双均匀分布定时器操作阶段触发冲突的概率求解问询

Alright, let's break this down step by step. I'll start by restating the problem to make sure we're aligned, then work through the probability calculation with clear, intuitive steps.

First, let's clarify the setup to avoid confusion:

• Two independent timers, $t_1$ and $t_2$, each follow a uniform distribution over the interval $[a,b]$ (written as $t_1 \sim U(a,b)$, $t_2 \sim U(a,b)$).
• Both start simultaneously. When one timer triggers first:
  1. We send a communication that takes $c$ time (and $c \ll b-a$, so it's negligible compared to the timer's range).
  2. We run an operation that takes $\varepsilon$ time (with $\varepsilon \ll c$, so it's even smaller relative to the communication time).
  3. We reset the triggered timer.
• We want to find the probability that the other timer triggers during the operation phase (triggers before the operation starts are totally acceptable, per the problem statement).

Since $t_1$ and $t_2$ are independent and identically distributed, we can calculate the probability for one scenario (e.g., $t_1$ triggers first) and double it (the $t_2$-first case is perfectly symmetric). Also, for continuous uniform distributions, the probability that $t_1 = t_2$ is 0, so we can safely ignore that edge case.

1. Case 1: $t_1 < t_2$ (probability = 0.5) #

We need to find the probability that $t_2$ falls exactly in $t_1$'s operation window. The operation window starts at $t_1 + c$ (after the communication finishes) and ends at $t_1 + c + \varepsilon$. So we're targeting:
$$P(t_1 + c < t_2 < t_1 + c + \varepsilon \mid t_1 < t_2)$$

The joint probability density function (PDF) of $t_1$ and $t_2$ is $\frac{1}{(b-a)^2}$ (since they're independent uniform variables). We'll integrate this over the valid region.

First, define the valid range for $t_1$:

• $t_1$ must be at least $a$.
• $t_1 + c < t_2 \leq b$, so $t_1$ can be at most $b - c$ (otherwise $t_1 + c > b$, and $t_2$ can't be larger than that).
• For $t_1 \leq b - c - \varepsilon$, the operation window fits entirely within $[a,b]$, so $t_2$ ranges over an interval of length $\varepsilon$.
• For $t_1$ between $b - c - \varepsilon$ and $b - c$, the operation window extends beyond $b$, so $t_2$ ranges from $t_1 + c$ to $b$ (length $b - (t_1 + c)$).

2. Compute the Integral #

The probability for the $t_1 < t_2$ case is:
$$
\frac{1}{(b-a)^2} \left[
\int_{a}^{b - c - \varepsilon} \varepsilon , dt_1 +
\int_{b - c - \varepsilon}^{b - c} (b - c - t_1) , dt_1
\right]
$$

Calculating each integral:

• First integral: $\varepsilon \cdot (b - c - \varepsilon - a) = \varepsilon \cdot [(b-a) - c - \varepsilon]$
• Second integral: Let $u = b - c - t_1$, so the integral becomes $\int_{0}^{\varepsilon} u , du = \frac{\varepsilon^2}{2}$

Adding these together, the probability for $t_1 < t_2$ is:
$$
\frac{\varepsilon[(b-a) - c - \varepsilon] + \frac{\varepsilon^2}{2}}{(b-a)2} = \frac{\varepsilon(b-a - c) - \frac{\varepsilon^2}{2}}{(b-a)2}
$$

3. Symmetric Case and Total Probability #

Since the $t_2 < t_1$ case is identical, we double the above result to get the total probability of the unwanted event:
$$
P = \frac{2\varepsilon(b-a - c) - \varepsilon^2}{(b-a)2}
$$

4. Simplification (Given $\varepsilon \ll c \ll b-a$) #

Because $\varepsilon$ is much smaller than $c$ and $b-a$, the $\varepsilon^2$ term is negligible compared to the other terms. We can also approximate $b-a - c \approx b-a$ (since $c \ll b-a$). This simplifies the probability to a much cleaner form:
$$
P \approx \frac{2\varepsilon}{b-a}
$$

This makes sense intuitively: the operation window is a tiny interval of length $\varepsilon$, and the total possible range for the time difference between the two timers is roughly $b-a$. Since either timer could be the one triggering first, we multiply by 2. The small $c$ term barely affects the result because it's tiny compared to the overall interval length.

内容的提问来源于stack exchange，提问作者Jose Javier Gonzalez Ortiz