Mean first-passage time ($\tau (x)={\mathbb{E}}_x{T}_A$)

The Mean first-passage time to a target $A$ is $u(x)={\mathbb{E}}_x{T}_A$, which can be infinite. For a diffusion with backward generator $L$, a finite sufficiently regular mean solves the Poisson equationoutside $A$, with $u=0$ on the absorbing target. Reflecting parts of the boundary require the reflecting condition of the generator, rather than an absorbing value. These conditions express different physical rules and cannot be exchanged.

For Brownian diffusion with generator $D{d}^2/d{x}^2$ on $[0,\ell ]$, $D>0$, absorption at both endpoints givesIf zero instead reflects while $\ell$ absorbs, the boundary conditions are $u^{\prime }(0)=0$, $u(\ell )=0$ and the solution is $u(x)=({\ell }^2-x^2)/(2D)$. In particular the mean time from the reflecting endpoint is ${\ell }^2/(2D)$, displaying the squared-distance diffusion scale.

Dynkin's formula explains the equation: before absorption, $u(X_{t\wedge T_A})+(t\wedge T_A)$ has constant expected value. For these bounded-interval nonnegative bounded solutions, this identity first bounds ${\mathbb{E}}_x(t\wedge T_A)$ by $u(x)$, proving integrability of $T_A$; passing to the limit then proves $u(x)={\mathbb{E}}_x{T}_A$. Boundary equations without a finiteness or minimality condition can have extraneous solutions on unbounded domains.

For a Markov chain with transition matrix $P$, fix a directed transition $(u,v)$ with $P_{uv}>0$. Its transition-completion time isIt counts the successful final step. Merely visiting $v$ by another route does not complete this event. In a finite irreducible chain its mean $g_i={\mathbb{E}}_i{T}_{uv}$ is finite and first-step analysis givesThe subtracted continuation term removes exactly the case where the designated transition has just succeeded. An equivalent construction sends that transition to a new absorbing success state and computes its Expected hitting time.

For the two-state chain $P_{01}=p$, $P_{10}=q$, with $0<p,q\le 1$, the waiting time for $0\to 1$ has means $g_0=1/p$ and $g_1=1/q+1/p$. Starting at one, the hitting time of state one is zero, whereas the transition-completion time must first wait for $1\to 0$ and then $0\to 1$. This example makes the difference between an edge event and a state event explicit.