The bullet problem with discrete speeds

Bullets are fired, one per second, with independent speeds sampled uniformly from a discrete set. Collisions result in mutual annihilation. We show that the second fastest bullet survives with positive probability, while a slowest bullet does not. This also holds for exponential spacings between firing times, and for certain non-uniform measures that place less probability on the second fastest bullet. Our results provide new insights into a two-sided version of the bullet process known to physicists as ballistic annihilation.


Introduction
The bullet process is a deceptively simple process for which we presently lack the tools to completely analyze. Each second, a bullet is fired from the origin along the positive real line with a speed uniformly sampled from (0, 1). When a faster bullet collides with a slower one, they mutually annihilate. The bullet problem is to show there exists s c > 0 such that if the first bullet has speed faster than s c it survives with positive probability, and if it has speed slower than s c it is almost surely annihilated. It is conjectured that s c ≈ 0.9. In this work, we prove an analogous transition occurs when speeds are instead sampled uniformly from a discrete set. Additionally, our results have applications to physics model ballistic annihilation [EF85,DRFP95,KRL95,TEW98,KS01,ST17].
Consider bullets b 1 , b 2 , . . . fired from the origin along the real line such that b i is fired at time i for all i ≥ 1. A deterministic delay between firings is convenient for our argument, but not needed. All of the results here hold for exponentially distributed firing times (see Remark 8). The speed of bullet b i is denoted by s(b i ). The bullets have independent and identically distributed (i.i.d.) speeds sampled according to a probability measure µ on a set of speeds S ⊆ (0, ∞). When two or more bullets collide, all of them are annihilated. We will refer to this as an (S, µ)-bullet process. To clean up our notation, we will write the probability of an atom as µ(s) rather than as µ({s}).
Let b i → b j denote the event of bullet b i and b j colliding with b i faster, thus resulting in their mutual annihilation. We say that b i catches b j . Note that this can only happen if i > j and s(b i ) > s(b j ). Defineτ to be the minimum index with bτ → b 1 . The minimum is to account for the possibility of a simultaneous collision of several bullets. If b 1 is never caught by another bullet, setτ = ∞. Whenτ = ∞, we say that b 1 survives. Whenτ < ∞, we say that b 1 perishes. Our main result is that, when the bullet speeds are uniformly sampled from a finite set, a second fastest bullet survives with positive probability, while the slowest bullet does not.
Theorem 1. Fix n ≥ 3 and 0 < s n < · · · < s 2 < s 1 < ∞. Let µ be the uniform measure on S = {s n , . . . , s 1 }. In an (S, µ)-bullet process it holds that The first three authors were undergraduates participating in the Summer 2016 University of Washington REU supervised by Matthew Junge.
(i) The second fastest bullet survives with positive probability: (ii) The slowest bullet perishes almost surely: The survival of b 1 when it has maximal speed is straighforward. No bullet can catch it. This is not the case with the second fastest bullet. There will a.s. be infinitely many faster bullets trailing it. So, its survival hinges on interference of slower bullets.
Theorem 1 solves the discrete analogue of the bullet problem. The coupling between two (S, µ)-bullet processes with bullet speeds (s(b i )) and ( i ) for i ≥ 2 has b 1 surviving for every realization in which b ′ 1 survives. This guarantees that, when S and µ are fixed, the probability the first bullet survives is nondecreasing with respect to its speed. This monotonicity combined with Theorem 1 implies that there is a speed at which an initial bullet with that speed will perish, while one with faster speed will survive with positive probability. An interesting further question, that relates back to the original bullet problem, is to locate where the phase transition occurs when S = {i/n : i = 1, . . . , n} and µ is uniform.
Observing a phase transition for survival of the second fastest particle as µ places less mass on it interests physicists and mathematicians who study ballistic annihilation. By adapting the proof of Theorem 1, we take a step towards addressing this question.
Theorem 2. Let S be as in Theorem 1. There exists a probability measure µ supported on S such that µ(s 2 ) < µ(s 1 ) and in an (S, µ)-bullet process.
We next explain how our results apply to ballistic annihilation.
