Is a Brownian Bridge less dispersed than a Brownian Motion?

Question

Consider the standard Wiener Process/Brownian Motion $W(t)$ on $[0,1]$ and the corresponding Brownian Bridge $B(t)=W(t)-\frac{t}{T}W(T)$. I am interested to know if the boundary crossing results for the Brownian motion provide a simple upper bound for the Brownian Bridge. Specifically, do we have for any function $f(t)$:

$$ \mathbb{P}(\exists t\in [0,1] \quad |W(t)|\geq f(t)) \geq \mathbb{P}(\exists t\in [0,1] \quad |B(t)|\geq f(t)) $$

More generally, if we consider two zero-mean Gaussian processes $S_1(t)$ and $S_2(t)$ with covariance functions $K_1(t,\tau)$ and $K_2(t,\tau)$, then if $K_1(t,\tau)\geq K_2(t,\tau)$ for all $t,\tau$, do we have: $$ \mathbb{P}(\exists t\in [0,1] \quad |S_1(t)|\geq f(t)) \geq \mathbb{P}(\exists t\in [0,1] \quad |S_2(t)|\geq f(t)) $$

The result seems very intuitive but somehow I dont know where I should even start.

Answer 1

Well, it's almost vice versa. Namely, Slepian's inequality says that if $K_1(t,\tau)\geq K_2(t,\tau)$ for all $t, \tau $ and $K_1(t,t) =K_2(t,t)$ for all $t$, then $$\mathbb{P}(\exists t\in [0,1] \quad |S_1(t)|\geq f(t)) \leq \mathbb{P}(\exists t\in [0,1] \quad |S_2(t)|\geq f(t)).$$

On the bright side, the inequality you're interested in is true whenever $K_1 - K_2 $ is positive semi-definite; in fact, in this case $\mathbb{P}(S_1\in C) \leq\mathbb{P}(S_2\in C) $ for any convex symmetric $C$ (I'm not sure if this fact has a name).

In particular, it is applicable to the Brownian bridge and the Wiener process, since $B(t)$ and $W(T)$ are independent.

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Here is the proof of the mentioned fact:

Since $\Delta K := K_1 - K_2$ is positive semidefinite, there exists a centered Gaussian process $X$ with covariance $\Delta K$. Since $K_1 = K_2 + \Delta K$, $S_1$ is distributed as $\widetilde S_2 + \widetilde X$, where $\widetilde S_2,\widetilde X$ are independent and have the same distributions as $S_2$ and $X$ respectively. Therefore, $$ \mathbb P (S_1 \in C) = \mathbb{P}(\widetilde S_2 + \widetilde X\in C) = \mathbb{E}\big[\mathbb{P}(\widetilde S_2 + h\in C)|_{h=\widetilde X_1}\big], $$ where the last equality holds thanks to independence of $\widetilde S_2$ and $\widetilde X$. Now by the Anderson inequality (see e.g. Lifshits Lectures on Gaussian Processes, Corollary 7.1), $\mathbb{P}(\widetilde S_2 + h\in C)\le \mathbb{P}(\widetilde S_2\in C)$, whence $$ \mathbb P (S_1 \in C)\le \mathbb{P}(\widetilde S_2\in C) = \mathbb{P}(S_2\in C), $$ as claimed.