
> 
> 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 $$Z(t)=B(t)-tB(1).$$



How to prove that a Brownian bridge $\mathbb{P}(M[0, 1/2]\geq s)\leq 2\mathbb{P}(B(1/2)\geq s/2)?$
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](https://www.researchgate.net/publication/236984395_On_the_maximum_of_the_generalized_Brownian_bridge) (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.


