This question appeared in a study guide for my graduate level written exam in physics. (It may have been the one from the University of Chicago.) I see that a similar question was asked here What is the average value of time since last collision in Drude model? but the discussion did not focus on the aspect that I'm going to present.

A particle in a gas undergoes random collisions with other gas particles. The inter-collision times are exponential based on the idea that the number of collisions per unit time is describable as a Poisson stochastic process. Consider a particle at time $t_0$. The distribution of times until the next collision is exponentially distributed $f(t | \tau) = (1/\tau)\exp(-(t-t_0)/\tau)$. Clearly, $\mathbb{E}[t-t_0] = \tau$. But by time reversal symmetry, this also describes the time since the last collision. So which is it? Is the mean time between collisions $\tau$ or $2\tau$?

The question always felt like a swindle $-$ in part because I was unable to track down a generally agreed upon answer. (This was long before stackoverflow existed.)

My take is that conditional on knowing that the particle just endured a collision, the mean time to the next collision is indeed $\tau.$ But unconditionally, the time since/until the last/next collision is properly $\tau$, so we compute the last-to-next time to be $2\tau.$

The problem is with the exact definition of the term "mean inter-collision time." As it can be defined to be 1) unconditionally "time to next collision", then it is $\tau$ (but it makes no reference to the last collision, so perhaps this definition is non-responsive to the name) or 2) conditional on a collision having just occurred, "time to next collision" = $\tau$, or 3) unconditionally, "time since last collision to next collision," for a randomly selected particle whose past is as unknown as its future, in which case it is in fact $2\tau.$

Is there a consensus on this point here?

  • $\begingroup$ I'm not sure I follow. Events before $t_0$ don't contribute/matter to the Poisson process, only those after $t_0$. Hence, $\tau$ is the result (not even counting the fact that one defines $\tau$ to be the mean, not $2\tau$). $\endgroup$
    – Kyle Kanos
    Sep 24, 2022 at 3:14
  • $\begingroup$ @Kyle Kanos The question is presented as a kind of paradox, which draws time reversal symmetry into the mix. How can the time to the next collision be $\tau$ and the time from the last collision be $\tau$ (due to time reversal symmetry) and the inter-collision time not be $2\tau$? There has to be a swindle here. What is it? That's the question, which disguises a misuse of words/idea, as in the case of all paradoxes. And yes, if we condition on a collision happening at $t=t_0$ (considering only those particles for which it is true) then the inter-collision time is $\tau$ and not $2\tau$. $\endgroup$ Sep 24, 2022 at 3:22
  • $\begingroup$ The Poisson process isn't backwards looking, it's forwards looking (typically one defines it such that the number of events at $t=0$ is 0). And the mean time is defined as $\tau$, which is why every source you will ever find says $\tau$ and not $2\tau$. So I'm not sure what you are trying to get at. $\endgroup$
    – Kyle Kanos
    Sep 24, 2022 at 15:59

1 Answer 1


The Poisson process is defined axiomatically as a counting process $N(t)$ for $t>0$ such that,

  1. $N(t=0)=0$.
  2. The increments are independent.
  3. The number of events at any time interval $t$ is Poisson distributed with mean $\lambda t$. That is, $$ \mathbb{P}(N(t)=n)=\mathrm{e}^{-\lambda t}\left(\lambda t\right)^n/n!$$ for $n\in\mathbb{N}_0$.

From Conditions #1 and #2, and the fact that the starting point $t$ is arbitrary, it should be clear that your first choice (being unconditioned) is the correct answer: $\tau$ is the average time from an arbitrary point to the next event, rather than a mean time between events (or even half that).

  • $\begingroup$ I agree with your analysis. Under your interpretation, $\tau$ is not the mean inter-collision time. It only becomes the inter-collision time if we condition on a collision having just occurred. What do you think is the mean time since the last collision? Does time reversal symmetry apply? Cam we make a similar argument for $\tau$, again unconditional on a collision just now occurring? Of course, the system is in a macroscopic equilibrium state, so there no macroscopic "expansion" or "contraction", except perhaps on time scales $\gg \tau.$ $\endgroup$ Sep 26, 2022 at 15:23
  • $\begingroup$ @peter: it is from any arbitrary starting point $t_0$, not conditioned on previous arrival (remember, events are independent!) or anything. Simply given $t_0$, the mean time before an event (collision of particles, in your case) is $\tau$. Can you show me how you would go about computing $\mathbb{E}(t_0-t)$? $\endgroup$
    – Kyle Kanos
    Sep 26, 2022 at 22:03
  • $\begingroup$ The basic time reversal symmetry argument stems from the fact that if you simply reversed the velocities of all particles in a system, it would evolve as if it were moving backwards in time, and the last collision would be the "next" collision under these circumstances. Given thermodynamic equilibrium, such a system would be indistinguishable from the original system, so all mean features (internal energy, temperature, pressure, mean free paths, etc.) would be the same. It is just an ensemble copy of the original system, where "running backwards in time" is an annotation, nothing more. $\endgroup$ Sep 26, 2022 at 23:04
  • $\begingroup$ Well if you want to consider it that way (which I'm too tired to consider the validity), then the same thing applies: $\tau$ is the mean time to an event from any arbitrary time $t_0$, as evident from the first two conditions. $\endgroup$
    – Kyle Kanos
    Sep 27, 2022 at 1:51
  • $\begingroup$ I'll accept your answer. :) I think we are in basic agreement. $\endgroup$ Sep 27, 2022 at 12:24

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