Lazy random walk ($P_{\mathrm{lazy}}=(I+P)/2$)

A lazy random walk remains at its current position with positive probability and otherwise makes an ordinary random walk step. A common convention replaces transition matrix $P$ by $P_{\mathrm{lazy}}=(I+P)/2$. This preserves every stationary distribution because $\pi P=\pi$ implies $\pi P_{\mathrm{lazy}}=\pi$. It also removes periodic alternation in an irreducible discrete-time chain: the self-loop permits return in one step. A lazy symmetric walk on the integers has increments $-1,0,1$ with probabilities $1/4,1/2,1/4$, so its step variance is $1/2$.

For a random walk with transition matrix $P$, its holding probability at state $x$ is $P(x,x)$, the conditional probability of staying at $x$ for the next step. It is the holding probability of a Markov chain specialized to the walk. A lazy random walk can be constructed byIts holding probability is $\alpha +(1-\alpha )P(x,x)$, which equals $\alpha$ only if the original walk cannot stay at $x$.

This change preserves every stationary distribution: $\pi P=\pi$ implies $\pi P_{\alpha }=\pi$. A positive holding probability permits return in one step, giving period one at that state; for an irreducible Markov chain, this removes periodicity throughout the class. Laziness does not connect previously disconnected classes. For a loop-free walk with $\alpha =1/2$, the conventional holding probability is therefore exactly $1/2$.

Consider a random walk on $\{0,1,\dots ,n\}$ with absorbing states $0,n$. At each interior state it stays in place with probability $q$, and moves one position in either direction with probability $(1-q)/2$, where $0\le q<1$. Its mean absorption time from $k$ isProof. A sequence of at most $n$ consecutive leftward moves ensures absorption, with probability at least $((1-q)/2{)}^n>0$. Dividing time into blocks of length $n$ gives a geometric series bound on the survival probabilities, so the expected value is finite. Let $e_k={\mathbb{E}}_k[T]$. The law of total expectation and first-step analysis giveThus $e_{k+1}-2e_k+e_{k-1}=-2/(1-q)$. A particular solution of this linear recurrence relation is $-k^2/(1-q)$; its homogeneous solutions are $A+Bk$. The two boundary conditions give $A=0$, $B=n/(1-q)$, proving both the formula and its uniqueness. Holding slows the usual symmetric random walk by the factor $1/(1-q)$.