Say I have a wiener process with $X(0) = X_0>0$ and the dynamics \begin{equation} dX(t) = \begin{cases} -\mu dt + \sigma X(t) dW(t)^{\mathbb{Q}} & \mathrm{for\ } X(t)>0\\ 0 & \mathrm{otherwise}\\ \end{cases} \end{equation} Where $\mu \geq 0$.

What can I say about the expected value of $E^{\mathbb{Q}}[X(T)]$ as $T \to \infty$? Naturally one would expect $X=0$ to happen eventually, but is it almost surely so?

  • $\begingroup$ yes. This is the classic problem gambler's ruin. Say we ignore the drift; then by symmetry, we have an equal change to go to $2X_0$ as to 0 - i.e. $50/50$. Once we're at $2X_0$, we again have an equal change to double our money as we do to hit zero (the change to hit zero after doubling once but before twice is 1/4). This series continues, and 0.5+0.25+0.125+... -> 1 - i.e. the probability to hit 0 goes to one. If you add in negative drift, it can only increase the rate of convergence. It's all in here. $\endgroup$ – will Jan 24 '17 at 23:47
  • $\begingroup$ The answer is actually on wikipedia, in a different page, it's caleld the risk of ruin. $\endgroup$ – will Jan 25 '17 at 9:24

Your Answer

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

Browse other questions tagged or ask your own question.