Hitting times, the reflection principle and the running maximum

SC · Chapter 212 min readAsked at Two Sigma, Citadel, Jane Street, Optiver

Assumes Brownian motion: the defining properties and what follows.

After this lesson you should be able to

  • State the reflection principle and use it on the running maximum.
  • Compute the probability of hitting one level before another.
  • Connect first-passage results to barrier options and gambler’s ruin.

Questions about whether a process ever reaches a level — rather than where it ends up — are first-passage questions, and one argument answers most of them. The reflection principle converts a statement about paths that touched a barrier into a statement about where paths finished, which is something you can compute.

Derivation 2.1

The reflection principle

Reflect the path after its first touch of the level.

  1. Take a path with max⁡s≤tWs≥a\text{Take a path with } \max_{s \le t} W_s \ge a

    It touched aa at some first time τ≤t\tau \le t.

  2. Reflect the portion after τ about the level a\text{Reflect the portion after } \tau \text{ about the level } a

    By symmetry the reflected path is equally likely.

  3. Pr⁡(max⁡s≤tWs≥a)=2Pr⁡(Wt≥a)\Pr\left(\max_{s\le t} W_s \ge a\right) = 2\Pr(W_t \ge a)

    Paths finishing above aa pair with paths that touched and came back.

Pr⁡(max⁡s≤tWs≥a)=2Φ ⁣(−at)\Pr\left(\max_{s\le t} W_s \ge a\right) = 2\Phi\!\left(-\frac{a}{\sqrt{t}}\right)
-3-2-10123Standard deviations at time T
Figure 2.2 · Why the reflection principle has a factor of two. The shaded right tail is the chance of finishing above two sigma. Every path that touched the level and came back has a mirror image that finished above it, so the chance of ever touching is twice the tail — about 4.6%4.6\% against 2.3%2.3\%.

Why the factor is exactly two. Every path that touched aa either finished above it or finished below. The reflection pairs each below-finishing touching path with an above-finishing one, and that pairing is exactly one-to-one — so the touching paths are twice the finishing-above paths. The factor of two is not an approximation; it is a bijection, and the same bijection is what makes the ballot problem and the Catalan count come out so cleanly.

Example 2.3

A stock at $100\$100 has 20%20\% annual volatility and zero drift. What is the chance it touches $110\$110 at some point in the next year?

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Worked solution

  1. Formula
    Pr⁡(max⁡≥a)=2Φ ⁣(−aσT)\Pr(\max \ge a) = 2\Phi\!\left(-\frac{a}{\sigma\sqrt{T}}\right)
  2. Substitute
    a=ln⁡(1.10)=0.0953,σT=0.20a = \ln(1.10) = 0.0953, \quad \sigma\sqrt{T} = 0.20
  3. Solve
    aσT=0.477\frac{a}{\sigma\sqrt{T}} = 0.477
  4. 2Φ(−0.477)=2×0.31682\Phi(-0.477) = 2 \times 0.3168
  5. Answer
    ≈63%\approx 63\%

Sanity check. The chance of finishing above $110\$110 is only about 32%32\%, so touching is twice as likely as ending there — which is the whole content of the reflection principle, and the reason a one-touch option costs about twice a digital.

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Hitting times, the reflection principle and the running maximum · QuantMax