The curiosity arose while reading the Ch.18 of Arbitrage Theory in Continuous Time 3/ed, dedicated to pricing Barrier Options.

Definition 18.1 For any $y\in R$, the hitting time of y, $\tau(X,y)$, sometimes denoted by $\tau(y)$ or $\tau_y$, is defined by


The $X$-process absorbed at y is defined by


where we have used the notation $\alpha\land\beta=\min[\alpha,\beta]$.

Further, Prof. Björk states the following proposition:

Proposition 18.3 The density $f_\beta(x;t,\alpha)$ of the absorbed process $X_\beta(t)$, where $X$ is defined by

$$dX(t)=\mu dt+\sigma d W_t,$$ $$X(0)=\alpha,$$

is given by

$$f_\beta(x;t,\alpha)=\varphi \left(x;\mu t+\alpha,\sigma \sqrt{t}\right)-\exp\left(-\frac{ 2 \mu (\alpha -\beta )}{\sigma ^2}\right)\varphi \left(x;\text{$\mu $t}-\alpha +2 \beta,\sigma \sqrt{t}\right)$$

The support of this density is the interval $(\beta,\infty)$ if $\alpha>\beta$, and the interval $(-\infty,\beta)$ if $\alpha<\beta$.

Where $\varphi \left(x;\mu,\sigma\right)$ was used to denote the density of a normal distribution with mean $\mu$ and variance $\sigma^2$.

Not that I don't trust Prof. Björk, but I wish very much to see a proof of this proposition, which I was unable (so far) to derive myself.

The question is related to the Transition function for absorbed Brownian motion.


I will try to summarize the derivation found in Harrison (1990):

Let $M_t\equiv \sup \left\{ X_s,0 \le s \le t \right\}$.

Basically the derivation is accomplished in 2 steps:

1 - Find the joint distribution $F_t(x,y)\equiv P \left\{X_t\leq x,M_t\leq y\right\}$, where $X_t$ is the standard Brownian motion

2 - Use the change of measure to calculate the joint distribution of $X_t$ and $M_t$ in generality

The first step goes like this:

Let $X_t$ be the standard Brownian motion, i.e. $\mu=0$ and $\sigma=1$. Then we have:

$$ \begin{align} F_t(x,y) & = P \left\{X_t\leq x,M_t\leq y\right\} \\ & =P \left\{X_t\leq x\right\}-P \left\{X_t\leq x,M_t>y\right\}\\ & =\Phi\left(\frac{x}{\sqrt{t}}\right)-P \left\{X_t\leq x,M_t>y\right\} \end{align} $$ with $\Phi(\cdot)$ denoting the standard normal distribution.

Now, the term $P\left\{X_t\le x, M_t > y \right\}$ can be calculated heuristically using the so-called reflection principle (note that the restriction $\mu=0$ is critical here): for every sample path of $X$ that hits level $y$ before time $t$ but finishes below level $x$ at time $t$, there is another equally probable path (shown by the dotted line in Figure 1) that hits $y$ before $t$ and then travels upward at least $y - x$ units to finish above level $y + (y - x) = 2y - x$ at time $t$.

enter image description here

Thus $$ \begin{align} P\left\{X_t\le x, M_t > y \right\} & = P\left\{X_t\ge 2y-x \right\} \\ & = P\left\{X_t\le x-2y \right\} = \Phi\left(\frac{x-2y}{\sqrt{t}}\right) \end{align} $$

(This argument can be made rigorous using the strong Markov property)

So we have:

$$ P \left\{X_t\leq x,M_t\leq y\right\}=\Phi\left(\frac{x}{\sqrt{t}}\right)-\Phi\left(\frac{x-2y}{\sqrt{t}}\right) $$

and as a corollary:

$$ P \left\{X_t\in d x,M_t\le y\right\}=g_t(x,y) d x $$ where $g_t(x,y)=\frac{1}{\sqrt{t}}\left(\phi\left(\frac{x}{\sqrt{t}}\right)-\phi\left(\frac{x-2y}{\sqrt{t}}\right)\right)$ and $\phi(\cdot)$ is the standard normal density.

Remembering the last result from above we can proceed to the next step where we want to calculate the joint distribution of $X_t$ and $M_t$ for general values $\mu$ and $\sigma$. For this we will need to perform the change of measure:

$$ P\left\{X_t\in d x,M_t\le y\right\}=f_t(x,y) d x $$ where $f_t(x,y)$ is obtained through the change of measure using the Radon-Nikodym derivative (through a necessary rescaling):

$$ f_t(x,y)=\frac{1}{\sigma}\exp\left(\frac{\mu x}{\sigma}-\frac{\mu^2 t}{2\sigma^2} \right) g_t\left(\frac{x}{\sigma},\frac{y}{\sigma}\right) $$ and $g_t(\cdot,\cdot)$ being defined above.

In fact, if we employ some simplifications we can get the $f_t(x,y)$ to look like the density from the question.

From there we can perform the integration to get the following result:

$$ F_t(x,y)=\Phi\left(\frac{x-\mu t}{\sigma\sqrt{t}}\right)-e^{2\mu y/\sigma^2}\Phi\left(\frac{x-2y-\mu t}{\sigma \sqrt{t}}\right) $$

It should be straightforward to refine it for the process starting at a level $\alpha$.


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