Q. 5.22 - A First Course in Probability

Question

Let $\mathrm{U}$ be a uniform $(0,1)$ random variable, and let $\mathrm{a}<\mathrm{b}$ be constants.

(a) Show that if$\mathrm{b}>0$, then $\mathrm{bU}$ is uniformly distributed on $(0, \mathrm{b})$, and if $\mathrm{b}<0$, then $\mathrm{bU}$is uniformly distributed on $(\mathrm{b}, 0)$.

(b) Show that $\mathrm{a}+\mathrm{U}$ is uniformly distributed on $(\mathrm{a}, 1+\mathrm{a})$.

(c) What function of $\mathrm{U}$ is uniformly distributed on $(\mathrm{a}, \mathrm{b})$ $?$

(d) Show that $\min (\mathrm{U},1-\mathrm{U})$ is a uniform $(0, 1/2)$ random variable.

(e) Show that $\max (\mathrm{U},1-\mathrm{U})$ is a uniform $(1/2,1)$ random variable.

Step-by-Step Solution

Answer

a. When $\mathrm{b}>0$, $\mathrm{bU}$ is uniformly distributed on $(0,\mathrm{b})$, and when $\mathrm{b}<0$, $\mathrm{bU}$ is uniformly distributed on $(\mathrm{b},0)$.

b. On$(\mathrm{a}, 1+\mathrm{a})$, $\mathrm{a}+\mathrm{U}$ is uniformly distributed.

c. $\mathrm{U}$is distributed on $\mathrm{a}+(\mathrm{b}-\mathrm{a})$.

d. $(0,\frac{1}{2})$is distributed on $(\mathrm{U},1-\mathrm{U})$.

e. $(\frac{1}{2},1)$is distributed on $(\mathrm{U},1-\mathrm{U})$.

Step 1: Introduction

Over$(1,-1)$,$\mathrm{U}$has a uniform distribution.

$\mathrm{a}=0$

$\mathrm{b}=1$

On the interval between the boundaries, the probability density function of a uniform distribution is the reciprocal of the difference of the boundaries.

${\mathrm{f}}_{\mathrm{U}}\left(\mathrm{x}\right)=\frac{1}{\mathrm{b}-\mathrm{a}}=\frac{1}{1-0}$

$=\frac{1}{1}=1$

The density function's integral.

${\mathrm{F}}_{\mathrm{U}}\left(\mathrm{x}\right)=\mathrm{P}(\mathrm{X}\le \mathrm{x})$

$={\int }_{-\infty }^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}$

$={\int }_0^{\mathrm{x}}1\mathrm{dx}$

$={\left.\left(\mathrm{x}\right)\right|}_0^{\mathrm{x}}$

$=\mathrm{x}-0$

$=\mathrm{x}$

Step 2: Explanation (part a)

a.

The random variable's cumulative distribution function $\mathrm{Ub}$.

$\mathrm{b}>0 {\mathrm{F}}_{\mathrm{bU}}\left(\mathrm{y}\right)=\mathrm{P}(\mathrm{bU}\le \mathrm{y})$

$=\mathrm{P}\left(\mathrm{U}\le \frac{\mathrm{y}}{\mathrm{b}}\right)$

$={\mathrm{F}}_{\mathrm{U}}\left(\frac{\mathrm{y}}{\mathrm{b}}\right)$

$=\frac{\mathrm{y}}{\mathrm{b}}$

$0<\mathrm{y}<\mathrm{b}$

$\mathrm{b}<0 {\mathrm{F}}_{\mathrm{bU}}\left(\mathrm{y}\right)=\mathrm{P}(\mathrm{bU}\le \mathrm{y})$

$=\mathrm{P}\left(\mathrm{U}\ge \frac{\mathrm{y}}{\mathrm{b}}\right)$

$=1-\mathrm{P}\left(\mathrm{U}\le \frac{\mathrm{y}}{\mathrm{b}}\right)$

$=1-{\mathrm{F}}_{\mathrm{U}}\left(\frac{\mathrm{y}}{\mathrm{b}}\right)$

$=1-\frac{\mathrm{y}}{\mathrm{b}}$

Finally, the derivative of the cumulative distribution function of the random variable,

$\mathrm{b}>0$${\mathrm{f}}_{\mathrm{bU}}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}{\mathrm{F}}_{\mathrm{bU}}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\frac{\mathrm{y}}{\mathrm{b}}=\frac{1}{\mathrm{b}}$

$0<\mathrm{y}<\mathrm{b}$

$\mathrm{b}<0$${\mathrm{f}}_{\mathrm{bU}}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}{\mathrm{F}}_{\mathrm{bU}}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\left(1-\frac{\mathrm{y}}{\mathrm{b}}\right)=-\frac{1}{\mathrm{b}}$

