Density of first hitting time of Brownian motion with drift

For $c=0$ this result is knows as reflection principle (see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 6) and follows from the Markov property and symmetry of Brownian motion. However, for $c>0$ the proof is more involved since we have to get rid of the drift term.

Since by definition

$$[H_a \leq t] = \left[ \sup_{s \leq t} X_s \geq a \right] \tag{1}$$

determining the distribution of $H_a$ is equivalent to finding the distribution of $\sup_{s \leq t} X_s$. In order to find the distribution of the latter, we need two ingredients: Girsanov's theorem and the joint distribution $(X_t,\sup_{s \leq t} X_s)$ for a Brownian motion $(X_t)_{t \ge 0}$.

Girsanov theorem: Let $c \in \mathbb{R}$ and $(B_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{P})$. Then $$X_t := B_t+ct, \qquad t \leq T,$$ is a Brownian motion with respect to the probability measure $$d\mathbb{Q} := d\mathbb{Q}_T := \exp \left( -c B_T - \frac{c^2}{2} T \right) d\mathbb{P}.$$

For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Chapter 18.

Joint distribution of $(X_t, \sup_{s \leq t} X_s)$: Let $(X_t)_{t \geq 0}$ be a Brownian motion on a probability space $(\Omega,\mathcal{A},\mathbb{Q})$. Then the joint distribution $(X_t,\sup_{s \leq t} X_s)$ equals $$\mathbb{Q} \left[ X_t \in dx, \sup_{s \leq t} X_s \in dy \right] = \frac{2 (2y-x)}{\sqrt{2\pi t^3}} \exp \left(- \frac{(2y-x)^2}{2t} \right) 1_{[-\infty,y]}(x) \, dx \, dy. \tag{2}$$

For a proof see e.g. René Schilling/Lothar Partzsch: Brownian Motion - An Introduction to Stochastic Processes, Exercise 6.8 (there are full solutions available on the web).

So let's finally put it all together: It follows from $(1)$ and the definition of the probability measure $\mathbb{Q}_T$ that

$$\begin{align*} \mathbb{P}[H_a \leq T] &= \mathbb{P} \left[ \sup_{s \leq T} X_s \geq a \right] = \int 1_{[a,\infty)} \left( \sup_{s \leq T} X_s \right) \, d\mathbb{P} \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right) \exp \left( c B_T + \frac{c^2}{2} T \right) \, d\mathbb{Q}_T \\ &= \int 1_{[a,\infty)}\left( \sup_{s \leq T} X_s \right)\exp\left(c X_T- \frac{c^2}{2} T \right) \, d\mathbb{Q}_T. \end{align*}$$

By Girsanov's theorem, $(X_t)_{t \leq T}$ is a Brownian motion with respect to $\mathbb{Q}_T$ and therefore $(2)$ gives

$$\begin{align*} \mathbb{P}[H_a \leq T] &= \exp\left(- \frac{c^2}{2} T \right) \int_{y \geq a} \int_{x \leq y} e^{cx} \frac{2 (2y-x)}{\sqrt{2\pi T^3}} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dx \, dy. \end{align*}$$

It remains to calculate the integral expression. First of all, by Fubini's theorem,

$$\begin{align*} \mathbb{P}[H_a \leq T] &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \left(\int_{x \geq a} e^{cx} I_1(x) \, dx + \int_{x \leq a} e^{-cx} I_2(x) \, dx \right) \\ &:=J_1+J_2 \tag{3} \end{align*}$$

where

$$\begin{align*} I_1(x):= \int_{y \geq x} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &= \left[ - \exp \left(- \frac{(2y-x)^2}{2T} \right) \right]_{y=x}^{\infty} \\ &= \exp \left(- \frac{x^2}{2T} \right) \\ I_2(x) := \int_{y \geq a} \frac{2(2y-x)}{T} \exp \left(- \frac{(2y-x)^2}{2T} \right) \, dy &=\exp \left(- \frac{(2a-x)^2}{2T} \right). \end{align*}$$

Hence, $$\begin{align*} J_1 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \geq a} e^{cx} I_1(x) \, dx \\ &= \frac{1}{\sqrt{2\pi T}} \int_{x \geq a} \exp \left(- \frac{(x-cT)^2}{2T} \right) \, dx \\ &= \frac{1}{\sqrt{\pi}} \int_{z \geq \frac{a-cT}{\sqrt{2T}}} \exp(-z^2) \, dz \tag{4} \end{align*}$$

and

$$\begin{align*} J_2 &= \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{x \leq a} e^{cx} I_2(x) \, dx \\ &\stackrel{u:=2a-x}{=} \frac{1}{\sqrt{2\pi T}} \exp \left(- \frac{c^2}{2} T \right) \int_{u \geq a} e^{c(2a-u)} \exp\left(-\frac{u^2}{2T} \right) \, du \\ &=\ldots = \frac{e^{2ac}}{\sqrt{\pi}} \int_{z \geq \frac{a+CT}{\sqrt{2T}}} \exp(-z^2) \, dz. \tag{5} \end{align*}$$

