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Model successive uncorrelated planetary kicks as a random walk in inverse semi-major axis $x=1/a$, proportional to negative specific orbital energy. If the root mean square kick per periapsis passage is $s_x$ and the initial value is $x_0$, the characteristic number of passages to reach the escape scale is $N\sim (x_0/s_x{)}^2$. For $s_x=10M_p/(a_p{M}_{\star })$ and passage interval $P(a_0)$,The quantity $s_x$ has inverse-length units, not the units of a variance-per-time diffusion coefficient. With $⟨(\Delta x{)}^2⟩=2D_{x}t$, that coefficient is $D_x=s_x^2/(2P)$. This estimate assumes many independent encounters; changing orbital periods and resonant correlations can invalidate the clock. It is not an exact mean first-passage time.

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.

Let $S_n$ have integrable, mean-zero independent and identically distributed random variables as increments, taking integer values at most one, with $\mathbb{P}(X_1=1)>0$. For a positive integer $b$, put $T_b=\inf \{n:S_n=b\}$ and $M(\tau )=\mathbb{E}{e}^{\tau X_1}$, $\tau >0$. Then $T_b$ is finite almost surely. Indeed, $b-S_{n\wedge T_b}$ is a nonnegative martingale; finite convergence on $\{T_b=\infty \}$ would force its integer increments eventually to vanish, contrary to the Borel-Cantelli lemmas applied to $\{X_n=1\}$.

The exponential martingale of a random walk $e^{\tau S_n}/M(\tau {)}^n$ stopped at $T_b$ is bounded by $e^{\tau b}$ because $S_{n\wedge T_b}\le b$ and $M(\tau )\ge 1$. The dominated convergence theorem therefore preserves its mean one at the limit and proves the displayed first-passage time transform. This argument accommodates arbitrarily large negative jumps; the restriction concerns upward jumps. For a simple symmetric random walk, $M(\tau )=\cosh \tau$, giving $\mathbb{E}{e}^{-\alpha T_b}=(e^{\alpha }-\sqrt{e^{2\alpha }-1}{)}^b$.

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 $a>0$, 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.

Model the polymerization Brownian ratchet by a rigid tip and a Brownian load. Let $a$ be the elementary load advance, ${\zeta }_L$ its drag coefficient, and $D=k_{B}T/{\zeta }_L$ its diffusion coefficient. For a spherical load, ${\zeta }_L=6\pi \mu R_L$, but no load radius is specified, so the rate is naturally expressed in terms of $D$. Assume addition is effectively instantaneous once the gap $h$ reaches $a$, and removal is negligible. The gap reflects at zero, first arrival at $a$ causes insertion, and insertion resets it to zero.

With no opposing force, the mean first-passage time satisfies $D{T}^{\prime \prime }=-1$, $T^{\prime }(0)=0$, $T(a)=0$. Integration gives $T(h)=(a^2-h^2)/(2D)$. One advance per passage givesThe advance-event rate is $2D/a^2$, while $V_0$ is a length per time.

A constant resisting force $F\ge 0$ produces drift $-u_F=-DF/(k_{B}T)$. The diffusion-limited Brownian ratchet with constant load now satisfiesAn integrating factor gives $T^{\prime }(h)=[1-e^{u_{F}h/D}]/u_F$. Integrating again from $h$ to $a$ gives $T(0)=D(e^z-1-z)/u_F^2$, where $z=Fa/(k_{B}T)$. ThereforeAt weak load the ratio is $1-z/3+O(z^2)$; at large load $V\sim (D/a)z^2{e}^{-z}$. Replacing the whole force dependence by $e^{-z}$ would discard the first-passage kinetics of this regime.

For two staggered strands that alternately advance the leading tip by half a monomer, $a=\delta /2$ and $V_0=4D/\delta$. Two aligned strands, as schematically drawn, do not automatically have that half-step: an advance may be $a=\delta$, with the other strand filling the same layer afterward. Use the same geometrical step in the passage problem and in converting event rate to velocity. The limiting formula assumes negligible removal, rigid tips, an overdamped load with constant drag, and fast insertion compared with $a^2/D$. With appreciable removal or additional internal strand states, the renewal reset model must be replaced by a coupled gap/reaction process.

