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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.

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    $\begingroup$ "wp" is an abbreviation for "with probability"? $\endgroup$ – Gerry Myerson Feb 12 '18 at 4:41
  • $\begingroup$ The difference of independent browinian bridges is also a Brownian bridge. As it is a gaussian process you can calculate covariances, but also, the bridge has the representation W(s)−sW(1) $\endgroup$ – user83457 Feb 12 '18 at 10:54
  • $\begingroup$ @GerryMyerson yes; the title was getting long (and I think wp would be understood by anyone in a probability-related field?) $\endgroup$ – David M Kaplan Feb 12 '18 at 16:01
  • $\begingroup$ @michael I agree $B_1(\cdot)-B_2(\cdot)$ is a Brownian bridge, so I think that's true if $F_1=F_2$ (like in the special case above), but otherwise the covariance is $F_{11}(1-F_{12})+F_{21}(1-F_{22})$ for $F_{ik}=F_i(t_k)$ for points $t_1<t_2$; it is a Gaussian process as you said, but maybe not directly related to a Brownian bridge? Should I call $B(F(\cdot))$ something else (what?) in the title, to clarify this? $\endgroup$ – David M Kaplan Feb 12 '18 at 16:06
  • $\begingroup$ my mistake, I see you made that point yourself. $\endgroup$ – user83457 Feb 12 '18 at 16:14
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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}


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. So, $P( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] >0 )=1$.

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  • $\begingroup$ Your (excellent) answer made me realize a small but significant typo in my question: the $\ge$ should be $>$ (as I have now edited). I feel bad changing it, but is there a way to extend your answer to accommodate this? For example, if the supremum has a continuous distribution, then $\Pr(\sup_{x\in\mathbb{R}} [\cdots] = 0) = 0$, which would be sufficient. It seems "obvious" that the $\sup$ indeed has a continuous distribution, but I don't know if that's a well-known fact (reference?) or else how to prove it? $\endgroup$ – David M Kaplan Feb 12 '18 at 3:38
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    $\begingroup$ The answer remains $1$ if $B_1$ and $B_2$ are independent. I'll try to write it down soon. $\endgroup$ – Iosif Pinelis Feb 12 '18 at 15:03
  • $\begingroup$ I'm still mid-way, but should $Z(x)=F_1(x)W_1(1)-F_2(x)W_2(1)$ (instead of +)? $\endgroup$ – David M Kaplan Feb 12 '18 at 16:38
  • $\begingroup$ @DavidMKaplan : You are of course right -- thank you for spotting this mistake, which is now fixed. Of course, what is important is that $Z(x)$ is negligible anyway as $x\to-\infty$. $\endgroup$ – Iosif Pinelis Feb 12 '18 at 16:42
  • $\begingroup$ 1) very helpful intuition! 2) if $F_1$ and $F_2$ have bounded support, then the $-\infty$ can be replaced by the smaller of the two lower bounds, right? 3) I'm accepting the answer [& will thank you in my paper] b/c it seems right, but how does the LIL work in the last step? I'm used to LIL as $n\to\infty$ for a scaled sum of iid rv's...or just a URL for some lecture notes to explain it? (I'm Googling it myself now...) $\endgroup$ – David M Kaplan Feb 12 '18 at 17:20
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(Edited: I added additional details)

The answer is yes also with the strict inequality. Denote by $a_j = \inf \{x : F_j(x) > 0\} \in [-\infty, \infty)$ the left endpoint of the support of $F_j$, $j = 1, 2$. With no loss of generality assume that $a_1 \leqslant a_2$.

With probability one, there is a decreasing sequence $(t_n)$ which converges to $0$ and such that $B_1(t_n) > 0$. Note that $t_n$ are random, but they depend only on $B_1$. Write $x_n = F_1^{-1}(t_n)$ and $s_n = F_2(x_n)$. Then $(x_n)$ converges to $a_1$. Furthermore, if $a_1 = a_2$, then $s_n > 0$ and $(s_n)$ converges to $0$; if $a_1 < a_2$, then $s_n = 0$ for $n$ large enough.

Conditioning on $B_1$, we obtain $$ \operatorname{Pr}(\sup_{x \in \mathbb{R}}[B_1(F_1(x)) - B_2(F_2(x))] > 0) \geqslant \operatorname{Pr}(\exists n : B_2(s_n) \leqslant 0) . $$ The right-hand side is one, because (as we prove below) for any decreasing sequence $(s_n)$ which converges to zero we have $\operatorname{Pr}(\forall n : B_2(s_n) > 0) = 0$. Note that we can consider the sequence $(s_n)$ to be fixed, because it only depends on $B_1$.