Now if we differentiate $(3)$ with respect to $T$, using $(4)$ and $(5)$, we get

$$\begin{align*} \frac{d}{dT} \mathbb{P}(H_a \leq T) &= \frac{-1}{\sqrt{\pi}} \exp \left( - \frac{(a-cT)^2}{2T} \right) \left( \frac{-c}{\sqrt{2T}} - \frac{a-cT}{2 \sqrt{2} T^{3/2}} \right) \\ &\quad + \frac{-1}{\sqrt{\pi}} \underbrace{e^{2ac} \exp \left( - \frac{(a+cT)^2}{2T} \right)}_{\exp(-(a-cT)^2/2T)} \left( \frac{c}{\sqrt{2T}} - \frac{a+cT}{2 \sqrt{2} T^{3/2}} \right) \\ &= \frac{a}{\sqrt{2\pi T^3}} \exp \left(- \frac{(a-cT)^2}{2T} \right). \end{align*}$$

There is a trick for this one, which alleviates you of having to use the joint distribution of $B_t$ and $\sup B_t$:

Start with using Girsanov to construct a change of measure $$ \frac{d\mathbb Q}{d \mathbb P}\bigg|_{\mathcal F_t} =\exp(c B_t -\frac{1}{2}c^2t) $$ so that $$ \hat B_t = B_t -c t \qquad (=X_t) $$ is a $\mathbb Q$-BM. Denote by $\mathbb P(H_a \in dt)$ the probability of $H_a$ being in some infinitesimal interval. We have $$ \mathbb P(H_a \in dt) = \mathbb E_{\mathbb P}[1_{H_a \in dt}] = \mathbb E_{\mathbb Q} \left[ \left(\frac{d\mathbb Q}{d \mathbb P}\bigg|_{\mathcal F_t} \right)^{-1}1_{H_a \in dt} \right] = \mathbb E_{\mathbb Q} \left[ \exp\lbrace-c B_t +\frac{1}{2}c^2t\rbrace 1_{H_a \in dt} \right] \\ = \mathbb E_{\mathbb Q} \left[ \exp\lbrace-c \hat B_t -\frac{1}{2}c^2t\rbrace 1_{H_a \in dt} \right] $$ Now the trick: Instead of invoking the joint density, notice that for $H_a \in dt$ we have that $\hat B_t = a$ since it is continuous. So the above is equal to $$ = e^{-c a -\frac{1}{2}c^2t}\mathbb Q ( H_a \in dt ) $$ That probability is simply the density of the hitting time of a standard brownian motion (since $X_t$ is driftless in $\mathbb Q$), which is know: Define $H^B_a = \inf \lbrace u:B_u \geq a \rbrace$ $$ \mathbb P(H^B_a \in dt) = \frac{a}{\sqrt{2 \pi t^3}} e^{-a^2/(2t)} dt $$ So $$ \mathbb P(H_a \in dt) = e^{-c a -\frac{1}{2}c^2t}\frac{a}{\sqrt{2 \pi t^3}} e^{-a^2/(2t)} dt = \frac{a}{\sqrt{2 \pi t^3}} e^{-(a-ct)^2/(2t)} dt $$

I will also use slightly different notation. The solution presented here uses the PDE approach to solve this problem and does not need the Girsanov theorem or the Joint distribution.

Theorem : Let the arithmetic Brownian motion process $X \left(t\right)$ be defined by the following Brownian motion driven SDE \begin{equation} \mbox{d}X \left(t\right) = \mu \mbox{d}t + \sigma \mbox{d}{W}\left(t\right). \end{equation} with initial value $X_0$. Let $\tau =\inf \left(u |X(u) \le B\right)$ denote the first passage time for the barrier $X_0 < B$. Then the first passage time $\tau$ is distributed as Inverse Gaussian Distribution \begin{equation} \tau \sim IG\left(\frac{B - X_0}{\mu}, \frac{\left(B - X_0\right)^2}{\sigma^2}\right),\label{abmFirstPassageDist} \end{equation} and for $t > 0$ the pdf of $\tau$ is \begin{equation} f(t) = \sqrt{\frac{(B - X_0)^2}{2 \pi \sigma^2 t^3}} \exp\left[-\frac{ \left(\mu t - B + X_0\right)^2}{2 \sigma^2 t}\right]\label{abmFirstPassageDensity}. \end{equation}

Proof: The Kolmogorov Forward Equation corresponding to the SDE is \begin{equation} \frac{\partial p\left(x, t\right)}{\partial t} = -\mu \frac{\partial p\left(x, t\right)}{\partial x} + \frac{1}{2} \sigma^2 \frac{\partial^2 p\left(x, t\right)}{\partial x^2}. \end{equation} To obtain the survival probability function, we first solve this PDE with the following initial and boundary value conditions \begin{equation} p\left(x, 0\right) = \delta\left(x - x_0\right); \qquad p\left(\infty, t\right) = p\left(x = B, t\right) = 0 \qquad (t > 0)\nonumber \end{equation} where $x = x_0$ is the starting point of the diffusive process, containing the initial concentration of the distribution and $B > x_0$ denotes the barrier.

