Question 11311: Row-Stochastic Qualification before Any Markov Step

Original question and givens

Use one fixed finite time-homogeneous transition table=[[11/13, 2/13], [6/7, 1/7]]. Rows denote the current state and columns the next state; row entries must be nonnegative and sum exactly to 1 before stepping. State order stays fixed. Use exact rational arithmetic, reduce fractions to positive denominators and omit /1 for integer outputs. No stationary, irreducibility, ergodicity or long-term convergence claim follows. This qualification family also admits negative or non-unit-row candidate entries for rejection. Give (size,normalized transition table,all entries nonnegative?,exact row sums,row-stochastic?,rows=current state, columns=next state).

• G001 — Use one fixed finite time-homogeneous transition table=[[11/13, 2/13], [6/7, 1/7]]. Rows denote the current state and columns the next state; row entries must be nonnegative and sum exactly to 1 before stepping. State order stays fixed. Use exact rational arithmetic, reduce fractions to positive denominators and omit /1 for integer outputs. No stationary, irreducibility, ergodicity or long-term convergence claim follows. This qualification family also admits negative or non-unit-row candidate entries for rejection. Give (size,normalized transition table,all entries nonnegative?,exact row sums,row-stochastic?,rows=current state, columns=next state).

Definition and scope

Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate.

• mko-finite-markov-chain

Step-by-step derivation

transition row0:raw rational pairs=[[11, 13], [2, 13]]

transition row0:normalized=[11/13, 2/13]

transition row0 total := sum(11/13, 2/13)

transition row0 total = 1

transition row0:weights=[11/13, 2/13];sum=1

nonnegative=Yes;total_one=Yes;qualified=Yes

transition row1:raw rational pairs=[[6, 7], [1, 7]]

transition row1:normalized=[6/7, 1/7]

transition row1 total := sum(6/7, 1/7)

transition row1 total = 1

transition row1:weights=[6/7, 1/7];sum=1

nonnegative=Yes;total_one=Yes;qualified=Yes

complete transition matrix=[[11/13, 2/13], [6/7, 1/7]];row sums=[1, 1]

nonnegative=Yes;row_stochastic=Yes;rows=current state, columns=next state

Reverse checks

• S007 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S006 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S005 · premise_and_result_audit — Recompute transition row1 total with exact rational arithmetic, not rounded decimals.
• S004 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
• S002 · premise_and_result_audit — Recompute transition row0 total with exact rational arithmetic, not rounded decimals.
• S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

(2, [[11/13, 2/13], [6/7, 1/7]], Yes, [1, 1], Yes, rows=current state, columns=next state)

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.