Subtract the boundary slope from the Brownian motion: $X_t=B_t-bt$. Then $T_{a,b}$ is the first-passage time of $X$ to the fixed level $a$. The zero-drift Brownian reflection principle givesHere $\Phi$ is the standard normal distribution function. To change the drift to $-b$, use the Girsanov theorem on each finite horizon, with density $L_t=\exp (-b{B}_t-b^{2}t/2)$. On $\{T_a\le t\}$, conditional expectation at the stopped time gives $\mathbb{E}[L_t\mid {\mathcal{F}}_{T_a}]=L_{T_a}$, because the density is a true martingale. Since $B_{T_a}=a$, the drifted Brownian first-passage density is consequentlyThere may be additional mass at infinite hitting time; the displayed density only describes finite passage. With $e^{-\theta \infty }=0$, apply the given zero-drift Brownian first-passage Laplace transform:The zero-discount limit is $1$ for $b\le 0$ and $e^{-2ab}$ for $b>0$, so the hitting probability is less than one precisely when the line moves away with positive slope.

For a direct verification of the linear-boundary Brownian first-passage distribution, defineBoth terms tend to zero as $t\downarrow 0$. Let $\varphi$ be the standard normal density. The identityfollows by expanding the two squares. Differentiating the two normal probabilities then givesSince this is the previously derived hitting density and both distribution functions start at zero,This proof applies to every real $b$, including the defective distribution when $b>0$.

For the barrier digital put, use the dividend-free Black-Scholes model with constant interest rate $\rho$ and spot volatility $\sigma >0$. Under the risk-neutral measure,Set $T=t_0$ andReaching the barrier $c$ is equivalent to $W_t^Q$ reaching $a+bt$. The contingent claim pays one only on the survival event $\{T_{a,b}>T\}$. Risk-neutral pricing therefore givesThis is a path-survival price, not the price of a terminal digital put option. Continuity of the stock paths and the absence of an atom in the finite-horizon maximum imply that alternative touching conventions change no price. As $c\to \infty$, the value tends to $e^{-\rho T}$; as $c\downarrow S_0$, it tends to zero. Both checks agree with the survival interpretation.

Translate the horizontal coordinate by $x$. Then $X_t+x$ is a standard Brownian motion starting at zero, and continuity identifies $\tau$ with its first-passage time to $x$. The vertical coordinate remains an independent standard Brownian motion starting at zero. Part (a) therefore gives the Cauchy exit law from a Brownian half-plane:The TeX transcription's initial value $-z$ and later $Y_T$ are both corrected here to the PDF's $-x$ and $Y_{\tau }$.

Write $x=1/a$ for the comet's inverse semi-major axis. Its specific orbital energy is $-G{M}_{\star }x/2$, so escape corresponds to $x=0$. The printed quantityhas dimensions of inverse length. It cannot, literally, be a standard variance-per-time diffusion coefficient. Interpret it as the characteristic root mean square step in $x$ per periapsis passage, as in a discrete comet energy diffusion model.

If successive kicks are unbiased and uncorrelated, after $N$ passages the root mean square displacement is $s_x\sqrt{N}$. Starting at $a_0$ gives $x_0=1/a_0$, hence $N_{\mathrm{ej}}\sim (x_0/s_x{)}^2$. Taking the characteristic interval to be the initial orbital period $P(a_0)$ givesHere $P_p=2\pi \sqrt{a_p^3/(G{M}_{\star })}$. In particular, for $a_0\sim a_p$ this is $P_p(M_{\star }/M_p{)}^2/100$. Without specifying the initial semi-major axis and the time per statistically independent encounter, the PDF cannot determine a unique time.

With the convention $⟨(\Delta x{)}^2⟩=2D_{x}t$, the actual diffusion coefficient would be $D_x=s_x^2/[2P(a_0)]$. The estimate requires $N_{\mathrm{ej}}\gg 1$, orbit crossing, and sufficient phase decorrelation. Near escape the orbital period grows, and resonant or secular correlations invalidate the simple constant-step clock. This is a characteristic diffusion scale, not an exact mean first-passage time: even an unbiased Brownian motion on an unbounded half-line has an infinite mean time to reach its absorbing endpoint.

For $a>0$, the Brownian reflection principle at the first-passage time of $a$ gives the Brownian running maximum identitywhere $\Phi$ is the distribution function of the standard normal distribution. Indeed, reflection pairs paths that have reached $a$ and end below $a$ with paths ending above $a$; $B_t$ has no atom at $a$.

As $t\to \infty$, this probability tends to one. The events $\{S_n\ge a\}$ increase with integer $n$, so with probability one the path reaches $a$ in finite time. Taking a countable intersection over positive integer $a$ givesHere $S_{\infty }={\sup }_{s\ge 0}{B}_s$. This also proves that every positive-level Brownian first-passage time is finite almost surely.