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Consider a one-dimensional Itô diffusion:

$$\mathrm{d} X_{t}=b\left(X_{t}\right) \mathrm{d} t+\sigma\left(X_{t}\right) \mathrm{d} B_{t}$$

where $X_0 = 0$ and $B_t$ is the standard Brownian Motion. The exit time $\tau$ is defined as:

$$\tau = \inf (t >0 : X_t \notin (a, b) )$$

Now I want to calculate the probability $P(X_{\tau} = a)$ and the expectation $E \tau$.

Previously, when I solved the questions about exit time, I used Optional Sampling Theorem. In this question, however, since I don't know the explicit expression of $X_t$, I cannot find a reasonable way to handle it. So if anybody can help me.? Thanks so much!

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    $\begingroup$ $f(x) = \mathbb{E}^x\tau$ satisfies $f(a) = f(b) = 0$ and $L f(x) = -1$, where $L = \sigma^2 \partial_{xx} + b \partial_x$ is the generator. The probability that $X_\tau = a$, $f(x) = \mathbb{P}^x(X_\tau = a)$, satisfies a similar relation: $f(a) = 1$, $f(b) = 0$, $L f(x) = 0$. This is pretty standard, you will find it in any textbook that discusses one-dimensional diffusions. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 13, 2020 at 6:51
  • $\begingroup$ @Mateusz Kwaśnicki Thanks so much! Your approach is very concise. I think $L$ should be $\sigma^2/2 \partial_{xx} + b \partial_{x}$. But when I try your method, I still cannot figure out why $L \mathrm{E}^x \tau = - 1$ and $L \mathrm{P}^x (X_{\tau} = a) = 0$? Could you do a favor and give me more details? Best. $\endgroup$
    – 香结丁
    Commented Apr 13, 2020 at 9:32
  • $\begingroup$ (1) Yes, I forgot $\tfrac{1}{2}$, sorry. (2) What textbook are you following? Once you identify $L$ with Dynkin's characteristic operator, that is $$L u(x) = \lim_{\substack{x_1 \to x^- \\ x_2 \to x^+}\} \frac{\mathbb{E}^x u(X(\tau_{(x_1, x_2)})) - u(x)}{\mathbb{E}^x \tau_{(x_1, x_2)}} ,$$ both properties follow directly from the strong Markov property. This is how Dynkin derives these kind of equations in his two-volume Markov processes. Modern textbooks usually take a different approach, though. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Apr 13, 2020 at 10:16
  • $\begingroup$ @Mateusz Kwaśnicki Appreciate your help!!! In fact, I am just taking a stochastic calculus class for undergraduate students, so it may take me a considerable amount of time to get it sussed. Thanks again! $\endgroup$
    – 香结丁
    Commented Apr 13, 2020 at 12:53

1 Answer 1

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You can solve this by reducing it to a problem of Brownian motion: Define the scale function

$\varsigma(x) = \int_{X_0}^x e^{-2\int_{X_0}^y \frac{b(z)}{\sigma^2(z)} dz} dy$

the process

$M_t = \varsigma(X_t)$

is a local martingale. Therefore, by Dambis-Dubins-Schwarz it is a time changed Brownian motion

$M_t = W_{\langle M, M\rangle_t}$.

Thus, it suffices to study a time-changed Brownian motion, starting at $\varsigma(X_0)$, in the interval $(\varsigma(a), \varsigma(b))$.

Note that to calculate the probability of hitting each boundary, you do not have to calculate $\langle M, M\rangle$ explicitly. As long as $\langle M, M\rangle< \infty$ a.s., the boundary that one path finally hits does not depend on the speed you let run the clock.

This seem to be no longer true for the expectation. Though here you could proceed similarly as outlined by Mateusz in its comments. If $g$ is a solution to

$$ Lg = 1, \quad g(X_0) = 0$$

you can use Dynkin's formula to conclude

$$\mathbb{E}[\tau] = \mathbb{E}\int_0^\tau Lg(s) ds = \mathbb{E}[g(X_\tau)] = g(a) \mathbb{P}[X_\tau=a] + g(b) \mathbb{P}[X_\tau=b]$$

To make the argument rigorous, you will have of course check technical conditions. In particular existence of a solution to an SDE is not enough, you have to make sure that it exits the interval in finite time a.s. (e.g., as extreme example consider $b=\sigma=0$ when the diffusion will never exit the interval.

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  • $\begingroup$ Thank you so much!!! Since I just studied Dambis-Dubins-Schwarz Theorem in the last class, this is the solution I really want. But I have another question: how to calculate $<M, M>_t$? (It seems to be somewhat difficult as it includes an integral.) $\endgroup$
    – 香结丁
    Commented Apr 13, 2020 at 8:51
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    $\begingroup$ I tried to address your question by expanding my answer. $\endgroup$
    – Stephan Sturm
    Commented Apr 14, 2020 at 6:46
  • $\begingroup$ Thanks very much! $\endgroup$
    – 香结丁
    Commented Apr 20, 2020 at 7:30
  • $\begingroup$ So sorry for accepting your answer so late. I am a new user, I just know I can accept one's answer. $\endgroup$
    – 香结丁
    Commented Aug 2, 2020 at 10:36
  • $\begingroup$ @香结丁 No issue. Thank you! $\endgroup$
    – Stephan Sturm
    Commented Aug 2, 2020 at 15:20

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