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Read the arrows in the original diagram as follows: $0$ moves to $1^{\prime }$ with probability one; from an unprimed $n\ge 1$, probability $q$ leads to $n-1$ and probability $p$ leads to $(n+1{)}^{\prime }$; from $n^{\prime }$ the corresponding destinations are $n$ and $(n+1{)}^{\prime }$. This is the two-lane Markov chain with a deterministic restart. It is irreducible: every state can reach zero by a finite path of $q$ transitions, and zero can reach every primed state by outward $p$ transitions and then every unprimed state by a $q$ transition. Thus all states have the same recurrence classification.

Let $a_n$ and $b_n$ be the hitting probabilities of zero starting from $n$ and $n^{\prime }$, with $a_0=1$. First-step conditioning givesSubtracting yields $b_n=(1+q)a_n-q{a}_{n-1}$. Eliminating $b_{n+1}$ givesIts characteristic roots are $1$ and $r=q/[p(1+q)]=q/(1-q^2)$. Put $q_c=(\sqrt{5}-1)/2$, so $r=1$ precisely when $q=q_c$.

For $q>q_c$, the general solution is $a_n=A+B{r}^n$. The bound $0\le a_n\le 1$ and $r>1$ force $B=0$, and $a_0=1$ then gives $a_n=b_n=1$. For $q=q_c$, the general solution is $a_n=A+Bn$; boundedness again gives $a_n=b_n=1$. In both cases, the return probability from zero is $b_1=1$, so the Markov chain is recurrent.

For $q<q_c$, $0<r<1$, and the pairis a nonnegative bounded solution of the first-step equations. To identify it as the actual hitting probability, use that hitting probability is the minimal nonnegative harmonic extension. This fact follows directly by iteration: a nonnegative solution with boundary value one at zero dominates the probability of hitting zero within $N$ steps, for every $N$. The actual bounded solution has form $a_n=A+(1-A)r^n$, where nonnegativity as $n\to \infty$ requires $A\ge 0$. Minimality bounds it above by $r^n$, while this form bounds it below by $r^n$. Hence it is exactly the displayed solution. Starting at zero, the first return time has finite-return probability $b_1=q^2/p<1$. The Markov chain is therefore transient.

It remains to distinguish positive and null recurrence. Write $u_n={\pi }_n$ and $v_n={\pi }_{n^{\prime }}$ for invariant probabilities. Balance at zero and at $1^{\prime }$ gives ${\pi }_0=q{u}_1$ and $v_1={\pi }_0$. For the finite cut consisting of zero and both lanes up to level $n$, invariance equates its outward and inward probability fluxes:Balance at the unprimed state $n$ also gives $u_n=q(u_{n+1}+v_n)$. Combining these identities yieldsTherefore every invariant probability distribution would necessarily have $u_n=({\pi }_0/q)t^{n-1}$ and $v_n={\pi }_0{t}^{n-1}$. Conversely, these weights satisfy all individual balance equations, including $v_{n+1}=p(u_n+v_n)$, so their summability is the only remaining condition.

When $q>q_c$, $t<1$, and normalization givesAn irreducible Markov chain has an invariant probability distribution exactly when it is positive recurrent; equivalently its mean return time is finite and ${\pi }_0$ is its reciprocal. Thus this case is positive recurrent. At $q=q_c$, $t=1$ and the nonzero invariant weights cannot be summed to one. Irreducibility would force ${\pi }_0>0$ in any invariant probability distribution. Since recurrence was already proved but no such distribution exists, this case is null recurrent.

The complete classification is therefore transient for $0<q<q_c$, null recurrent at $q=q_c$, and positive recurrent for $q_c<q<1$. A drift argument using only the unprimed arrows would miss the primed lane and give the wrong threshold.