Past exam of the options II course of the University of Oxford / 2004 / N (Probability (Donnelly))

Reading positive entries as directed edges, the communicating class decomposition isFor the first class, paths $1\to 5\to 8\to 9\to 1$ establish mutual reachability, and every transition from one of these states stays in this set. States 2 and 3 communicate, but state 2 can enter the first class and never return. States 4,6,7 communicate through $4\to 6\to 7\to 4$, but state 6 can enter state 2 with no route back. These paths also exclude any larger communicating class. In a finite Markov chain, the closed class is recurrent and the two open classes are transient.

Assume $K\ge 1$ is an integer, $0<p<1$, and put $q=1-p$. There is a discrepancy between the printed state definition and the formula requested later. We first follow the stated cumulative count, which retains previous good items when a defective item appears during phase A. The transition probabilities arewith all others zero. The balance equations imply $q{\pi }_i=q{\pi }_{i-1}$ for $1\le i<K$ and $(p/r){\pi }_K=q{\pi }_{K-1}$. ThusFor comparison, if clearance requires $K$ consecutive good items, each defect in phase A resets the state to zero. Replace its transitions by $P_{i,0}=p$, $P_{i,i+1}=q$, combining contributions if destinations coincide. Phase B retains its two transitions. For this consecutive-clearance inspection model, balance yields ${\pi }_i=q^i{\pi }_0$ and $(p/r){\pi }_K=q^K{\pi }_0$. With $D=1+(r-1)q^K$, normalization givesThis second model is the one giving the printed formula in part (iii). It is a different transition rule, rather than an algebraic simplification of the cumulative model.

For an irreducible positive recurrent Markov chain, the Markov chain ergodic theorem gives a unique stationary distribution $\pi$, with almost-sure long-run occupation fraction ${\pi }_j$ in state $j$. More generally, time averages of an integrable state reward converge to its stationary mean. If the chain is also aperiodic, then ${\mathbb{P}}_i(X_n=j)\to {\pi }_j$ for every initial state $i$; the aperiodicity assumption is needed for this last distributional conclusion.

For either transition rule in part (i), $0<p<1$ permits progression through the phase-A states and a return from phase B to zero. Both chains are finite and irreducible, hence positive recurrent. State $K$ has a positive self-loop, so both are aperiodic. Thereforeusing respectively the cumulative or consecutive stationary distribution given in part (i), for every starting distribution.

The excluded endpoint cases are simple: at $p=0$, phase B is eventually reached and then absorbing, so the limiting law is the point mass at $K$. At $p=1$, the cumulative rule traps the chain at its current phase-A count (or eventually at zero if started in phase B), so the limit can depend on the initial count; under the consecutive rule zero is eventually absorbing. Irreducibility and the preceding stationary formulas must not be asserted unchanged at these endpoints.

During phase A every item is inspected, while during phase B inspection has probability $1/r$, independently of item quality. The long-run inspected fraction is therefore the stationary rewardThis can be justified by the Markov chain ergodic theorem for state occupation and the strong law for the independent inspection coins (or the corresponding bounded martingale differences).

For the printed cumulative count this givesIt does not equal the claimed expression in general. For instance $K=2$, $p=1/2$, $r=2$ gives $I=3/4$, whereas the printed expression gives $4/5$.

For the consecutive-clearance rule, substituting ${\pi }_K=r{q}^K/D$ instead givesThus the claimed formula is valid for consecutive clearance, and false for the literal cumulative count. For $p=0$ the limiting inspected fraction is $1/r$ in either model; for $p=1$ it is 1 because phase A never clears.

An item escapes detection only in phase B, when it is defective and its inspection coin says no. Since the current item's quality and inspection choice are independent of the phase determined by previous items, the long-run fraction of all production that is defective and undetected isConsequently the two interpretations giveIf the denominator population is defective items alone, divide by their long-run fraction $p$: the corresponding missed fractions are $(r-1)q/(Kp+rq)$ and $(r-1)q^K/[1+(r-1)q^K]$. Stating the denominator matters; the phrase “proportion of defective items” can mean either this conditional fraction or the unconditional production fraction. At $p=0$ no defective item exists, so the unconditional fraction is zero and the conditional fraction is undefined; at $p=1$ eventual full inspection gives zero misses.