Chapter 2 Section 4 Question 1
Question
U ~ $Uniform[0, 1]$. Compute each of the following.
1. $P(U\leq 0)$
2. $P(U=\frac{1}{2})$
3. $P(U < -\frac{1}{3})$
4. $P(U\leq \frac{2}{3})$
5. $P(U < \frac{2}{3})$
6. $P(U < 1)$
7. $P(U\leq 17)$
Author: Mohammad-Ali Bandzar| Date:Oct 21 2020
From example 2.4.3 in the book we know: $$\textit{f}(x)=\begin{cases}\frac{1}{R-L}&L\leq x\leq R \\ 0& otherwise\end{cases}$$ In this case $R=1,L=0$
(a) $P(U\leq 0)$
since our probability will be equal to the area of a rectangle (base*height) where $U\leq 0$. In this case our base will be $0-0=0$ therefore $P(U\leq 0)=0$
(b) $P(U=\frac{1}{2})$
By textbook definition 2.4.1: 'A random variable X is continuous if $P(X=x)=0$ for all $x\in\mathbb{R}$'
By this definition $P(U=\frac{1}{2})=0$ since uniform distribution is a form of continuous distribution
(c) $P(U < -\frac{1}{3})$
since our probability will be equal to the area of a rectangle (base*height) where $U < -\frac{1}{3}$. In this case our height will be zero since $f(x)=0$ for all $x\in (-\infty ,-\frac{1}{3}]$ therefore $P(U < -\frac{1}{3})=0$
(d) $P(U\leq \frac{2}{3})$
since our probability will be equal to the area of a rectangle (base*height) where $U\leq \frac{2}{3}$. In this case our base will be the intersection of $(-\infty ,\frac{2}{3}]\cap [0,1]=[0,\frac{2}{3}]$ the length of our base is therefore $\frac{2}{3}-0=\frac{2}{3}$. Our height will be $\frac{1}{1-0}=\frac{1}{1}$ $$area=base \times height$$ $$P(U\leq \frac{2}{3})=\frac{2}{3}*1$$ $$P(U\leq \frac{2}{3})=\frac{2}{3}$$
(e) $P(U < \frac{2}{3})$
since our probability will be equal to the area of a rectangle (base*height) where $U < \frac{2}{3}$. In this case our base will be the intersection of $(-\infty ,\frac{2}{3})\cap [0,1]=[0,\frac{2}{3})$ the length of our base is therefore $\frac{2}{3}-0=\frac{2}{3}$. Our height will be $\frac{1}{1-0}=\frac{1}{1}$ $$area=base \times height$$ $$P(U < \frac{2}{3})=\frac{2}{3}*1$$ $$P(U < \frac{2}{3})=\frac{2}{3}$$
(f) $P(U < 1)$
since our probability will be equal to the area of a rectangle (base*height) where $U < 1$. In this case our base will be the intersection of $(-\infty ,1)\cap [0,1]=[0,1]$ the length of our base is therefore $1-0=1$. Our height will be $\frac{1}{1-0}=\frac{1}{1}$ $$area=base \times height$$ $$P(U < 1)=1*1$$ $$P(U < 1)=1$$
(g) $P(U\leq 17)$
since our probability will be equal to the area of a rectangle (base*height) where $U \leq 17$. In this case our base will be the intersection of $(-\infty ,17]\cap [0,1]=[0,1]$ the length of our base is therefore $1-0=1$. Our height will be $\frac{1}{1-0}=\frac{1}{1}$ $$area=base \times height$$ $$P(U \leq 17)=1*1$$ $$P(U \leq 17)=1$$