Probability & Statistics Ch.2 Section 4 Question 1

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

$U \sim \text{Uniform}[0, 1]$ . Compute each of the following:

a) $P(U \leq 0)$
b) $P(U = \frac{1}{2})$
c) $P(U < -\frac{1}{3})$
d) $P(U \leq \frac{2}{3})$
e) $P(U < \frac{2}{3})$
f) $P(U < 1)$
g) $P(U \leq 17)$

Solution

From Example 2.4.3 in the textbook:

f(x) = \begin{cases}\frac{1}{R-L} & L \leq x \leq R \\ 0 & \text{otherwise}\end{cases}

In this case R = 1, L = 0, so the density is 1 on [0,1] and 0 elsewhere.

(a) $P(U \leq 0)$

The probability equals the area of a rectangle where $U \leq 0$ . The base is $0 - 0 = 0$ .

P(U \leq 0) = 0

(b) $P(U = \frac{1}{2})$

By Definition 2.4.1: “A random variable X is continuous if $P(X=x)=0$ for all $x \in \mathbb{R}$ .”

Since uniform distribution is continuous:

P(U = \frac{1}{2}) = 0

(c) $P(U < -\frac{1}{3})$

The height is zero since $f(x)=0$ for all $x \in (-\infty, -\frac{1}{3}]$ .

P(U < -\frac{1}{3}) = 0

(d) $P(U \leq \frac{2}{3})$

Base = intersection of $(-\infty, \frac{2}{3}] \cap [0,1] = [0, \frac{2}{3}]$ , so length = $\frac{2}{3}$

Height = $\frac{1}{1-0} = 1$

P(U \leq \frac{2}{3}) = \frac{2}{3} \times 1 = \frac{2}{3}

(e) $P(U < \frac{2}{3})$

Base = intersection of $(-\infty, \frac{2}{3}) \cap [0,1] = [0, \frac{2}{3})$ , length = $\frac{2}{3}$

Height = 1

P(U < \frac{2}{3}) = \frac{2}{3}

(f) $P(U < 1)$

Base = intersection of $(-\infty, 1) \cap [0,1] = [0, 1)$ , length = 1

Height = 1

P(U < 1) = 1

(g) $P(U \leq 17)$

Base = intersection of $(-\infty, 17] \cap [0,1] = [0, 1]$ , length = 1

Height = 1

P(U \leq 17) = 1