1.1. Applications to ballistic annihilation. If time and space are interchanged the bullet process is a one-sided version of ballistic annihilation. This model received considerable attention from physicists in the 1990's. There are very precise conjectures that still lack satisfactory justification. The probability measure on speeds in ballistic annihilation is typically assumed to be symmetric, but not necessarily uniform. Sidoravicius and Tournier establish survival in ballistic annihilation for such measures [ST17]. A corollary of our main theorem is survival of the second fastest particle for asymmetric three-element sets with the uniform measure. This is proven for one-sided ballistic annihilation in the discussion following [ST17, Proposition 4.1]. However, our main theorem allows us to extend to the usual two-sided setting. Also, our secondary result provides an upper bound for where the conjectured phase transition occurs in the canonical symmetric three-speed ballistic annihilation.
Ballistic annihilation is a physics model that was introduced to try to isolate intriguing features observed in more complicated systems, such as irreversible aggregation [BNRL93]. Particles are placed on the real line according to a unit intensity Poisson point process. Each particle is assigned a speed from a measure ν on R. Particles move at their assigned speed and mutually annihilate upon colliding.
Although it appears to have arisen independently, the bullet problem is equivalent to one-sided ballistic annihilation on [0, ∞). If one considers the graphical representation of   bullet locations, it is easy to see that inverting time and space coordinates makes the process into ballistic annihilation with inverted speeds (see Figure 1).
In Section 3 we describe how to make the bullet process two-sided, so that it is equivalent to the usual ballistic annihilation.
Ballistic annihilation is conjectured to exhibit more interesting behavior when ν is atomic [BNRL93]. The canonical example is when ν is a symmetric measure on {−1, 0, 1}: and p is the probability a particle has speed-0. Symmetry and ergodicity ensure that no speed ±1 particles can survive. However, it is not so clear what happens with speed-0 particles. By analyzing a complicated differential equation, Krapivsky et al. infer that a speed-0 particle survives if and only if p > .25 [KRL95]. Providing a probabilistic proof of this remains an important question. Currently, there is no proof that a speed-0 particle perishes almost surely for any p.
An application of Theorem 1 (i) is that ballistic annihilation with the uniform measure on any three speeds from R has the middle speed surviving with positive probability. Typically the measure in ballistic annihilation is assumed to be symmetric about 0 (as in [ST17]). Our result implies that the second fastest particle survives with positive probability for asymmetric speeds.
Corollary 3. Let −∞ < r 3 < r 2 < r 1 < ∞ and ν be the uniform measure on {r 3 , r 2 , r 1 }. For ballistic annihilation with either unit or exponential spacings, a particle with speed-r 2 will survive with positive probability.
As a corollary to Theorem 2 we consider ballistic annihilation with ν from (1) and give concrete bounds for when a speed-0 particle survives in the process with either unit or exponential(1) spacings.
Corollary 4. In a ν-ballistic annihilation with ν from (1) and particles started at each site of Z, a speed-0 particle survives with positive probability for p ≥ .3325. If the spacings are according to a unit intensity Poisson point process, then p ≥ .3313 suffices.
Note that [ST17] establishes a better bound p ≥ .3280 (with exponential spacings). We include Corollary 4 to illustrate the proof of Theorem 2, and because it lays a foundation that can be further optimized. The latter is pursued in a followup work with Burdinsky, Gupta, and Junge in [BGJ18].
1.2. History. The IBM problem of the month from May in 2014 credits a version of the problem to an engineer named David Wilson. The question there is to fire exactly 2m bullets with independent uniform(0, 1) speeds and compute the probability of the event E m = {no bullets survive}. There is an unpublished result of Fedor Nazarov that It is surprising that changing one bullet speed out of the 2n total bullets affects the exponent. One would naively expect it only changes P[E m ] by a constant factor. This conjecture comes from simulations performed by Kostya Makarychev.If one could prove that c s > 1 for some value of s, then a Borel-Cantelli style argument would imply b 1 survives when it has speed at least s. Thus, understanding P[E m,s ] would lead to a solution to the bullet problem. Makarychev's simulations suggest that the critical value is approximately 0.9.
