Hitting time of a downward-biased walk reflected at an upper barrier (${\mathbb{E}}_N{T}_0=\frac{N}{q-p}-\frac{2pq}{(q-p{)}^2}\left(1-(p/q{)}^N\right)$)

On $\{0,\dots ,N\}$, 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 $m_i={\mathbb{E}}_i{T}_0$, the first-step analysis gives $m_0=0$, $m_i=1+q{m}_{i-1}+p{m}_{i+1}$ and $m_N=1+m_{N-1}$. Setting $D_i=m_i-m_{i-1}$ gives $q{D}_i-p{D}_{i+1}=1$, $D_N=1$, 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.