Suppose I have three independent Brownian motions started at positions $-a$, $0$, and $a$ with $a>0$.
What is the asymptotic distribution of the time $\tau$ of the first collision between any pair of these Brownian motions?
In particular, what is the suppression in time: can $\mathbb{P}(\tau > t)$ be upper-bounded by a decaying exponential in $t$, like $\mathbb{P}(\tau > t) \leq c_1 e^{-c_2 \frac{t}{a^2}}$ for an appropriate pair of constants $c_1, c_2 >0$?
Here are my thoughts. First, the first collision will necessarily involve the Brownian motion started at $0$ and either of the motions started at $a$ or $-a$.
As a first approximation, consider the case where the walkers at $a$ and $-a$ are stationary. The collision time then corresponds to the distribution of the first exit time of a Brownian motion, which obeys $\mathbb{P}(\tau_{\rm exit} >t ) \leq c_1 e^{-c_2 \frac{t}{a^2}}$ for an appropriate pair of constants $c_1$ and $c_2$, as noted in an answer by Chris Jangjigian to another question.
I suspect that the suppression in time in our original question at the top is actually weaker (has a heavier tail) so that the answer to my question is no. My reason for this is that there will be rare events where the Brownian motion started at $-a$ drifts even more negative while the Brownian motion started at $a$ drifts even more positive for an extended period of time. Though these trajectories may be rare, I suspect they are sufficient to change the tail properties of $\mathbb{P}(\tau > t)$ at large $t$.
My motivation for this question came from a comment from kodlu on another question. There, I considered the case of simple symmetric random walks on the integer line started at all integer multiples of $a$, and I was curious about the tail bounds of the first collision time of the walk started at $0$ with any other walk.
While the questions are similar, I actually suspect the answers will be different, and that the case with three walkers may have heavier tails than the case with an infinite number of walkers. My rough reason is that it is harder to imagine the kinds of rare events that could enhance the tail probability in the case of an infinite number of walkers.
I realized I can sketch out an answer that shows that $$\mathbb{P}(\tau > t) \propto \frac{1}{t^{3/2}}$$ asymptotically at large $t$. While this is indeed a heavier tail than an exponential, it is also heavier than I appreciated, as it's a power law and not a stretched exponential.
To see this result, we can use the relationship between Brownian motion and solutions to the heat equation. I will actually interpret the heat equation as an imaginary time Schrodinger equation.
The $\mathbb{P}(\tau > t)$ then corresponds to an appropriate integral of the imaginary time propagator (i.e. a matrix element of the thermal density matrix with an inverse temperature scaling with time) for three hard-core particles starting from the initial state $|-a\rangle |0\rangle |a\rangle$ and ending in a final state $|x_1\rangle |x_2\rangle |x_3\rangle$ with $-\infty<x_3<x_2<x_1<\infty$. Specifically, we have $$ \mathbb{P}(\tau > t) \propto \int_{-\infty}^\infty dx_1 \int_{-\infty}^{x_1} dx_2 \int_{-\infty}^{x_2} dx_3 \langle x_3, x_2, x_1|e^{-Ht}|-a, 0, a\rangle $$
Note that the proportionality in the equation above is in fact an exact equality, but I will be cavalier about purely numerical proportionality constants in the following. We can fix the purely numerical proportionality constants by noting that $\mathbb{P}(\tau > t) \to 1$ as $t \to 0$.
We can calculate the propagator by inserting a resolution of the identity made of eigenvectors for this Hamiltonian. Note again that this is a free particle Hamiltonian up to the hard-core constraint, and the hard-core constraint is trivially satisfied for fermions. We have distinguishable particles, but this just means that we get several eigenfunctions for each fermionic wavefunction corresponding to appropriate multiplications by $\text{sign}(x_i - x_j)$. Each such wavefunction will contribute the same to the integral above, as the integral above is over a restricted domain. That is,
$$ \mathbb{P}(\tau > t) \propto \int_{-\infty}^\infty dx_1 \int_{-\infty}^{x_1} dx_2 \int_{-\infty}^{x_2} dx_3 \int_{-\infty}^\infty dk_1 \int_{-\infty}^\infty dk_2 \int_{-\infty}^\infty dk_3 e^{-\frac{k_1^2 + k_2^2 +k_3^2}{2}t}\langle x_3, x_2, x_1|k_1,k_2,k_3\rangle_A \langle k_1,k_2,k_3 |-a, 0, a\rangle $$
where $|k_1,k_2,k_3\rangle_A = |k_1, k_2, k_3\rangle +|k_2, k_3, k_1\rangle + |k_3, k_1, k_2\rangle-(|k_2, k_1, k_3\rangle +|k_3, k_2, k_1\rangle + |k_1, k_3, k_2\rangle)$ is completely antisymmetrized.
