Past exam of the probability course of the University of Oxford / 2021 / 3 / d

Let $d=\gcd (a,b)>1$ and put $a^{\prime }=a/d$, $b^{\prime }=b/d$. Then $a^{\prime },b^{\prime }$ are coprime integers, and $p=b^{\prime }/(a^{\prime }+b^{\prime })$. The six requested revisions are as follows.

(b)(i): communicating classes. Every increment is divisible by $d$, so the residue class modulo $d$ is preserved. Conversely, if $j-i$ is divisible by $d$, Bézout's identity applied to $a^{\prime },b^{\prime }$ gives a path just as in (b)(i). Thus the original Markov chain is reducible on $\mathbb{Z}$, with exactly the $d$ closed communicating classesThe specified walk starting at zero stays in $d\mathbb{Z}$.

(b)(ii): period of a state in a Markov chain. The return equation $au=bv$ reduces to $a^{\prime }u=b^{\prime }v$. Hence $u=b^{\prime }k$, $v=a^{\prime }k$. For every state, the period of a state in a Markov chain is thereforeThis is still at least two, so no state is aperiodic.

(b)(iii): stationary distribution. The conclusion is unchanged: there is no stationary distribution on $\mathbb{Z}$. The maximum-mass proof in (b)(iii) applies verbatim, without using coprimality. Equivalently, each closed communicating class, after translating and dividing its state labels by $d$, is the same centered random walk with step sizes $a^{\prime },-b^{\prime }$ and has no stationary distribution.

(b)(iv): occupation of zero. On $d\mathbb{Z}$, the rescaled walk $S_n/d$ has exactly the transition probabilities of the coprime walk with parameters $a^{\prime },b^{\prime }$. The return-probability argument in (b)(iv) therefore makes this class null recurrent. The renewal process proof of the occupation limit applies unchanged:Starting in another residue class would make every visit to zero impossible, also giving occupation proportion zero.

(c)(i): modified communicating classes. These are unchanged: $C=\{-c-b,\dots ,c+a\}$ is one finite closed communicating class, and all other states are transient singleton classes. The original proof used consecutive reset destinations and never used coprimality. Resetting to the $b$ consecutive integers $c-b+1,\dots ,c$ connects different residue classes, so it would be incorrect to split $C$ according to residue classes modulo $d$.

(c)(ii): limiting distribution. The column-sum calculation, the irreducible Markov chain property on a finite class, positive self-transition probabilities, and geometric entrance argument also required no coprimality. Thus for every initial state the same convergence in distribution holds:In particular, the denominator $2c+a+b+1$ is unchanged; it is not divided by $d$.