This BVP can be solved using the standard method of images technique.

The free-space fundamental solution (Green’s function) of this PDE is \begin{equation} \phi\left(x, t\right) = \frac{1}{\sqrt{2\pi \sigma^2 t}}\, \exp\left[{-\frac{\left\lbrack x - \mu t \right\rbrack^{2}}{2\sigma^2 t}}\right] \end{equation} hence, given initial condition the normalized solution for an unrestricted process, starting from $x_0$ can be obtained as

\begin{equation} \phi_{x_0}\left(x, t\right) = \frac{1}{\sqrt{2\pi \sigma^2 t}}\, \exp\left[{-\frac{\left\lbrack x - \mu t - x_0\right\rbrack^{2}}{2\sigma^2 t}}\right] \end{equation}

To solve this problem with the method of images, the barrier at $B$ is replaced by a mirror source located at a generic point $x = m$, with $m > B$ such that the solutions of equation emanating from the original and mirror sources exactly cancel each other at the position of the barrier at each instant of time. This implies the initial conditions in must now be changed to \begin{equation} \qquad p\left(x, 0\right) = \delta\left(x - x_0\right) - \exp\left(- \eta\right)\delta\left(x - m\right),\nonumber \end{equation} where $\eta$ determines the strength of the mirror image source. Due to the linearity of the PDE, a solution of this BVP is provided by \begin{equation} \qquad p\left(x, t\right) = \phi_{x_0}\left(x, t\right) - \exp\left(- \eta\right)\phi_{m}\left(x, t\right), \tag{1} \end{equation} where $\eta$ determines the strength of the mirror image source and $m > B$ is the location of this source.

The boundary condition requires for $x = B$, $p\left(x, t\right) = 0$ for all $t > 0$, which yields \begin{eqnarray} \nonumber \frac{\left\lbrack x - \mu t - x_0 \right\rbrack^{2}}{2 \sigma^2 t} &=& \eta + \frac{\left\lbrack x - \mu t - m\right\rbrack^{2}}{2 \sigma^2 t} \\ \Leftrightarrow\;\; \left\lbrack x - \mu t - x_0\right\rbrack^{2} &=& 2 \eta \sigma^2 t + \left\lbrack x - \mu t - m\right\rbrack^{2}. \tag{2} \end{eqnarray} Upon substituting $x = B$ and $t = 0$, we get \begin{equation} \left\lbrack B - x_0 \right\rbrack^{2} = \left\lbrack B - m\right\rbrack^{2}\nonumber \end{equation} upon recalling $m > b$, we see that $m = 2 B - x_0$. Upon resubstituting the value of $m$ and $x = B$ in equation $(2)$ we obtain $\eta = \frac{2 \mu (x_0 - B)}{\sigma ^2}$. With these choices of $m$ and $\eta$, $(1)$ gives the solution of the BVP as \begin{eqnarray}\label{p(x,t)} p\left(x, t\right) &=& \frac{1}{\sqrt{2\pi \sigma^2 t}}\, \left\{\exp\left[{-\frac{\left\lbrack x - \mu t - x_0\right\rbrack^{2}}{2\sigma^2 t}}\right] - \exp\left[-\frac{2 \mu (x_0 - B)}{\sigma ^2}\right] \exp\left[{-\frac{\left\lbrack x - \mu t - 2 B + x_0\right\rbrack^{2}}{2\sigma^2 t}}\right]\right\}. \end{eqnarray}

Under the condition $B > x_0$, the survival probability can be obtained as \begin{eqnarray}\nonumber S(t) &=& \int_{- \infty}^{b } {p\left(x, t\right){\kern 1pt} \,dx} \\\nonumber &=&\frac{1}{2} \left\{\text{erfc}\left[\frac{-B + x_0 + \mu t}{\sqrt{2} \sigma \sqrt{t}}\right] - \exp\left[\frac{2 \mu (B-x_0)}{\sigma ^2}\right] \text{erfc}\left[\frac{B - x_0 +\mu t}{\sqrt{2} \sigma \sqrt{t}}\right]\right\} \end{eqnarray} where $\text{erfc}\left(z\right)$ denotes the complementary error function. The first passage density can now be obtained as \begin{eqnarray} f\left(t\right) &=& - \frac{\mbox{d} S\left(t\right)}{\mbox{d}t}\\\nonumber &=& \frac{(B - x_0)}{\sqrt{2 \pi } \sigma t^{3/2}} \exp\left[ -\frac{( x_0 - B + \mu t)^2}{2 \sigma ^2 t}\right]. \end{eqnarray} In particular, a Brownian motion with drift $\mu$ reaches the level B with probability one if and only if $\mu$ and $B$ have the same sign and $B > x_0$. If $\mu$ and $B$ have opposite signs, this density is "defective" in the sense that $\mathbb{P}\left(\tau_B < \infty\right) < 1$. To put it another way the drift must orient the process towards the barrier, for a non-defective density to exist.