Sharpening Doob’s upcrossing inequality for Brownian motion

Question

Note: This question is heavily related to a series of posts ([1], [2]) by user GJC20.

Provided a martingale $X$ in continuous-time, Doob's upcrosssing inequality states:

If $U(a,b)$ denotes the number of up-crossings of $X$ through $(a,b)$ up to time $1$, then

$$\mathbb E \big[U(a,b)\big] \leq \frac{1}{b-a}\mathbb E\big[(a-X_1)^+\big].$$

This inequality should essentially never be sharp for a martingale with continuous sample paths, except for those almost surely constant in time, or those that never hit $a$ to begin with.

Suppose now $X$ is a Brownian motion. How much can we sharpen this bound?

Answer 1

The goal is to obtain sharper/more explicit bounds for the upcrossing number of Brownian motion compared to the naive bound given by Doob’s upcrossing inequality.

Notation:

• For $b > a$, we set $\nu_{0}(a, b) = 0$ and write, for all $i \in \mathbb Z_+$,

$$\tau_i(a, b) := \inf \{t \geq \nu_{i-1}(a, b) | W_t = a\}$$

$$\nu_i(a, b) := \inf \{t \geq \tau_i(a, b)| W_t = b\}$$

We shall assume the Brownian motion starts at $c \in \mathbb R$.

We have

$$U(a, b) = \sum_{i = 1}^\infty \mathbf 1(\nu_i (a, b) \leq 1),$$

so that

$$\mathbb E[U(a, b)] = \sum_{i = 1}^\infty \mathbb P(\nu_i (a, b) \leq 1),$$

But, writing $M$ for the supremum over $[0, 1]$ of a Brownian motion starting at $0$, we have by iterated use of the reflection principle,

$$\mathbb P(\nu_i (a, b) \leq 1) = \mathbb P (M \geq |c-a| + (2i-1)(b - a)).$$

Recalling that $M \overset{d}{=} |B_1|$, we get the exact series expression

$$\mathbb E[U(a, b)] = \sum_{i = 1}^\infty \mathbb P (|B_1| \geq |c-a| + (2i-1)(b - a)).$$

We can obtain useful bounds from this expression. Indeed, by a simple Chebyshev inequality,

$$\mathbb E[U(a, b)] \leq \sum_{i = 1}^\infty \frac{1}{(b-a)^2(2i-1)^2},$$

so that

$$\mathbb E[U(a, b)] \leq \frac{\pi^2}{8(b-a)^2},$$

which is already more effective than Doob’s inequality once $b-a$ is large.