5
$\begingroup$

Let $X$ be a geometric Brownian motion, satisfying the SDE

$$dX_t = \sigma X_t \, dW_t, X_0 = 1.$$

for $W$ a standard one dimensional Brownian motion, and $\sigma > 0$ a constant.

Define the drawdown process $D$ by

$$D_t := \max_{0 \leq s \leq t} X_s - X_t.$$

and for some fixed $a >0$, let $\tau$ be the stopping time

$$\tau := \inf \{t > 0 \, | \, D_t =a\}.$$

Question: Can we obtain, or characterise the joint distribution of $(\tau, X_\tau)$?

$\endgroup$

1 Answer 1

2
+200
$\begingroup$

Before you get no response i think i can provide you with some limited insight into the distribution. Firstly with $X_0 = 1 \implies X_\tau \leq a+1$ and since $X_t > 0, \forall t > 0$, we can also say that $X_\tau > a$.

From these two observations we can see that $\tau$ is bounded by two other stoppage times: $$ \tau_- = \inf\{t > 0|X_t = a \} \leq \tau \leq \inf\{t>0 |X_t = a +1 \} = \tau_+ $$

Now $\tau_{-, +}$ can be transformed to be about $W_t$, as in $\tau_- = \inf\{t>0|W_t = \frac{\log(a)}{\sigma}\}$, $\tau_+ = ...$. Professor Steve Lalley calls this a first-passage time in this course: http://galton.uchicago.edu/~lalley/Courses/312/ (relevant subsections, 2nd Random walk and last Brownian Motion)

The expectation value of such passage times for Wiener Processes are $\infty$ because they are Martingales and this a type of Gamblers Ruin: https://en.wikipedia.org/wiki/Optional_stopping_theorem#Applications

So that's pretty much it. $X_\tau \in (a, a+1]$ and $\mathbb E(\tau) = \infty$. I hope it helps.

$\endgroup$
1
  • $\begingroup$ Thanks for your help! I have awarded you the bounty so it doesn’t go to waste. $\endgroup$
    – Nate River
    Dec 28, 2022 at 7:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.