The bullet process with n bullets fired was recently studied by Broutin and Marckert [BM17]. They consider arbitrary non-atomic speed distributions on [0, ∞) and find that the distribution q n for the number of surviving bullets is invariant for several different spacings and acceleration functions for the bullets. The distribution shows up in other contexts such as random permutations and random matrices. It is characterized by the following recurrence relation: q 0 (0) = 1, q 1 (1) = 1, q 1 (0) = 0, and for n ≥ 2 and any 0 ≤ n, This formula generalizes (2), which describes q 2m (0). The equation for q n can be analyzed to prove a central limit theorem that says ≈ log n bullets survive (see [BM17, Proposition 2]). Unfortunately, this does not imply survival with infinitely many bullets. Although the number of surviving bullets is growing like log n, we cannot rule out the possibility that the number of bullets alive at time n in the process is 0 infinitely often. Indeed, there are instances of q n for which this happens and others where it does not. These results suggest that it is equally challenging to analyze variants of the bullet problem.
1.3. Overview of proofs. Let τ be distributed asτ conditioned on the event {s(b 1 ) = s 2 }. Letting τ 1 , . . . , τ 5 be independent copies of τ we find an event The behavior this captures is that if s(b 2 ) = s 1 then b 1 is caught no matter what. However, if s(b 2 ) = s 2 , then b 1 survives "twice" as long as it would have otherwise. If the second bullet is slower than s 2 , then it acts as a shield for b 1 -thus increasing the survival time of b 1 . These arguments hinge on the renewal properties described in Lemma 5 and Lemma 6, and a fortuitous dependence that makes fast bullets less likely to appear behind the bullet that catches b 2 when s(b 2 ) < s 2 . All of this is made rigorous in Proposition 7.
We prove Theorem 1 (ii) via contradiction. If the slowest bullet survives with positive probability, then monotonicity implies that the second slowest bullet also survives with positive probability. When we extend the bullet process to be two-sided, the two slowest speeds become the two fastest speeds from the perspective of bullets fired before them. Theorem 1 then implies that both speeds survive with positive probability in the two-sided process. Because the two-sided process is ergodic, the Birkhoff ergodic theorem gives a positive density of both speeds that survive. This is a contradiction since these surviving bullets with different speeds must eventually meet, and thus cannot survive.

Survival of a second fastest bullet
Write s 2 < s 1 for the two largest elements of S. Let τ to be the minimum index with b τ → b 1 in this process with b 1 deterministically set to have s(b 1 ) = s 2 . The goal of this section is to prove that P[τ = ∞] > 0.
2.1. Obtaining a recursive inequality. We start with two lemmas describing a renewal property in the (S, µ)-bullet process satisfying our hypotheses. The first states that the bullet speeds behind a maximal speed bullet are independent of any event involving this bullet.
Proof. The bullet b γ has the fastest speed, so the bullets behind it do not interfere. Thus A longer range renewal property holds for other annihilations where, outside of a particular window, the bullet speeds become independent.
such that, conditional on E, the bullet speeds s(b i+a ), s(b i+a+1 ), . . . are independent of one another and have distribution µ.
Proof. Given i, j, s(b i ), and s(b j ), let a be such that a maximal speed bullet fired at time i+a cannot reach b i before b i → b j . This is the latest time at which b i could be prevented from catching b j . The event b i → b j is thus unaffected by the bullet speeds s(b i+a ), s(b i+a+1 ), . . .. The independence claim follows.
Because we will need it later, we write down an explicit formula for a. A collision between b i and b j would occur at time t 0 and location x 0 given by The last firing time k at which a bullet with speed s 1 could prevent this is We then set a equal to (4)−i.
We will occasionally refer to the interval [j + 1, a] as the window of dependence of E. This is because, as described more precisely above in Lemma 6, the bullet speeds in this interval are influenced by E, while those beyond it are again i.i.d.
Recall that one of the several equivalent forms of stochastic dominance X Y is that there is a coupling with marginals X ′ ∼ X and Y ′ ∼ Y such that X ′ ≥ Y ′ almost surely. We let 1 {·} denote an indicator function.
