# Given particle undergoing Geometric Brownian Motion, want to find formula for probability that max-min > z after n days

Consider a particle undergoing geometric brownian motion with drift $\mu$ and volatility $\sigma$ e.g. as in here. Let $W_t$ denote this geometric brownian motion with drift at time $t$. I am looking for a formula to calculate:
$$\mathbb{P}\big(\max_{0 \leq t \leq n} W_t - \min_{0\leq t \leq n} W_t > z\big)$$ The inputs to the formula will be $\mu$, $\sigma$, $z$, and $n$.

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Here is a result for (nongeometric) Brownian motion with drift for a related quantity. –  cardinal Mar 12 '11 at 21:57
I am going through that paper and it seems to be plain wrong. E.g.: try calculating G(0). It should give 1 since Prob(max-min>=0) = 1, but it leads to problems. Also, in the paper: E[D] = \int dhG(h). This is plain wrong. E[D] = \int hg(h)dh. –  morpheus Mar 12 '11 at 23:17
That's entirely possible. I haven't read that paper carefully. When I saw your question, I remembered having seen it maybe five or six years ago, so I typed it in to Google and linked to it. Caveat lector, I suppose. (Note, you can use standard TeX notation even in comments by wrapping math with $. – cardinal Mar 13 '11 at 1:21 Also, let$B_t = \log W_t$be a brownian motion with drift. Then if you know$\mathbb{P}(\min_{0 \leq t \leq n} B_t > a, \max_{0 \leq t \leq n} B_t < b)$, then you could conceivably back out what you desire by transformation of variables. Now, the quantity I have written is equivalent to the distribution of the first exit time of brownian motion with drift from the strip$(a,b)$for$a < b$. I imagine some fairly strong statement about this must be known, since the distribution of$(B_t, \max_{u \leq t} B_u)$is known, has closed form, and is closely related to the first hitting time of$b$. – cardinal Mar 13 '11 at 1:30 I am adding back my comment (sorry I deleted it). For RV subjected to brownian motion, equation 4.10 in this [paper](www-sop.inria.fr/members/Etienne.Tanre/publication/jtp.pdf) gives the answer – morpheus Mar 14 '11 at 17:56 show 2 more comments ## 2 Answers Let's start with the joint probability density of Brownian motion$W_t$and its maximum$M_t = \max_{0 \le u \le t} W_u$$f(m,w) = \frac{2(2m-w)}{T\sqrt{2 \pi T}} \exp \left\{ - \frac{(2m-w)^2}{2T} \right\}$$$We have to integrate$f(\mu,\omega)$over$\omega \in(-\infty,m), \mu \in (0,m), \mu \ge \omega$to get the distribution function of the maximum.$P[M_t \le m] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\frac{m}{\sqrt{t}}} e^{-\frac{-x^2}{2}} dx -\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{-\frac{m}{\sqrt{t}}} e^{-\frac{-x^2}{2}} dx $Differentiating we'll obtain the pdf of the maximum$ f(m) = \frac{2}{\sqrt{2 \pi t}} e^{-\frac{m^2}{2t}} dmm \in (0,\infty)$is understandable, since$W_0 = M_0 =0$By the reflection principle, the minimum$\min_{0 \le u \le t} W_u$has the same pdf, except that$m \in (-\infty,0)$By the reflection principle we also know that$\forall x \ge0: P[\min_{0 \le u \le t} W_u \le x] = P[\max_{0 \le u \le t} W_u \ge -x]therefore \begin{align} P[\max_{0 \le u \le t} W_u - \min_{0 \le u \le t} W_u \ge x] &= 2 P[\max_{0 \le u \le t} W_u \ge \frac{x}{2}]\\ &= 2 P[\min_{0 \le u \le t} W_u \le -\frac{x}{2}] \\ &= 2 \frac{2}{\sqrt{2 \pi}}\int_{-\infty}^{-\frac{x}{2}} e^{\frac{-m^2}{2}}{d}m\\ \end{align}\forall x \in -\infty, 0)$the only problem with this solution is that the probability adds together to 2 as opposed to 1. I guess the easiest way to solve this would be by eliminating the 2 in front, but is can't see any mathematical rationale behind doing so. Any suggestions? - add comment The following horrible formula for the joint distribution of max, min and end value of a Brownian motion was copied without guarantees from the Handbook Of Brownian Motion (Borodin/Salminen), 1.15.8, p.271. First, for simplicity, this is only written for$\sigma=1,t=1$, and the more general case comes directly from scaling. If we shorten W as the Brownian Botion at t=1, m as the minimum and M as the maximum over$[0,1]$, then for$a < min(0,z) \le max(0,z) < b$it holds $$P(a < m, M < b, W \in dz) = \frac{1}{\sqrt{2\pi}}e^{(\mu z-\mu^2/2)} \cdot \sum_{k =-\infty}^{\infty} \Bigl(e^{-(z+2k(b-a))^2/2} - e^{(z-2a + 2k(b-a))^2/2} \Bigr) dz\; .$$ (Apologies for using z here in a different context.) If one really wants to, one can compute from this an even more horrible formula for the above probability. It is now in principle possible to derive from this a formula for what you want, by finding the density function$p_{m,M,W}\$, and using $$P(e^M-e^m\le r) = \int_{(x,y,z)\ :\ e^x \le e^z \le e^y \le e^x + r} p_{m,M,W}(x,y,z) d(x,y,z)\;,$$ but I shudder at the monster I expect to fall out from this. It might be better to give up and simulate the probability in question, and find some asymptotics.

However, if you would like to proceed with it, I suggest you look not into the Handbook Of Brownian Motion, but rather into this paper, as it is much more readable.

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