Hitting probability

A hitting probability is the probability of reaching one set before another and solves a discrete harmonic boundary problem.

Outside a target set, first-step analysis for a Markov chain gives $h_i=P_{ii}{h}_i+{\sum }_{j\ne i}{P}_{ij}{h}_j$. Moving the holding term and dividing gives the displayed average over the next distinct state. Thus inserting or removing self-loops preserves eventual hitting probabilities, provided the state is not made absorbing and the relative probabilities of its exits are unchanged. Boundary values are one on the target and zero in closed classes unable to reach it. The waiting time changes, but the sequence of distinct visited states does not.

For $Y_t=-vt+\sqrt{2D}{B}_t$ with $v,D,\ell >0$ and standard Brownian motion, the one-time occupation probability is $\frac{1}{2}\mathrm{erfc}[(\ell +vt)/\sqrt{4Dt}]$. It is maximized at $t=\ell /v$, giving $\frac{1}{2}\mathrm{erfc}(\sqrt{v\ell /D})$. The actual hitting probability is $e^{-v\ell /D}$. To derive it, solve $D{h}^{\prime \prime }-v{h}^{\prime }=0$ on $(-A,\ell )$ with $h(-A)=0$, $h(\ell )=1$, obtaining $h(0)=(1-e^{-vA/D})/(e^{v\ell /D}-e^{-vA/D})$, and let $A\to \infty$. A path that reached the level and returned contributes to the latter event but not to a given one-time tail.

With three evenly matched players, the previous winner plays the previous spectator. A pair of successive distinct winners $ij$ moves to $jj$ or $jk$ with equal probabilities, where $k$ is the remaining player. The transient states form two directed 3-cycles and each step has absorption probability one half. Starting with a game between two specified players, their overall winning probabilities are each $5/14$, while the original spectator wins with probability $2/7$.

The hitting probability of a set is one on that set and transition-harmonic elsewhere by the Markov property. Finite-horizon hitting probabilities begin at the boundary indicator and increase through the same positive transition recursion. Any nonnegative solution dominates every iterate; monotone passage to infinite time shows that it dominates the hitting probability. This proves minimality without requiring almost-sure hitting.

A splitting probability is the probability that a stochastic process reaches one target before competing targets. For a diffusion it is harmonic for the Kolmogorov backward equation, with boundary values one on the selected target and zero on the others.

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.

For a diffusion with backward generator $L$, the mean first-passage time satisfies $L\tau =-1$ in the interior, with the backward forms of the process's boundary conditions.

The waiting time for a specified directed edge to be traversed differs from the waiting time to visit its terminal state. First-step equations include no continuation term when the designated transition succeeds, but still count that last step. They can also be derived by augmenting the chain with an absorbing success state.

The survival probability is $Q(x,t)={\mathbb{P}}_x(T_A>t)$. When absorption eventually occurs,

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.

Suppose an urn starts with $m$ green balls and $m+2$ red balls. Drawing green removes one green ball, while drawing red adds one green and one red ball. The difference remains two, and $h(g)=1/(g+1)$ is harmonic for the green-ball chain. Stopping $h$ on hitting $0$ or $N$ and then letting $N$ tend to infinity gives termination probability