= Solution
{c}
Start at any integer $x\geq2$, for example two. Propose $x+1$ or $x-1$ with <probability> $1/2$ each; if the proposal is one, reject it immediately. Between distinct valid neighboring states the proposal is symmetric. The <Metropolis–Hastings acceptance probability> for an upward proposal is
$$
\boxed{\alpha_+(x)=\min\{1,\tfrac{x}{x-1}(1-p)\}.}
$$
For a downward proposal from $x\geq3$, it is
$$
\boxed{\alpha_-(x)=\min\{1,\tfrac{x-2}{(x-1)(1-p)}\}.}
$$
These are exactly the ratios of the supplied <negative binomial distribution> probabilities; the factor $p^2$ cancels. The rejection at the boundary supplies a self-loop at two. All neighboring positive states communicate, so the chain is <irreducible> and aperiodic. Detailed balance gives the desired stationary law. Rejecting the invalid move, rather than forcing an upward move at two, preserves the proposal symmetry used in these ratios.
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