Proposition 7. At least one of the following holds: • τ is infinite with positive probability.
• Let τ 1 , . . . , τ 5 be i.i.d. copies of τ . There exists an event F ⊆ {s(b 2 ) < s 2 } independent of the τ i with P[F ] = ǫ = ǫ(S) > 0 so that Proof. We will establish each line of the above by conditioning on the value of s(b 2 ). When s(b 2 ) = s 1 as in (5), we have b 2 → b 1 deterministically. Although τ = 2 on this event, it will simplify our calculations later to use the indicator function as a lower bound.
The pivotal case is (7), when s(b 2 ) < s 2 . The idea is that b 2 acts as a shield, and causes an ǫ-bias for the bullets close behind it to have speed-s 2 . The reasoning in (6) then ensures that b 1 will survive twice as long on this ǫ-likely event. To see this rigorously, suppose that b γ is the earliest bullet catching b 2 . If γ is infinite with positive probability, then so is τ . Indeed, b 1 cannot be caught until b 2 is destroyed. In this case the first condition of the proposition is met and we are done. Now, let us suppose that γ is a.s. finite. We will start by describing the ǫ-likely event F for which we obtain an extra copy of τ . When b 2 is caught, there is a finite window of dependence behind the catching bullet (see Lemma 6). With positive probability this window contains only bullets with speed-s 2 .
A minor nuisance is showing that there is enough room in the window behind b γ for a speed-s 2 bullet. We start by restricting to the event that s(b 2 ) = s n and show that P[γ > M ] > 0 for all M > 0. Let m ≥ 2. With positive probability, there are alternating fastest and slowest bullets from index 3 up to 2m, and then a speed-s 2 bullet. Call this event On the event A, we have γ = 2m + 1 and s(b γ ) = s 2 so long as nothing catches b γ before it reaches b 2 . We track the size of the window of dependence behind b γ with the function h(m) = a(s 2 , s n , 2m + 1, 2), m ≥ 2.
Here a(s 2 , s n , 2m+1, 2) ≥ 1 is as in Lemma 6. It is the index distance behind 2m+1 at which bullets resume being i.i.d. conditioned on the event {b 2m+1 → b 2 , s(b 2m+1 ) = s 2 , s(b 2 ) = s n }. We remark that, because we are fixing the indices and speeds in a, the function h is deterministic. Plugging our conditions into the explicit formula at (4), we have t 0 → ∞ as m → ∞, and also α = s 2 /s 1 < 1. Thus, h(m) is non-decreasing with lim m→∞ h(m) = ∞. Let m 0 = min{m ≥ 2 : h(m) > 1}. As bullet speeds are between s n and s 1 , we must have m 0 < ∞ and thus 1 < h(m 0 ) < ∞. Let B be the event that all of the bullets in this window have speed-s 2 . Formally, Let F = A ∩ B. This event specifies the speeds of 2m 0 + h(m 0 ) − 1 bullets, and by independence we have where p i = µ(s i ).
Conditioned on F , all of b 2 , . . . , b 2m+1 mutually annihilate. Moreover, s(b 2m+1+i ) = s 2 for i = 1, . . . , h(m 0 ) − 1. The trailing bullets speeds (s(b 2m+1+I )) I≥h(m0) are i.i.d. uniform. The reasoning that yields the additional copy of τ in (6) then gives h(m 0 ) − 1 ≥ 1 additional copies of τ when F occurs. We take only one of them and set ǫ = P[F ] as in (8). This accounts for the term 1 {F } (τ 3 + τ 4 ) in (7). Now that we have constructed the ǫ-likely event to have b 1 survive for at least two copies of τ . It remains to show that b 1 survives for at least a τ -distributed amount of time on the event {s(b 2 ) < s 2 } ∩ F c . This will give the term 1 {F c } τ 5 in (7).
Let a = a(s(b γ ), s(b 2 ), γ, 2) be the largest index for which b γ+a could catch b γ before b γ catches b 2 . Bullets with indices in the set I = {γ + 1, . . . , γ + a} are dependent upon s(b γ ), s(b 2 ), and γ. In particular, bullets faster than s(b γ ) can survive to intercept b γ . By Lemma 6, the bullets with indices larger than γ + a are once again independent (see Figure  3).
