Two signals arrive at a receiver, and every moment of a time…

Two signals arrive at a receiver, and every moment of a time interval of length $T$ is equally likely for the arrival of each signal. The receiver is jammed if the difference between the arrival times of the two signals is less than $\tau$ $(\tau\lt T)$. Determine the probability that the receiver will be jammed.

Explanation of Method

Let $x$ and $y$ be the arrival times of the two signals. The pair $(x,y)$ is a random point in the square $[0,T]\times[0,T]$, uniformly distributed, so probabilities are areas divided by $T^2$. The receiver is jammed when $|x-y|\lt\tau$ — a band around the diagonal. The easiest way to find its area is to subtract the two corner triangles where $|x-y|\ge\tau$.

Step 1: The model

Measure time from the start of the interval. The arrival times $x$ and $y$ are independent and uniformly distributed on $[0,T]$, so the point $(x,y)$ is uniformly distributed over the square

\[ \Omega = [0,T]\times[0,T], \qquad \text{area } T^2. \]

(The statement's "at any moment" refers to this interval of length $T$; without a finite interval the probability would not be defined.)

Step 2: The favourable region

Jamming: $|x-y|\lt\tau$, i.e. $x-\tau\lt y\lt x+\tau$ — the band between the lines $y=x-\tau$ and $y=x+\tau$ (shaded).

$\tau$ $T$ $\tau$ $T$ $x$ $y$ $\tfrac12(T-\tau)^2$ $\tfrac12(T-\tau)^2$

jamming: $|x-y|\lt\tau$

The square of arrival times. The band around the diagonal is the jamming region; the two white corner triangles have legs $T-\tau$.

Step 3: The area

Outside the band there are two right triangles with legs $T-\tau$, of total area $2\cdot\frac12(T-\tau)^2=(T-\tau)^2$. So the band has area $T^2-(T-\tau)^2$ and

\[ \boxed{P = \frac{T^2-(T-\tau)^2}{T^2} = 1-\left(1-\frac{\tau}{T}\right)^2 = \frac{\tau(2T-\tau)}{T^2}.} \]

Step 4: Example

$T=60$ s, $\tau=5$ s: $P = 1-\left(\frac{55}{60}\right)^2\approx 0.160$. Even though $5$ s is only $\frac1{12}$ of the interval, jamming happens about one time in six, because either signal may come first.

Step 5: Check the extremes

$\tau\to0$: $P\to0$ (the signals practically never coincide exactly).
$\tau\to T$: $P\to1$ (any two times in the interval differ by less than $T$).

Result

	Answer
Favourable region	band $\|x-y\|\lt\tau$ in the square $[0,T]^2$
Probability	$P=1-(1-\tau/T)^2=\tau(2T-\tau)/T^2$