If $a_1 < a_2$, then $s_n = 0$ for $n$ large enough, and so obviously $\operatorname{Pr}(\forall n : B_2(s_n) > 0) = 0$. Hence, we suppose that $a_1 = a_2$, so that $s_n > 0$ and $(s_n)$ converges to $0$.

We choose a (random) subsequence $s_{k_n}$ recursively in such a way that $$ \operatorname{Pr}(B_2(s_{k_{n+1}}) \geqslant 0 | B_2(s_{k_1}), \ldots, B_2(s_{k_n})) < \tfrac{3}{4} \tag{1}$$ (see below for a remark why this is possible). Then we have $$ \operatorname{Pr}(B_2(s_{k_1}) \geqslant 0, \ldots, B_2(s_{k_n}) \geqslant 0) < (\tfrac{3}{4})^n , $$ and so $$ \operatorname{Pr}(\forall n : B_2(s_{k_n}) \geqslant 0) = 0 , $$ as desired.

To see why it is possible to choose $s_{k_{n+1}}$ so that (1) is satisfied, note that for $m > k_n$, the conditional distribution of $B_2(s_m)$ given the values of $B_2(s_{k_1}), \ldots, B_2(s_{k_n})$ is normal with mean $(s_m / s_{k_n}) B_2(s_{k_n})$ and variance $s_m (s_{k_n} - s_m)$. It follows that the conditional probability in (1) is equal to $$ \Phi\left(\frac{(s_m / s_{k_n}) B_2(s_{k_n})}{\sqrt{s_m (s_{k_n} - s_m)}} \right) , $$ where $\Phi$ is the CDF of the standard normal distribution. As $m \to \infty$, the above expression converges to $\Phi(0) = \tfrac{1}{2}$, so it is less than $\tfrac{3}{4}$ when $m$ is large enough.

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  • $\begingroup$ The same argument seems to suggest that $B_2$ takes at least one strictly negative value over any interval on $[0,1]$ (i.e., any $t_n$) with probability 1, which I don't think is true? Since $B_1$ is a process on $[0,1]$, then $t_n$ is a decreasing sequence within some bounded interval, so the "increments" $t_{n+1}-t_n$ must be approaching zero...so even as $m\to\infty$, $s_{k_n}-s_m \to 0$ as $n\to\infty$, I think? The $\Phi(0)$ suggests approximate independence, but I think the correlation approaches 1 as $s_{k_n}-s_m \to 0$ (as $n\to\infty$)? $\endgroup$ – David M Kaplan Feb 12 '18 at 15:58
  • $\begingroup$ @DavidMKaplan: No, the argument only works if $s_n$ converges to $0$. $\endgroup$ – Mateusz Kwaśnicki Feb 12 '18 at 20:33
  • $\begingroup$ Ah...meaning, $\lim_{n\to\infty} s_n = \lim_{n\to\infty} F_2(F_1^{-1}(t_n)) = 0$? Would that rule out cases like if $F_2$ has unbounded support but $F_1$ has bounded support, so $F_2(F_1^{-1}(0))>0$? Or if they both have unbounded support but $\lim_{n\to\infty} t_n > 0$? (Or can it be proved that the sequence $t_n$ can always be chosen to have a limit of zero?) $\endgroup$ – David M Kaplan Feb 13 '18 at 3:39
  • $\begingroup$ @DavidMKaplan: Oh, I thought $F_1$ and $F_2$ are strictly positive. If the left endpoints of supports of $F_1$ and $F_2$ are $a_1$ and $a_2$, respectively, then we may assume that $a_1 \leqslant a_2$. If $a_1 = a_2$, we apply the argument given in the answer. If $a_1 < a_2$, then $s_n = 0$ for $n$ large enough and the argument is even simpler (because with probability one $B_2(s_n) \geqslant 0$ for $n$ large enough). $\endgroup$ – Mateusz Kwaśnicki Feb 13 '18 at 7:26
  • $\begingroup$ Yes, $a_1<a_2$ is easier (I presume you mean $B_2(s_n) \le 0$ for large enough $n$); but I still don't see why $\lim_{n\to\infty} t_n = 0$ wp 1 (which seems necessary for $s_n \to 0$, which your earlier comment said is required)? $\endgroup$ – David M Kaplan Feb 14 '18 at 13:06

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