In order for b γ → b 2 to occur, all of the bullets b 3 , . . . , b γ−1 must mutually annihilate. We can then ignore them for the remainder of the argument. When s(b γ ) = s 1 , it resets the model just as in the s(b 2 ) = s 2 case, and b 1 survives until a bullet with index distributed as τ + γ destroys it. The process has i.i.d. bullet speeds for indices after γ + a. Let us restrict our attention to just the bullets with indices in I. That is, consider a bullet model with only |I| bullets, with speeds conditioned so that b γ → b 2 with s(b 2 ) < s 2 . Since b γ → b 2 , no bullets with speed s 1 in I can survive. Otherwise such a bullet would catch b γ before b γ catches b 2 . These slower bullets only prolong the survival of b 1 .
Returning to the bullet process with infinitely many bullets, before b 1 is destroyed all of the surviving bullets in I must be destroyed by bullets with indices at least γ+a. Upon being destroyed, each of the surviving bullets from I generates its own window of dependence that contains no surviving speed-s 1 bullets. Either these windows keep spawning new windows, in which case b 1 is never destroyed, or all of the bullets in these windows of dependence are destroyed. In the first case we have τ is infinite with positive probability. In the second, we have b 1 is again trailed by bullets with i.i.d. uniform speeds. Once this occurs, it takes a τ distributed number of bullets to catch b 1 .
Remark 8. The same recursive inequality as in Proposition 7 holds for exponential spacings. Let (ζ i ) be i.i.d. unit exponential random variables and consider an (S, µ)-bullet process where we fire b 1 at time t 1 = ζ 1 , and b i at time t i = t i−1 + ζ i for i ≥ 2. As before, let τ be the random index of the first bullet to catch b 1 conditional on s(b 1 ) = 2. We claim that τ still satisfies Proposition 7, but with a different event F ⊆ {s(b 2 ) < s 2 }.
As before if s(b 2 ) = s 1 then τ = 2. So, (5) still holds. Next, if s(b 2 ) = s 2 , then b 1 survives twice as long in the same sense as (6). This is because a bullet with speed s 1 must catch b 2 , and the bullets trailing it have independent speeds and firing times that keep the exponential spacings just as in Lemma 5. Figure 2. The picture when s(b 2 ) = s 2 . The bullet b 2 is annihilated by, b σ , a bullet that is fired a τ -distributed number indices after it. The bullets trailing b σ are i.i.d. and thus b 1 is caught by, b σ ′ , a bullet another τ -distributed indices behind b σ . Lastly, if s(b 2 ) < s 2 we let γ be the index b γ → b 2 . The construction is simpler than before. Just as in Lemma 6 the event b γ → b 2 induces a finite window of dependence t γ + a. Let N be the number of bullets fired in the window of dependence. We take to be the event that b 2 is caught by b 3 when it has speed-s 2 The conditions N = 1, s(b 4 ) = s 2 ensure that there is one speed-s 2 bullet in the window of dependence and no others. It is important that the spacings have the memoryless property, otherwise the times bullets are fired after t γ + a would not have the same distribution as at the start of the process.
We will see in the next section that satisfying the recursive distributional inequality in Proposition 7 is sufficient to deduce a nonnegative random variable places some mass at ∞. So, our results extend to exponential spacings.

Analyzing the recursive inequality.
Our goal now is to show that any random variable satisfying the recursive distributional inequality in Proposition 7 must be infinite with positive probability. With ǫ as in Proposition 7, we introduce an operator A = A(µ) that acts on probability measures supported on the positive integers. It will be more convenient to represent such a measures by a random variable T with that law. To define A we let s ∈ S be sampled according to µ, and X ǫ ∼Bernoulli(ǫ), both independent of one another. Take T 1 , . . . , T 5 to be i.i.d. copies of T that are also independent of X ǫ and s. We obtain a new distribution By Proposition 7, we have τ Aτ.
The operator A is monotone.
Lemma 9. If T T ′ then AT AT ′ .
