Uniform depletion Markov chain

A finite stock decreases by choosing a uniformly random positive number of its remaining items at every step. The remaining stock is a Markov chain with $P(i,j)=1/i$ for $0\le j<i$ and $P(0,0)=1$. State $0$ is an absorbing state, and every positive state is a transient state. If $e_i$ denotes the expected number of positive-stock steps before absorption, a first-step decomposition gives $e_0=0$ and $i{e}_i=i+{\sum }_{j<i}{e}_j$. Subtracting consecutive equations yields $e_i-e_{i-1}=1/i$, so $e_i=H_i$, the $i$th harmonic number.

Refill the stock to a fixed positive size $N$ immediately whenever a step empties it. The resulting Markov chain has state space $\{1,\dots ,N\}$ and transitions $P(i,j)=1/i$ for $j<i$ and $P(i,N)=1/i$. It is irreducible and aperiodic, since every state reaches $N$, $N$ reaches every state, and $P(N,N)>0$. Its unique stationary distribution is ${\pi }_N=1/H_N$ and ${\pi }_k=1/((k+1)H_N)$ for $k<N$. Normalization is the harmonic number identity. For $k<N$, stationarity follows from ${\sum }_{j=k+1}^{N-1}1/(j(j+1))+1/N=1/(k+1)$; the same telescoping sum starting at $1$ gives stationarity at $N$. The finite-state convergence theorem therefore makes this stationary distribution the limiting distribution from every initial state.