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Let $\dot{X_{t}} = b(X_{t}) + \sigma(X_{t})\dot{W_{t}}$ where $X_{0} = x \in \mathbb{R}$ and $W_{t}$ is a Wiener process. Let $\tau = \min\{t \mid X_{t} \not \in G\}$, where $G = (M, N) \subset \mathbb{R}$. Let $\beta = 0$ and $\sigma = 1$. Then, calculate $\mathbb{E}[\tau^{2}]$.

A solution is provided here, which uses martingales: Squared hitting time expectation including Brownian motion

I was wondering whether it is possible to solve this problem without the use of martingales. I've tried for a couple of hours, and I can't come up with anything. I've seen a derivation somewhere of $\mathbb{E}[\tau]$ that didn't use martingales, and I was wondering if something similar is possible.

Thank you

  • In your case, $\tau$ is the hitting time of a two sided barrier for the Brownian motion. It is trivial to show that $\tau = \tau_a\wedge\tau_b$ where $\tau_x$ is hitting time of a level $x$ of the BM. I recall the the Laplace transform of $\tau$: \begin{align} E[\exp\left(-\mu\tau\right)] = \frac{\cosh\left(\sqrt{2\mu}(a+b)/2\right)}{\cosh\left(\sqrt{2\mu}(b-a)/2\right)} \end{align} In order to compute $E[\tau^2]$, you can derivate the above expression twice and choose wisely $\mu$. – Sesame May 08 '19 at 17:28
  • Thanks for your reply. I didn't learn Laplace transform either. How else can I do this problem? –  May 08 '19 at 22:05
  • You can get just the expectation by solving a certain differential equation obtained by renewal theory. The result is basically that the boundary hitting time in an SDE with generator $L$ satisfies the differential equation $Lu=-1$ with the boundary condition $u=0$. This method is nontrivial to generalize to get higher moments. – Ian May 08 '19 at 22:13
  • Yes, I have learned about the differential operator $L$, and I've also learned stochastic differential equations. So I think that's the way I'm supposed to do it. This is a sample problem I've had for a final exam I'm studying for. I've been trying this problem for two days, and I can't get it still. I've been reading through these notes here: http://math.stanford.edu/~ryzhik/STANFORD/STANF227-10/notes227-09.pdf. But I still cannot get it –  May 08 '19 at 22:34
  • One way you can do it is to use a PDE to derive the distribution of $\tau$ as a function of $x$. The idea of this is to define a suitable PDE with the terminal condition $p(x,T)=0$ and the side boundary conditions $p(a,t)=p(b,t)=1$, where $p(0,x)$ is the probability to hit the boundary starting from the point $x$ before time $T$ elapses, i.e. $P(\tau_x \leq T)$. Here the spatial differential operator is just $L$ but as I recall you need a temporal differential operator too, I think it's just $\frac{\partial}{\partial t}$ but I'd have to play with it to be sure. – Ian May 08 '19 at 23:09
  • @Ian I am really unsure about how to do this. I don't have many examples in my book, and I am trying to learn this material for an exam. Can you please show me how to solve this problem? It will really, really help me a lot. –  May 09 '19 at 04:16

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