11
$\begingroup$

If an increasing process $X_t$ has a known Laplace transform $\mathbb{E} e^{-s X_t} = m_t(s)$, define its hitting time $\tau$ to some level $B$ to be $$ \tau = \inf\{ u > 0 : X_u \geq B \}. $$ Can we express the Laplace transform (or its CDF) of $\tau$ in terms of $m$?

In particular, I am interested in the hitting time of the integrated CIR process $$ X_t = \int_0^t V_s \ ds $$ where $$ dV_t = (\alpha V_t + \beta) dt + \gamma \sqrt{V_t} dW_t. $$ The Laplace transform of $X_t$ is known in closed-form in this case, and given by $$ m_t(s) = \mathbb{E} e^{-s X_t} = \left( \frac{e^{-\alpha t/2}}{\cosh(Pt/2)-\frac{\alpha}{P}\sinh(Pt/2)} \right)^{2\beta/\gamma^2} \exp\left(-\frac{s V_0}{P} \frac{2 \sinh(Pt/2)}{\cosh(Pt/2)-\frac{\alpha}{P} \sinh(Pt/2)}\right) $$ where $P = \sqrt{\alpha^2 + 2 \gamma^2 s}$.

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ You may be able to use Dynkan's formula and then numerically integrate the resulting ODE. I doubt there is a closed form solution. $\endgroup$
    – user9403
    Oct 21, 2015 at 13:58

1 Answer 1

Reset to default
-1
$\begingroup$

The Laplace transform of the integrated process CIR process is given by, see e.g. Dufresne (2001). you can download it

  1. The integrated square-root process
Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ Nice paper. But how does this relate to the Laplace transform of the hitting time? $\endgroup$
    – Hans
    Jan 10, 2017 at 9:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.