= Solution
Take $d=1$ and start from the lower endpoint. The stationary distribution of this birth-death chain satisfies detailed balance with
$$
\frac{\pi(k+1)}{\pi(k)}=\frac{1/3}{1/6}=2,
$$
so it is concentrated within $O(1)$ of the upper endpoint $n$. Before reaching that region the walk has drift $1/6$. The <weak law of large numbers> and exponential concentration therefore imply that its hitting time of $n-O(1)$ is
$$
6n+O_{\mathbb P}(\sqrt n).
$$
At time $(6-\varepsilon)n$ the chain is still macroscopically below the stationary region with probability tending to one, so its total-variation distance tends to one. Under the monotone coupling from part (b), by time $(6+\varepsilon)n$ the extremal copies have coalesced with probability tending to one, so the distance tends to zero. Therefore the family has
$$
\boxed{\text{cutoff at }6n\text{ with an }O(\sqrt n)\text{ window}.}
$$
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