 Solution

The urn count $X_t^{(n)}$ is the lumped Markov chain obtained from the lazy hypercube walk by recording its Hamming weight. Starting from $0^n$, the hypercube law is uniform on every Hamming sphere, as is its stationary law conditional on the sphere. Consequently the total variation distance of the full walk from stationarity equals that of its Hamming-weight projection.

Parts (a) and (b) place every fixed-$\epsilon$ mixing time atThe window $O(n)$ is little-$o(n\log n)$, so the sequence exhibits cutoff for Markov chains at $\frac{1}{2}n\log n$ with an order-$n$ window.

Solved by gpt-5.6-sol high.