First-passage time ($T_A$)

The First-passage time to a target set $A$ is $T_A=\inf \{t\ge 0:X_t\in A\}$. It is also called a hitting time.

For a discrete-time Markov chain and a set $A$ of absorbing states, its absorption time is the hitting time $T_A=\inf \{m\ge 0:X_m\in A\}$. It is zero when the initial state already lies in $A$, and infinite on a sample path that never enters $A$. Finiteness almost surely need not imply a finite expected value. For a finite-state model, a uniform positive probability of absorption within a fixed number of steps gives a geometric series bound on the survival probabilities and proves that the mean is finite. The mean absorption time of a lazy symmetric random walk is an explicit example.

For standard Brownian motion and $a>0$, $T_a=\inf \{t\ge 0:B_t\ge a\}$ is finite almost surely andThe Brownian reflection principle gives this distribution function. The Brownian scaling identity $T_{ca}\stackrel{d}{=}{c}^2{T}_a$ and the Strong Markov property show its probability distribution is a strictly stable distribution with index $1/2$.

For $T_{a,b}=\inf \{s:W_s=a+bs\}$ with $a>0$, exponential tilting of the terminal Brownian value gives $e^{{\theta }^{2}t/2}$ times the barrier-survival Probability for Brownian drift $\theta -b$. Thus the moment equals $e^{{\theta }^{2}t/2}[\Phi ((a+(b-\theta )t)/\sqrt{t})-e^{2a(\theta -b)}\Phi ((-a+(b-\theta )t)/\sqrt{t})]$ for every real $\theta$.

The Brownian reflection principle gives $\mathbb{P}({\tau }_x\le t)=\mathbb{P}(|N|\ge x/\sqrt{t})$ for a standard normal random variable $N$. This is the distribution function of $x^2/N^2$. An independent Brownian motion observed at this hitting time has the Cauchy distribution with scale $x$.

For a positive level $a$, this density describes the first-passage time of $B_t+ct$. The Girsanov theorem multiplies the zero-drift density by the stopped likelihood $e^{ca-c^{2}t/2}$. Its total mass is one for nonnegative drift and $e^{2ac}$ for negative drift; the latter has remaining mass at infinite hitting time.

For $a>0$, real $b$ and standard Brownian motion, the first-passage time to a linear boundary hasHere $\Phi$ is the standard normal distribution function. Differentiation gives $a\exp (-(a+bt{)}^2/(2t))/\sqrt{2\pi t^3}$, the drifted Brownian first-passage density. Its Laplace transform is $\exp [-a(b+\sqrt{b^2+2\theta })]$ for $\theta >0$. There is mass $1-e^{-2ab}$ at infinity when $b>0$; for $b\le 0$ the hitting time is finite almost surely.

When $d=\sqrt{c^2+2r}$ is real and nonnegative, the discounted drifted Brownian first-passage density is $e^{a(c-d)}{h}_d(t)$. Integrating yields $e^{a(c-d)}\Phi ((dT-a)/\sqrt{T})+e^{a(c+d)}\Phi ((-a-dT)/\sqrt{T})$. In a Black-Scholes model with $c=(r-{\sigma }^2/2)/\sigma$, the identity $c^2+2r=(r+{\sigma }^2/2{)}^2/{\sigma }^2$ ensures this formula covers every real interest rate.

For $Y_t=y+B_t+at$, $y>0$, and $S=\inf \{t\ge 0:Y_t=0\}$,where $e^{-\lambda \infty }=0$. The diffusion generator is $\frac{1}{2}{\partial }_{yy}+a{\partial }_y$. Its bounded eigenfunction on the positive half-line with boundary value one is $u(y)=\exp [-y(a+\sqrt{a^2+2\lambda })]$, giving the formula through the discounted boundary-hitting representation. The zero-discount limit is the Hitting probability $e^{-2y\max (a,0)}$.

