → reject — At (2,0) the mass of J is 0 while p_X(2)p_Y(0)=5/72.
  • The table with masses 0.166, 0.333, 0.25, 0.083, 0.166 is a valid finite joint distribution → reject — The total is 0.998≠1; the MKO requires the total to be exactly 1.
  • P(X∈{1,2}, Y=1)=P(X∈{1,2})·P(Y=1)=7/24 → reject — Aggregating the joint law gives 1/12+1/6=1/4=6/24≠7/24.
  • Listing (0,1) twice with mass 1/6 each is still a joint table in the sense of the MKO → reject — The MKO requires the support to list distinct pairs; a repeated pair must be merged into one row (mass 1/3) before the table qualifies.

Result and limitations

(S,p) is a valid two-dimensional finite joint distribution (distinct pairs, positive exact rational masses, total exactly 1). From this one joint law, p_X(0)=1/2, p_X(1)=1/3, p_X(2)=1/6, p_Y(0)=5/12, p_Y(1)=7/12, the rectangle event P(X∈{1,2}, Y=1)=1/4 and the single-pair mass P(X=2,Y=0)=0. The product law Q=p_X p_Y has the same marginals but Q(2,0)=5/72≠0, so two marginals do not determine the joint distribution.

  • Regrouping of finite sums (finite_sum_regroup) and joint_event_sum are premises of this guide; they are not derived here from more basic axioms.
  • N14 and N15 check the numbers in the error controls and are valid for those numbers only.

Verification and source binding

15 declared checks for this guide; only the listed expressions are covered.

  • source commit: b28a5edbd46279b1db2a3f3f48bb3f447e18899f
  • adopted source SHA-256: 5b995a039b001aab1d5b78c71b0b7780f68598a4f9bf05748dd25e86da4430d9
  • original source SHA-256: 5b995a039b001aab1d5b78c71b0b7780f68598a4f9bf05748dd25e86da4430d9
  • Lean: NOT_RUN · independent AI: NOT_MEASURED

Same-source full JSON (formulas, ASTs, controls and revision log)