How to calculate $\mathbb{P}(|U_{1} - U_{2}| < \frac{1}{12})$, where $U_{i} \sim \mathcal{U}(0, \frac{1}{2})$ and $U_{i}$ are independent.

Question

We have a Poisson process of intensity $\lambda = 4$. We have the following event:

$A$: "Two marks appear with a separation of $\frac{1}{12}$ or less".

We need to calculate the probability of exactly two "marks" appearing between $0$ and $\frac{1}{2}$, and $A$ ocurring at the same time. That is, the probability of exactly two marks ocurring in $(0, \frac{1}{2})$ with a separation of $\frac{1}{12}$ or less.

Here's how far I've gotten:

Let $N$: Number of marks between $0$ and $\frac{1}{2}$. We want: $$\mathbb{P}(N=2, A) = \mathbb{P}(A|N=2)\mathbb{P}(N=2) $$

$N\sim Poi(\frac{1}{2}4)$, so: $$\mathbb{P}(N=2) = \frac{2^2}{2!}e^{-2} = 2e^{-2}$$

The arrival times for a conditioned number of marks (in this case two) is $U_{1}$, and $U_{2}$, where $U_{i} \sim \mathcal{U}(0, \frac{1}{2})$, and $U_{i}$ are independent, so:

$$\mathbb{P}(A|N=2) = \mathbb{P}(\max(U_{1}, U_{2}) - \min(U_{1}, U_{2}) < \frac{1}{12}) = \mathbb{P}\left(|U_{1} - U_{2}| < \frac{1}{12}\right)$$

This is where I'm stuck. The answer is supposed to be $\frac{11}{18}e^{-2}$, so $\mathbb{P}\left(|U_{1} - U_{2}| < \frac{1}{12}\right)$ should be $\frac{11}{36}$.

Thanks.

Answer 1

Hint: saying that $U_1,U_2$ are independent uniform $[0,1/2]$ is equivalent to saying that $(U_1,U_2)$ is uniformly distributed over the square $[0,1/2]\times [0,1/2]=\{(x,y):0\le x\le 1/2,0\le y\le 1/2\}$. Draw this square, draw the region where $|x-y|\le \frac1{12}$, and compute the fraction of this area over the total area.

Answer 2

You can proceed e.g. as follows: Let $V_1=2U_1, V_2=2U_2$. Then $Z = |V_1-V_2|$ follows a triangular distribution with parameters $(a,b,c)=(0,1,0)$ , and you want $$ \mathbb{P}\left\{ |U_1-U_2| < \frac{1}{12}\right\} = \mathbb{P}\left\{ Z < \frac{1}{6}\right\} = \int_0^{1/6} f_Z(z)dz = \int_0^{1/6} (2-2z)dz = \boxed{\frac{11}{36}} $$ as claimed.