Survival probability ($Q(x,t)$)

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.