Using $\langle x | k\rangle = e^{ikx}$, we can integrate over $k_1, k_2,k_3$ to find
$$\mathbb{P}(\tau > t) \propto \left( \frac{1}{\sqrt{t}} \right)^3 \int_{-\infty}^\infty dx_1 \int_{-\infty}^{x_1} dx_2 \int_{-\infty}^{x_2} dx_3 \left[ e^{-\left( \frac{(x_1-a)^2}{2t}+ \frac{(x_2)^2}{2t} +\frac{(x_3+a)^2}{2t} \right)} \right]_A$$
Here $A$ denotes completely antisymmetrizing on $x_1,x_2,x_3$. We can equivalently write the complete antisymmetrization through an unnormalized Slater determinant.
Note that when $t$ is small, the integrand is highly peaked about its maxima. From the bounds on the integral, only the original term $e^{-\left( \frac{(x_1-a)^2}{2t}+ \frac{(x_2)^2}{2t} +\frac{(x_3+a)^2}{2t} \right)}$ and not the terms with permutations exchanging the $x$'s will have the maxima within the integration region. That is, we see that to have $\mathbb{P}(\tau > t) \to 1$ as $t \to 0$,
$$\mathbb{P}(\tau > t) = \left( \frac{1}{\sqrt{2 \pi t}} \right)^3 \int_{-\infty}^\infty dx_1 \int_{-\infty}^{x_1} dx_2 \int_{-\infty}^{x_2} dx_3 \left[ e^{-\left( \frac{(x_1-a)^2}{2t}+ \frac{(x_2)^2}{2t} +\frac{(x_3+a)^2}{2t} \right)} \right]_A$$
For a slightly more compact form that will help with the asymptotics, note that we can use a $u$-substitution to absorb factors of $\frac{1}{\sqrt{t}}$:
$$\mathbb{P}(\tau > t) = \left( \frac{1}{\sqrt{2 \pi}} \right)^3 \int_{-\infty}^\infty dx_1 \int_{-\infty}^{x_1} dx_2 \int_{-\infty}^{x_2} dx_3 \left[ e^{-\left( \frac{\left(x_1-\frac{a}{\sqrt{t}}\right)^2}{2}+ \frac{\left(x_2\right)^2}{2} +\frac{\left(x_3+\frac{a}{\sqrt{t}}\right)^2}{2} \right)} \right]_A$$
For large $t >> a^2$, we can Taylor expand the integrand in powers of $\frac{a}{\sqrt{t}}$. This yields
$$\mathbb{P}(\tau > t) \sim \left( \frac{a}{\sqrt{2 \pi t}} \right)^3 \int_{-\infty}^\infty dx_1 \int_{-\infty}^{x_1} dx_2 \int_{-\infty}^{x_2} dx_3 (x_2-x_1) (x_3-x_1) (x_2-x_3) e^{-\frac{x_1^2}{2}-\frac{x_2^2}{2}-\frac{x_3^2}{2}}$$
which means
$$\mathbb{P}(\tau > t) \sim \frac{1}{2\sqrt{\pi}}\left( \frac{a}{\sqrt{t}} \right)^3 $$
The steps above appear to have relatively straightforward extensions to calculating the probability there are no pairwise collisions in a set of $N$ particles in a time $t$, which may be useful in answering my related question. There, I am only interested in the collision time of the walker at the origin, but I suspect that will be equal in distribution to $\lim_{N \to \infty} N \tau_N$ where $\tau_N$ is the first collision time for any pair in $N$ particles.