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.

  • 3
    $\begingroup$ "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

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

| cite | improve this answer | |
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

(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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.