Proof. This follows from the canonical coupling which sets each T i ≥ T ′ i .
Additionally, A has a unique fixed distribution.
Lemma 10. Let τ be as in Proposition 7, and let A n denote n iterations of A. It holds that A n τ → τ * with τ * d = Aτ * .
Proof. Let F n (k) = P[A n τ ≤ k] be the cumulative distribution function of A n τ . By the previous lemma and (10), we have A n τ A n+1 τ for all n ≥ 0. The definition of stochastic dominance implies that {F n (k)} ∞ n=0 is an increasing bounded sequence. Let F (k) denote its limit. The function F (k) is non-decreasing and belongs to [0, 1]. Thus, F (k) is the cumulative distribution function of some random variable τ * . The limiting distribution must be fixed by A since an additional iteration A(A ∞ τ ) will not change the distribution.
Next we observe that τ * couples to the return time to zero of a lazy biased random walk on the integers.

Proof. Consider the partition of events
Since the two events in the union forming A 2 are disjoint, it does not affect the distribution of Aτ * if we set τ * 3 = τ * 1 and τ * 4 = τ * 2 . This lets us rewrite the equality τ * d = Aτ * as . This RDE describes the number of leftward steps to reach 0 of a discrete-time lazy random walk on Z started at 1. The walk moves left with probability P[A 1 ], moves right with probability P[A 2 ], and stays put with probability P[A 3 ]. The formulas at (11) and (12) along with our hypothesis that p 1 ≤ p 2 ensures that this walk has a rightward drift. Such a biased random walk does not return to 0 with positive probability.
To relate this back to τ * note that any random variable T d = AT is unique. One way to see this is to precisely compute the generating function f (x) := Ex T = Ex AT . This gives a quadratic equation in f (x) that can be solved for explicitly. Choosing the proper branch is straightforward since f (0) = 0. Since the probability generating function uniquely specifies the distribution of a random variable, we have τ * is equivalent to the return time of the lazy biased random walk just described. Hence P[τ * = ∞] > 0.
We are now ready to establish survival for the second fastest bullet.
Proof of Theorem 1 (i) . By (10) and Proposition 11, τ is stochastically larger than a random variable that is infinite with positive probability. Hence τ is infinite with positive probability.
Proof of Theorem 2. Suppose that |S| = n. Recall that s n is the smallest element of S and that p i = µ(s i ). By (10) and Proposition 11 we have survival of a second fastest bullet so long as The formula ǫ = p m0−1 is derived in (8). The constant m 0 > 0 and function h are deterministic. So, any measure µ satisfying will have a second fastest bullet surviving with positive probability. To see that there is a solution with p 2 < p 1 let 0 < δ < n −1 be a small, yet to be determined constant and define the measure Using µ δ in (14) and letting δ → 0 gives the inequality n −1 < n −1 + n −2m0−h(m0) (n − 2). Thus, for small enough δ 0 > 0, an (S, µ δ0 )-bullet process has a second fastest bullet surviving with positive probability.

The slowest bullet does not survive
In this section we assume that S is finite with at least three elements and µ is the uniform measure. In the usual bullet process the bullet b i has position s(b i )(t − i). We can extend this definition all integers i ∈ Z to make the two-sided (S, µ)-bullet process. In this process bullets are removed the first time their position coincides with another. Now bullets can be destroyed from both sides. We will say that b i survives + if the position of b i never coincides with the position of any other b j for j > i. Alternatively, we say that b j survives − if its position never coincides with the position of a b j for j < i. If both occur, we say that b j survives +,− .
Survival + only depends on bullets fired after a given bullet, so it describes whether a bullet catches the survivor. So, survival + favors faster bullets. On the other hand, survival − favors slower bullets since it describes whether a bullet catches one fired before it. As bullet speeds are independent, we can describe survival +,− as a product of the probabilities of one-sided survival.
The advantage of the two-sided process is that it is ergodic, and so there cannot be two different bullet speeds that survive with positive probability.
Proposition 13. Only one bullet speed can survive +,− with positive probability in the twosided (S, µ)-bullet process.
