= Hitting time of a downward-biased walk reflected at an upper barrier
{title2=$\mathbb E_NT_0=\frac N{q-p}-\frac{2pq}{(q-p)^2}\left(1-(p/q)^N\right)$}
On $\{0,\ldots,N\}$, stop at zero, move left with probability $q>1/2$ and right with probability $p=1-q$ at interior states, and move deterministically from $N$ to $N-1$. For $m_i=\mathbb E_iT_0$, the <first-step analysis> gives $m_0=0$, $m_i=1+qm_{i-1}+pm_{i+1}$ and $m_N=1+m_{N-1}$. Setting $D_i=m_i-m_{i-1}$ gives $qD_i-pD_{i+1}=1$, $D_N=1$, hence $D_i=[1-2p(p/q)^{N-i}]/(q-p)$. Summing these differences proves the displayed formula, and more generally $m_i=i/(q-p)-2pq(p/q)^{N-i}[1-(p/q)^i]/(q-p)^2$. A downward block has a uniformly positive probability of reaching zero, which supplies a geometric tail bound and justifies finiteness of these expectations.
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