Hitting probability

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

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 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 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.

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,

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