给定随机事件时间间隔分布，求解等待时间分布ρ(t_wait)

Great question! The key here is accounting for length-biased sampling—random arrivals are more likely to land in longer inter-arrival gaps, which changes the distribution compared to the raw gap distribution. Let's break this down step by step:

Key Definitions #

First, let's formalize the terms we're working with:

• Let $\rho(t_{\text{gap}})$ be the probability density function (PDF) of inter-arrival gaps (time between consecutive events).
• Let $S(t) = P(t_{\text{gap}} > t) = \int_{t}^{\infty} \rho(\tau) d\tau$ be the survival function of the gap distribution (probability a gap lasts longer than $t$).
• Let $\langle t_{\text{gap}} \rangle = \int_{0}^{\infty} t \rho(t) dt$ be the average inter-arrival gap duration.
• $t_{\text{wait}}$ is the time a random arrival waits until the next event.

Core Intuition #

When you arrive randomly (uniformly across time), your arrival point is evenly distributed within whatever gap you land in. Longer gaps take up more total time, so you're more likely to end up in them. This means we can't just average the gap distribution directly—we have to weight gaps by their length.

Step-by-Step Derivation #

1. Survival Function for Waiting Time
   The probability that your waiting time exceeds $t$ ($P(t_{\text{wait}} > t)$) is the weighted average of:

   • For a gap of length $\tau$, the chance you land in the first $\tau - t$ portion of the gap (so you wait more than $t$), which is $\frac{\tau - t}{\tau}$ (only valid when $\tau > t$; if $\tau \leq t$, this probability is 0).
   • Weighted by the probability of that gap length, scaled by the gap's length (since longer gaps are more likely to contain your arrival).

   Mathematically, this gives:
   $$
   P(t_{\text{wait}} > t) = \frac{1}{\langle t_{\text{gap}} \rangle} \int_{t}^{\infty} (\tau - t) \rho(\tau) d\tau
   $$
   The $\frac{1}{\langle t_{\text{gap}} \rangle}$ term normalizes the weight to a valid probability.

2. PDF of Waiting Time
   To get the PDF $\rho(t_{\text{wait}})$, take the negative derivative of the survival function (since $\rho(t) = -\frac{d}{dt} P(t_{\text{wait}} > t)$):
   $$
   \rho(t_{\text{wait}}) = -\frac{d}{dt} \left( \frac{1}{\langle t_{\text{gap}} \rangle} \int_{t}^{\infty} (\tau - t) \rho(\tau) d\tau \right)
   $$
   Evaluating the derivative simplifies this to:
   $$
   \rho(t_{\text{wait}}) = \frac{S(t)}{\langle t_{\text{gap}} \rangle}
   $$
   That's it! The waiting time PDF is just the gap survival function divided by the average gap length.

Verify with Poisson Process #

For a Poisson process, $\rho(t_{\text{gap}}) = \lambda e^{-\lambda t}$, so:

• $\langle t_{\text{gap}} \rangle = \frac{1}{\lambda}$
• $S(t) = e^{-\lambda t}$

Plugging in, we get $\rho(t_{\text{wait}}) = \frac{e^{-\lambda t}}{1/\lambda} = \lambda e^{-\lambda t}$, which matches the gap distribution—exactly the memoryless property we expect.

Example: Uniform Gap Distribution #

Suppose gaps are uniformly distributed between 0 and $T$: $\rho(t_{\text{gap}}) = \frac{1}{T}$ for $0 \leq t \leq T$.

• $\langle t_{\text{gap}} \rangle = \frac{T}{2}$
• $S(t) = 1 - \frac{t}{T}$ for $0 \leq t \leq T$

Then the waiting time PDF is:
$$
\rho(t_{\text{wait}}) = \frac{1 - t/T}{T/2} = \frac{2(T - t)}{T^2}
$$
This is a linearly decreasing distribution, which makes sense—you're more likely to wait a short time (since you could land near the end of a long gap) but still have a chance of waiting almost $T$.

内容的提问来源于stack exchange，提问作者Dwagg