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First-hitting-time problem



A hitting time is the time when a system described by a random process first crosses a certain threshold, like the Earth’s climate when it makes a transition from one locally stable state to another locally stable state, from today’s climate to a climate closer to Snowball Earth, for example. The hitting time in this case is the time when the system or process reaches a tipping point. Another example is the rate of a chemical reaction, the tipping point is the point when the reaction starts, and the hitting time contains information about the rate at which the reaction takes place.

First-hitting-time problems or first-passage-time problems or first-escape-time problems try pose questions and to infer information about the probability distribution of the hitting time. This information may be used to predict and control the system under scrutiny.


Stationary diffusion processes using Fokker-Planck equations

The problem

Disclaimer for the mathematically inclined reader: We silently assume that all functions are smooth, all integrals are defined and finite, and everything converges absolutely, in short: We assume we are physicists.

Let’s assume we have a one dimensional diffusion process - a particle - described by an autonomous Fokker-Planck equation that starts at x 0x_0 \in \mathbb{R}. “Autonomous” means that the Fokker-Planck operator has no explicit time dependency, and the stochastic process is therefore stationary.

Let’s say the particle is absorbed at a,ba, b with a<x 0<ba \lt x_0 \lt b, and let

p(x t,t|x 0,t 0=0) p(x_t, t| x_0, t_0 = 0)

denote the probability to find the particle at time tt at x tx_t, when it was at time t 0=0t_0 = 0 at x 0x_0. That is, pp is the solution of the Fokker-Planck equation with the initial condition

p(x,t 0=0)=δ(xx 0) p(x, t_0 = 0) = \delta(x- x_0)

In this paragraph we will show what we can find out about the distribution of the hitting time, which in this example is the time of the absorption of the particle.

Formula for the moments of the hitting time

From the formulation of our problem we already know that the probability that the particle is not absorbed at time t, that is that the particle is still in (a,b)(a, b) at time t, is

a bp(x,t|x 0,0)dx=:G(x 0,t) \int_a^b p(x, t | x_0, 0) d x =: G(x_0, t)

When we call the time of absorption TT, then we have for the probability distribution of TT:

(Tt)=G(x 0,t) \mathbb{P}(T \ge t) = G(x_0, t)

Therefore, the mean for any function f(T)f(T) is

f(T)= 0 f(t)dG(x 0,t)= 0 f(T) tG(x 0,t)dt \langle f(T) \rangle = - \int_0^{\infty} f(t) \; d G(x_0, t) = - \int_0^{\infty} f(T) \; \partial_t G(x_0, t) \; d t

For the mean of the hitting time itself we therefore get:

T= 0 t tG(x 0,t)dt= 0 G(x 0,t)dt \langle T \rangle = - \int_0^{\infty} t \; \partial_t G(x_0, t) \; d t = \int_0^{\infty} G(x_0, t) \; d t

The last part follows from integration by parts.

For the higher moments we get of course

T n(x):=T n= 0 t n1G(x 0,t)dt T_n(x) := \langle T^n \rangle = \int_0^{\infty} \; t^{n-1} \; G(x_0, t) \; d t

G satisfies the backward equation

Now that we have a formula for the expectation and all higher moments of the hitting time involving the function G(x 0,t)G(x_0, t), we need more information about G(x 0,t)G(x_0, t).

To this end, we will see that the function G(x,t)G(x, t) satisfies the backward Fokker-Planck equation. Note that we treat the function argument xx now as a variable, not as a constant starting point x 0x_0.

Since our system is homogeneous in time, we have

p(x,t|x 0,0)=p(x,0|x 0,t) p(x, t| x_0, 0) = p(x, 0| x_0, -t)

which implies that we can state the backward Fokker-Planck equation via symmetry as a forward equation:

tp(x,t|x 0,0)=(f(x)x+12g 2(x,t) 2x 2)p(x,t|x 0,0) \frac{\partial}{\partial t} p(x, t | x_0, 0) = ( - f(x) \frac{\partial}{\partial x} + \frac{1}{2} g^2(x, t) \frac{\partial^2}{\partial x^2} ) p(x, t | x_0, 0)

When we assume that all involved functions are “nice” enough so that we may interchange differentiation and integration, we see that G(x,t)G(x, t) satisfies this backward equation, too. The initial condition for G(x,t)G(x, t) is obviously

G(x,t)={1, for axb 0, else G(x, t) = \begin{cases} 1, & \text{for } \; a \le x \le b \\ 0, & \text{else } \end{cases}

Recursion formula for the moments

We can integrate the backwards equation over time from 00 to \infty, noting that

0 tG(x,t)=G(x,)G(x,0)=1 \int_0^{\infty} \partial_t \; G(x, t) = G(x, \infty) - G(x, 0) = -1

and get a differential equation for the expectation value T(x)T(x) and similarly for all higher moments T x(x)T_x(x):

f(x)ddxT(x)+12g(x)d 2dx 2T(x)=1 f(x) \frac{d}{ d x} T(x) + \frac{1}{2} g(x) \frac{d^2}{ d x^2} T(x) = -1

with boundary conditions T(a)=T(b)=0T(a) = T(b) = 0 and for the higher moments

f(x)ddxT n(x)+12g(x)d 2dx 2T n(x)=nT n1(x) f(x) \; \frac{d}{ d x} \; T_n(x) + \frac{1}{2} \; g(x) \; \frac{d^2}{ d x^2} \; T_n(x) = -n \; T_{n-1}(x)

Therefore, in our example, we are able to calculate the moments of the hitting time recursively.


The following textbook is a general introduction to stochastic models for practitioners, including first-hitting-time problems:

First hitting time problems have been a research topic in mathematics of their own, for some time now, here is a textbook that provides an introduction at an elementary level, for time independent diffusion processes, using electrostatics as an analogy: