1

I am stuck with this question. Suppose a stock price follow Brownian motion starting at price $p$. Denote by $\tau_r$ the passage time reaching level $r$ with $r<p$. What would be the distribution of the stock at time $t$ given that $\tau_r \geq t$ (given that the stock's price never went below price $r$)?

FARRAF
  • 187

1 Answers1

0

Once one has the joint distribution, we can obtain the conditional one $${\displaystyle f_{Y\mid X}(y\mid x)={\frac {f_{X,Y}(x,y)}{f_{X}(x)}}\qquad .}$$

For the case of undrifted BM, one can just use the reflection principle see The Maximum of Brownian Motion and the Reflection Principle.

By definition

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

and so determining the distribution of hitting time $\tau_a$ is equivalent to finding the distribution of $\sup_{s \leq t} X_s$.

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).

For the case of drifted BM $X_{t}=B_{t}+c t$, there is a generalization here Density of first hitting time of Brownian motion with drift.

Thomas Kojar
  • 7,349