Question

Given a, b such that both a and b are real numbers between 0 and 15, what is the probability for |a-b|

Answer 1

I guess you're asking about the probability density for the random variable $|A-B|$ where $A,B$ are independent and identically distributed uniformly on the interval (0, 15). The PDF of e.g. $A$ is

$\mathrm{Pr}(A=a) = \begin{cases}\dfrac1{15} & \text{if } 0 < a < 15 \\\\ 0 & \text{otherwise}\end{cases}$

It's easy to see that the support of $|A-B|$ is the same interval, (0, 15), since $|x|\ge0$, and

• at most, if $A=15$ and $B=0$, or vice versa, then $|A-B|=15$

• at least, if $A=B$, then $|A-B|=0$

Compute the CDF of $C=|A-B|$ :

$\mathrm{Pr}(C\le c) = \mathrm{Pr}(|A - B| \le c) = \mathrm{Pr}(-c \le A - B \le c)$

This probability corresponds to the integral of the joint density of $A,B$ over a subset of a square with side length 15 (see attached). Since $A,B$ are independent, their joint density is

$\mathrm{Pr}(A=a,B=b) = \begin{cases}\dfrac1{15^2} & \text{if } (a,b) \in (0,15) \times (0,15) \\ 0 &\text{otherwise}\end{cases}$

The easiest way to compute this probability is by using the complementary region. The triangular corners are much easier to parameterize.

$\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \mathrm{Pr}(|A-B| > c) \\\\ ~~~~~~~~ = 1 - \int_0^{15-c} \int_{a+c}^{15} \frac{db\,da}{15^2} - \int_c^{15} \int_0^{a-c} \frac{db\,da}{15^2} \\\\ ~~~~~~~~ = 1 - \frac1{225} \left(\int_0^{15-c} (15 - a - c) \, da + \int_c^{15} (a - c) \, da\right)$

In the second integral, substitute $a=15-a'$ and $da=-da'$, so that

$\displaystyle \int_c^{15} (a-c) \, da = \int_{15-c}^0 (15-a'-c) (-da') = \int_0^{15-c} (15 - a' - c) \, da'$

which is the same as the first integral. This tells us the joint density is symmetric over the two triangular regions.

Then the CDF is

$\displaystyle \mathrm{Pr}(|A-B|\le c) = 1 - \frac2{225} \int_0^{15-c} (15 - a - c) \, da \\\\ ~~~~~~~~ = 1 - \frac2{225} \left((15-c) a - \frac12 a^2\right) \bigg|_{a=0}^{a=15-c} \\\\ ~~~~~~~~ = \begin{cases}0 & \text{if } c < 0 \\\\ 1 - \dfrac{(15-c)^2}{225} = \dfrac{2c}{15} - \dfrac{c^2}{225} & \text{if } 0 \le c < 15 \\\\ 1 & \text{if } c \ge 15\end{cases}$

We recover the PDF by differentiating with respect to $c$.

$\mathrm{Pr}(|A-B| = c) = \begin{cases}\dfrac2{15} - \dfrac{2c}{225} & \text{if } 0 < c < 15 \\\\ 0 & \text{otherwise}\end{cases}$