Let $B_t$ be standard Brownian motion and $\tau_a=\inf\{t\geq 0 : B_t \geq a\}$ be the stopping time where $B_t$ exceeds some value $a$.

Is there an analytic form for $\mathbb{E}\left[\frac{1}{\tau_a}\right]$ ?


Suppose $B_t$ represents the price of a financial instrument. If an investor buys this instrument today at a price of $p$ and plans to sell at a profit of $a$, then the expected (continually-compounded) return $r$ of the investor is given by

$$r = \log\left(1+\frac{a}{p}\right)\mathbb{E}\left[\frac{1}{\tau_a}\right]$$

I wish to find an analytic form for $r$ so that I may then maximise $r$ over all $a$. That is, I wish to find the optimal price at which the investor should sell the instrument in order to maximise her return.


1 Answer 1


First, note that for all $t\ge0$, we have the equality of events : $$\{\tau_a\le t\} = \left\{\sup_{0\le s \le t} B_s \ge a\right\}$$ (In words, this says that the Brownian Motion $B$ reaches level $a$ before time $t$ if and only if its supremum on $[0,t]$is at least $a$).

By the reflection principle and the above equality, we thus have $$\mathbb P\left(\tau_a\le t\right)=\mathbb P\left(\sup_{0\le s \le t} B_s \ge a\right)=2\mathbb P\left(B_t\ge a\right) $$ The above is true for any $t>0$ and in particular it is true for $t'\equiv 1/t$, so we have

$$\begin{align} F_a(t)\equiv\mathbb P\left(\frac {1}{\tau_a}\le t\right) &= \mathbb P\left(\tau_a\ge \frac 1 t\right)\\ &= 1 - \mathbb P\left(\tau_a\le \frac 1 t\right)\\ &= 1 - 2\mathbb P\left(B_{1/t}\ge a\right) \\ &= 1 - 2\mathbb P\left(tB_{1/t}\ge ta\right)\\ &= 1 - 2\mathbb P\left(B_t\ge ta\right)\tag1\\ &= 1 - 2\mathbb P \left(Z\ge a\sqrt t\right)\tag2\\ \end{align}$$ Where I used the time inversion property in $(1)$ and the fact that $Z\equiv B_t/\sqrt t\sim \mathcal N(0,1)$ in $(2)$.

At this point, it is quite straightforward to express the CDF of $\frac {1}{\tau_a}$ $F_{a} $ in terms of the standard normal CDF. In particular, if you want to compute $\mathbb E\left[\frac {1}{\tau_a}\right] $, you have at least two options :

  1. By a well-known formula, you can integrate $1-F_a$ directly : $$\mathbb E\left[\frac {1}{\tau_a}\right] = \int_0^\infty (1-F_a(t))\ dt = \int_0^\infty \left[1-\text{erf}\left(\frac{a\sqrt t}{\sqrt 2}\right)\right]\ dt $$
  2. You can compute the PDF $f_a(t) :=dF_a(t)/dt$ and integrate $$\mathbb E\left[\frac {1}{\tau_a}\right] = \int_0^\infty tf_a(t)\ dt $$

Both of these integrals involve very standard functions and by digging up the literature you should be able to find a way to express them in terms of $\text{erf}, \text{erfc}$ or other well-known quantities, which should allow you to approximate $\mathbb E\left[\frac {1}{\tau_a}\right] $ well enough for your purposes.


Your Answer

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

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