Proof. Notice that the two-sided process is translation invariant with i.i.d. speeds and thus ergodic. If two or more different speeds survived +,− with positive probability, then by the Birkhoff ergodic theorem, we would have a positive fraction of surviving +,− bullets of each speed. Suppose that b i is one of these surviving bullets. For some j, k > 0 there almost surely are surviving +,− bullets b i+j and b i−k with the same speed as one another, but different speed than b i . With different speeds, one of these must collide with b i , or perhaps some other surviving +,− bullet. In either case, this contradicts that these bullets survive +,− .
Proof of Theorem 1 (ii). If b 1 survives + then b 1 survives in the usual bullet process. So it suffices to prove that P[b 1 survives + | s(b 1 ) = s n ] = 0. To show a contradiction suppose this probability is equal to q > 0. A bullet with speed-s n is the easiest to catch for bullets fired at times after it, but it is uncatchable by bullets fired before it. Thus, P[b 1 survives − | s(b 1 ) = s n ] = 1.
Let s ′ 2 be the second slowest speed in S (possibly s ′ 2 = s 2 if |S| = 3). The monotonicity for survival of bullets discussed in the introduction following the statement of Theorem 1 ensures that P[b 1 survives + | s(b 1 ) = s ′ 2 ] ≥ q. Moreover, a bullet with speed s ′ 2 is the second fastest bullet from the perspective of bullets fired before it. Since µ is uniform, we can apply Theorem 1 (i) and deduce P[b 1 survives − | s(b 1 ) = s ′ 2 ] = p > 0. The one-sided survival probabilities above are all positive. By Lemma 12, a bullet with speed-s n or s ′ 2 survives +,− with positive probability. This contradicts Proposition 13.

Applications to ballistic annihilation
Corollary 3 follows from Theorem 1 (i) and Lemma 12.
Proof of Corollary 3. Start with ballistic annihilation with the uniform measure on three speeds: r 3 < r 2 < r 1 . If r 1 > 0, then this is equivalent to a two-sided bullet process with speeds s i = 1/r i . If r 1 ≤ 0 we can use the fact that the manner in which collisions happen in ballistic annihilation is translation invariant (this is referred to as the linear speed-change invariance property in [ST17, Section 2]). Namely, the same particle collisions will occur (although at different times) in ballistic annihilation with shifted-speeds r ′ i = r i −r 1 +1. The r ′ i are positive and, so this process is equivalent to a two-sided bullet process with speeds s i = 1/r ′ i . In both cases we have s n < s 2 < s 1 and µ the uniform measure. In the two-sided (S, µ)-bullet process, a bullet with speed s 2 is the second fastest from the perspective of bullets fired before and after it. So, Theorem 1 (i) guarantees that both P[b 1 survives + | s(b 1 ) = s 2 ], P[b 1 survives − | s(b 1 ) = s 2 ] > 0, Note that these probabilities are positive, but may not be equal. Combine this with Lemma 12 and we have P[b 1 survives +,− | s(b 1 ) = s 2 ] > 0. We conclude by noting that equivalence of the two processes ensures that a speed-s 2 bullet surviving with positive probability is the same as a speed-r 2 particle surviving in ballistic annihilation.
We can make the estimate in Theorem 2 more concrete by considering the canonical example of three-speed ballistic annihilation with speed law ν from (1).
We can also consider exponential(1) spacings between firing times. A quick calculation shows that if the gap between firing b 2 and b 3 with s(b 2 ) = 1 and s(b 3 ) = 3/2 is ξ, then the window of dependence also has size ξ. We can exactly compute the probability of F from (9). Recall, we require that s(b 4 ) = 3/2 with b 4 fired within ξ time units of b 3 , and then no other bullets fired inside the window of dependence. This probability is easy to compute since there are N = Poi(ξ) many bullets fired in this window. So we have P[F ] = P[N = 1, s(b 2 ) = 1, s(b 3 ) = 3/2, s(b 4 ) = 3/2] = (1/4)p 2 (1 − p)/2.
Plugging this into (13) and solving numerically gives a speed-3/2 bullet survives so long as p ≥ .3313.