EDIT: the original $\ge$ is now $>$ (sorry for the typo!)


Let $B\_1(\cdot)$ and $B\_2(\cdot)$ denote independent, standard Brownian bridges, i.e., they are mean-zero Gaussian processes on $[0,1]$ with $B(0)=B(1)=0$ and covariance function $t\_1(1-t\_2)$ for $t\_1 \le t\_2$. 
Let $F\_1(\cdot)$ and $F\_2(\cdot)$ each be a cumulative distribution function (CDF) that is continuous. 
(CDFs are also non-decreasing and satisfy $\lim\_{x\to-\infty}F(x)=0$ and $\lim\_{x\to\infty}F(x)=1$.) 


What is $\Pr( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] > 0 )$?




---


My guess/special case: I think the answer should be one. 
For example, if $F\_1(\cdot)=F\_2(\cdot)=F(\cdot)$, then $B\_1(F\_1(\cdot))-B\_2(F\_2(\cdot)) = B\_1(F(\cdot)) - B\_2(F(\cdot))$, so 
\begin{align}
 \Pr( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] > 0 )
&=\Pr( \sup\_{x\in\mathbb{R}} [B\_1(F(x)) - B\_2(F(x))] > 0 ) \\
&=\Pr( \sup\_{t\in[0,1]} [B\_1(t) - B\_2(t)] > 0 ) \\
&=\Pr( \sup\_{t\in[0,1]} \sqrt{2} B(t) > 0 ) \\
&= 1 .
\end{align}
The second-to-last line follows since $B\_1(\cdot)$ and $B\_2(\cdot)$ are independent, their difference has the distribution $\sqrt{2} B(\cdot)$ for some other standard Brownian bridge $B(\cdot)$. 
The final line follows from, e.g., Theorem 2 in Smirnov (1939) or equation (1.1) in Birnbaum and Tingey (1951).


However, if $F\_1 \ne F\_2$, then I don't think the difference $B\_1(F\_1(\cdot))-B\_2(F\_2(\cdot))$ is technically a Brownian bridge (it is some other sort of "Gaussian bridge"?), so I do not know how to rigorously show that the probability is one (if indeed it is!), nor do I have a reference for such a result.


Also: the answer below (to the original version with my typo...) shows that the probability of being $\ge0$ is one, so the question is now equivalent to the question whether the probability that the supremum equals zero is zero.



Supremum of difference of Brownian bridges: strictly positive wp 1?
The answer is $1$, even if $B\_1$ and $B\_2$ are not independent. Indeed, take any $u>0$. Then 
\begin{align\*}
 P( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] \ge -u )
 &\ge \sup\_{x\in\mathbb{R}}P(B\_1(F\_1(x)) - B\_2(F\_2(x)) \ge -u ) \\ 
 &\ge \lim\_{x\to-\infty}P(B\_1(F\_1(x)) - B\_2(F\_2(x)) \ge -u ) \\ 
 &=P(B\_1(0) - B\_2(0) \ge -u )=P(0-0 \ge -u )=1,
\end{align\*}
because $B\_i(\cdot)$ is continuous and therefore $B\_i(t)\underset{t\downarrow0}\longrightarrow B\_i(0)=0$ almost surely and hence in probability and in distribution. 
So, 
\begin{equation}
 P( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] \ge 0 )=
 \lim\_{u\downarrow0}P( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] \ge -u )
 =\lim\_{u\downarrow0}1=1. 
\end{equation}




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The OP has changed the question, now asking if $P( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] > 0 )=1$. In general, the answer to this question is no: consider e.g. the case when $B\_1=B\_2$ and $F\_1=F\_2$. 


However, the answer remains yes if, as is assumed in the OP's question, $B\_1$ and $B\_2$ are independent. The main idea here is that the Brownian bridge is close to the corresponding Brownian motion on any small interval of the form $[0,t]$. 


Indeed, we can write 
$B\_i(t)=W\_i(t)-tW\_i(1)$, where $W,W\_1,W\_2$ are independent standard Brownian motions. 
Let 
\begin{gather\*}
 X(x):=B\_1(F\_1(x)) - B\_2(F\_2(x)),\quad Y(x):=W\_1(F\_1(x)) - W\_2(F\_2(x)),\\
 Z(x):=Y(x)-X(x)=F\_1(x)W\_1(1)-F\_2(x)W\_2(1). 
\end{gather\*}
Note that the process $(Y(x))\_{x\in\mathbb{R}}$ equals the process $(\sqrt2\,W(\frac{F\_1(x)+F\_2(x)}2))\_{x\in\mathbb{R}}$ in distribution and 
\begin{equation}
 \lim\_{x\to-\infty}\frac{Z(x)}{\sqrt{F\_1(x)+F\_2(x)}}=0. 
\end{equation}
So, the probability in question is 
\begin{align\*}
 P( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] >0 )
 &\ge P( \limsup\_{x\to-\infty} \frac{X(x)}{\sqrt{F\_1(x)+F\_2(x)}} =\infty) \\
 &=P( \limsup\_{x\to-\infty} \frac{Y(x)-Z(x)}{\sqrt{F\_1(x)+F\_2(x)}} =\infty) \\ 
 &=P( \limsup\_{x\to-\infty} \frac{Y(x)}{\sqrt{F\_1(x)+F\_2(x)}} =\infty) \\ 
 &=P( \limsup\_{x\to-\infty} \sqrt2\,\frac{W(\frac{F\_1(x)+F\_2(x)}2)}{\sqrt{F\_1(x)+F\_2(x)}} =\infty)=1 
\end{align\*}
by the law of the iterated logarithm -- see e.g. Corollary 5.3 in [M\"orters and Peres](https://www.stat.berkeley.edu/~peres/bmbook.pdf). 
So, $P( \sup\_{x\in\mathbb{R}} [B\_1(F\_1(x)) - B\_2(F\_2(x))] >0 )=1$. 