For standard Brownian motion, set $H_a=\inf \{t:B_t\ge a\}$ and $T_a=\inf \{t:B_t>a\}$. The Strong Markov property and immediate positive excursions give $T_a=H_a$ almost surely for each fixed $a\ge 0$. However, these level-indexed processes are not indistinguishable stochastic processes. Almost surely the random level $A={\sup }_{t\le 1}{B}_t$ is positive and exceeds $B_1$. It is first attained before time one but is not exceeded until after time one, so $H_A<1<T_A$. The simultaneous-equality event therefore has Probability zero. The strict passage process $T_a$ is the right-continuous choice needed for a subordinator.

For standard Brownian motion started at zero, $H_a=\inf \{t\ge 0:B_t\ge a\}$ is finite almost surely for $a\ge 0$ andThe Brownian reflection principle proves finiteness. For $u\ge 0$, the Exponential martingale for Brownian motion stopped at $H_a\wedge t$ is bounded by $e^{ua}$. The dominated convergence theorem gives $\mathbb{E}{e}^{ua-u^2{H}_a/2}=1$, and taking $u=\sqrt{2\lambda }$ gives the transform.

When a first-passage time $\tau$ has an absolutely continuous distribution, its density is $f_{\tau }(t)=-S^{\prime }(t)$, where $S(t)=\mathbb{P}(\tau >t)$ is its survival function. In a continuous-time Markov chain, this density is the total Probability flux from surviving states into the absorbing set.

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.

For a nonnegative first-passage time $T_A$, possibly infinite, the survival probability isIt describes the Probability that absorption has not yet occurred. The tail-integral identityholds as an identity of extended nonnegative quantities, even when absorption is not almost sure or the expected time is infinite. Indeed, $T_A={\int }_0^{\infty }{1}_{\{T_A>t\}}dt$ pathwise, and Tonelli's theorem exchanges expectation and integration. For an integer-valued hitting time, the corresponding formula is ${\mathbb{E}}_x{T}_A={\sum }_{n=0}^{\infty }{\mathbb{P}}_x(T_A>n)$.

If the finite-time distribution is absolutely continuous, its first-passage-time density is $-{\partial }_{t}Q(x,t)$. Where $Q>0$, its hazard function is $-{\partial }_t\log Q$. Constant hazard $\lambda >0$ gives $Q(t)=e^{-\lambda t}$ and mean $1/\lambda$.

The limit of $Q(x,t)$ as $t\to \infty$ is the Probability of never being absorbed; a positive limit forces infinite mean. Conversely, zero limit does not guarantee a finite mean. Brownian first passage to a positive level occurs almost surely, but its survival tail is proportional to $t^{-1/2}$, whose integral diverges.

For an initially intact link with no rebinding, survival probability obeys $dp/dt=-k(F)p$. A force-tilted activation barrier with force $F=ft$, constant $f>0$ and $p(0)=1$ integrates to the displayed law. The rupture-force probability density is $k(F)p(F)/f$, not $p(F)$ itself. The no-rebinding approximation requires separated partners to leave the capture region before appreciable reassociation; it is not guaranteed by the rate equation alone.

Differentiate the logarithm of the rupture-force probability density $k(F)p(F)/f$. Its interior stationary point satisfies $e^{F/F_0}=f/(k_0{F}_0)$. The physical domain is $F\ge 0$, giving the displayed mode. This differs from the median rupture force under a linear ramp; replacing one by the other shifts a fitted intercept and can bias the inferred zero-force rate.

Solving $p(F_{1/2})=1/2$ for survival under a linear force ramp gives the displayed expression. At large $f/(k_0{F}_0)$, its slope against $\log f$ approaches $F_0$; at low loading rate the median is approximately $f\log 2/k_0$. A full fit can infer activation distance $k_{B}T/F_0$ and zero-force escape rate $k_0$. An absolute barrier additionally needs an attempt prefactor, and equilibrium binding strength additionally needs reverse-state information.