How to prove that a Brownian bridge $\mathbb{P}(M[0, 1/2]\geq s)\leq 2\mathbb{P}(B(1/2)\geq s/2)?$

Question

Consider a Brownian bridge $B: [0,1]\to \mathbb{R}$ with $B(0)=B(1)=0$. Let $M[0, 1/2]=\max_{x\in[0,1/2]}B(x)$. How to prove that $$\mathbb{P}(M[0, 1/2]\geq s)\leq 2\mathbb{P}(B(1/2)\geq s/2)?$$

Actually, here is a "no big max" argument that could be used in the proof of construction Airy line ensemble. The "no big max" means the top curve between $(a, b)$ cannot get too high.

(Definition of Brownian bridge) If $\{B(t): t\geq 0\}$ is standard Brownian motion, then $\{Z(t): 0\leq t\leq 1\}$ is a Brownian bridge process when $$Z(t)=B(t)-tB(1).$$

Answer 1

Let $B_t:=B(t)$. We have to show that \begin{equation} P(M_{1/2}\ge s)\le2P(B_{1/2}\ge s/2) \tag{1} \end{equation} for $s\ge0$, where $M_T:=\max_{0\le t\le T}B_t$ for $T\in(0,1)$.

We shall prove (1) by first obtaining an explicit expression for $P(M_T\ge s)$.

Indeed, for $t\in[0,1]$, we can write $$B_t=W_t-tW_1,$$ where $W$ is a standard Wiener process. For each real $x$, consider \begin{equation} C^{-x}_t:=W_t-tx, \end{equation} so that $C^{-x}$ is a Wiener process with drift $-x$. Then \begin{align*} P(M_T\ge s|W_1=x)&=P(\max_{0\le t\le T}(W_t-tW_1)\ge s|W_1=x) \\ &=P(\max_{0\le t\le T}(W_t-tx)\ge s|W_1=x) \\ &=P(\max_{0\le t\le T}C^{-x}_t\ge s|C^{-x}_1=0) \\ &=e^{-2s^2}G\Big(\frac{(1-2T)s}{\sqrt{(1-T)T}}\Big)+G\Big(\frac{s}{\sqrt{(1-T)T}}\Big) \end{align*} by Theorem 3.1 (with $-x,s,1,T,0$ in place of $\mu,\beta,u,t,\eta$, respectively), where $1-G$ is the standard normal cdf. Note that $P(M_T\ge s|W_1=x)$ does not depend on $x$ (which of course should have been expected, since the Brownian bridge $B$ is independent of $W_1$). We conclude that \begin{equation} P(M_T\ge s) =e^{-2s^2}G\Big(\frac{(1-2T)s}{\sqrt{(1-T)T}}\Big)+G\Big(\frac{s}{\sqrt{(1-T)T}}\Big). \end{equation} In particular, \begin{equation} P(M_{1/2}\ge s) =\tfrac12\,e^{-2s^2}+G(2s)=:l(s). \end{equation} On the other hand, since $B_{1/2}$ equals $\tfrac12\,W_1$ in distribution, we have $P(B_{1/2}\ge s/2)=P(W_1\ge s)=G(s)$.

So, it remains to show that \begin{equation} d(s):=l(s)-2G(s)\le0. \end{equation} For real $s>0$, let \begin{equation} d_1(s):=d'(s)e^{2 s^2} =\sqrt{\tfrac{2}{\pi }} e^{3 s^2/2}-2 s-\sqrt{\tfrac{2}{\pi }}. \end{equation} Then $d_1'(s)=3 \sqrt{\frac{2}{\pi }} e^{3 s^2/2} s-2$ and hence $d_1$ is clearly $-+$; that is, $d_1$ may change its sign at most once (on $(0,\infty)$) and only from $-$ to $+$. So, $d_1$ is down-up -- that is, there is some $c\in[0,\infty]$ such that $d_1$ is decreasing on $(0,c]$ and increasing on $[c,\infty)$. Also, $d_1(0)=0$ and $d_1(\infty-)=\infty$. So, $d_1$ is $-+$ and hence $d$ is down-up. Also, $d(0)=0=d(\infty-)$. So, $d<0$ (on $(0,\infty)$), as desired.

Answer 2

You can prove it via standard computations. See (2.1), (2.7), and (2.8) of On the maximum of the generalized Brownian bridge.

Answer 3

Let $B_t:=B(t)$. For $t\in[0,1]$, we can write $$B_t=W_t-tW_1,$$ where $W$ is a standard Wiener process. We have to show that $$P(M_{1/2}\ge s)\le2P(B_{1/2}\ge s/2)$$ for $s\ge0$, where $M_{1/2}:=\max_{0\le t\le1/2}B_t$. The Brownian bridge $B$ is independent of $W_1$. Therefore and by the reflection principle, $$\tfrac12\,P(M_{1/2}\ge s)=P(M_{1/2}\ge s,W_1\ge0) =P(\max_{0\le t\le1/2}(W_t-tW_1)\ge s,W_1\ge0) \le P(\max_{0\le t\le1/2}W_t\ge s) =2P(W_{1/2}\ge s)=2P(W_1\ge s\sqrt2)=2P(B_{1/2}\ge s/\sqrt2);$$ here we also used the fact that $B_{1/2}$ equals $\tfrac12\,W_1$ in distribution. So, $$P(M_{1/2}\ge s)\le4P(B_{1/2}\ge s/\sqrt2).$$ The upper bound $4P(B_{1/2}\ge s/\sqrt2)$ on $P(M_{1/2}\ge s)$ is better than your desired upper bound $2P(B_{1/2}\ge s/2)$ for all $s\ge0.992$.