Step 3: Explanation (part b)

Cumulative function is,

${\mathrm{F}}_{\mathrm{a}+\mathrm{U}}\left(\mathrm{y}\right)=\mathrm{P}(\mathrm{a}+\mathrm{U}\le \mathrm{y})$

$=\mathrm{P}(\mathrm{U}\le \mathrm{y}-\mathrm{a})$

$={\mathrm{F}}_{\mathrm{U}}(\mathrm{y}-\mathrm{a})$

$=\mathrm{y}-\mathrm{a}$

$\mathrm{a}<\mathrm{y}<1+\mathrm{a}$

density function is,

${\mathrm{f}}_{\mathrm{a}+\mathrm{U}}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}{\mathrm{F}}_{\mathrm{a}+\mathrm{U}}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}(\mathrm{y}-\mathrm{a})=1$

$\mathrm{a}<\mathrm{y}<1+\mathrm{a}$

Step 4: Explanation (part c)

c.

On the interval between the borders, the probability density function of a uniform distribution is the reciprocal of the difference of the boundaries:

${\mathrm{f}}_{\mathrm{g}\left(\mathrm{U}\right)}\left(\mathrm{x}\right)=\frac{1}{\mathrm{b}-\mathrm{a}}$

Cumulative function is,

${\mathrm{F}}_{\mathrm{g}\left(\mathrm{U}\right)}\left(\mathrm{x}\right)=\mathrm{P}\left(\mathrm{g}\right(\mathrm{U})\le \mathrm{x})$

$={\int }_{-\infty }^{\mathrm{x}}{\mathrm{f}}_{\mathrm{g}\left(\mathrm{U}\right)}\left(\mathrm{x}\right)\mathrm{dx}$

$={\left.\left(\frac{\mathrm{x}}{\mathrm{b}-\mathrm{a}}\right)\right|}_{\mathrm{a}}^{\mathrm{x}}$

$=\frac{\mathrm{x}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}$

${\mathrm{F}}_{\mathrm{g}\left(\mathrm{U}\right)}\left(\mathrm{x}\right)=\mathrm{P}\left(\mathrm{g}\right(\mathrm{U})\le \mathrm{x})=\mathrm{P}\left(\mathrm{U}\le {\mathrm{g}}^{-1}\left(\mathrm{x}\right)\right)={\mathrm{F}}_{\mathrm{U}}\left({\mathrm{g}}^{-1}\left(\mathrm{x}\right)\right)={\mathrm{g}}^{-1}\left(\mathrm{x}\right)$

${\mathrm{g}}^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{x}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}$

$\mathrm{y}=\frac{\mathrm{x}-\mathrm{a}}{\mathrm{b}-\mathrm{a}}$

$\mathrm{a}+(\mathrm{b}-\mathrm{a}) \mathrm{y}=\mathrm{x}$

$\mathrm{g}\left(\mathrm{y}\right)=\mathrm{a}+(\mathrm{b}-\mathrm{a}) \mathrm{y}$

Step 5: Explanation (part d)

d.

Cumulative function is,

${\mathrm{F}}_{\min (\mathrm{U},1-\mathrm{U})}\left(\mathrm{y}\right)=\mathrm{P}\left(\min \right(\mathrm{U},1-\mathrm{U})\le \mathrm{y})$

$=\mathrm{P}(\mathrm{U} \le \mathrm{y})+\mathrm{P}(1-\mathrm{U}\le \mathrm{y})$

$=\mathrm{P}(\mathrm{U}\le \mathrm{y})+\mathrm{P}(-\mathrm{U}\le \mathrm{y}-1)$

$=\mathrm{P}(\mathrm{U}\le \mathrm{y})+\mathrm{P}(\mathrm{U}\ge 1-\mathrm{y})$

$=\mathrm{P}(\mathrm{U}\le \mathrm{y})+1-\mathrm{P}(\mathrm{U}\le 1-\mathrm{y})$

$=\mathrm{y}+1-1+\mathrm{y}=2\mathrm{y}$

$0<\mathrm{y}<\frac{1}{2}$

Density function is,

${\mathrm{f}}_{\min (\mathrm{U},1-\mathrm{U})}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}{\mathrm{F}}_{\min (\mathrm{U},1-\mathrm{U})}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}2\mathrm{y}=2$

$0<\mathrm{y}<\frac{1}{2}$

Step 6: Explanation (part e)

e.

Cumulative function,

${\mathrm{F}}_{\min (\mathrm{U},1-\mathrm{U})}\left(\mathrm{y}\right)=\mathrm{P}\left(\min \right(\mathrm{U},1-\mathrm{U})\le \mathrm{y})$

$=\mathrm{P}(\mathrm{U}\le \mathrm{y})+\mathrm{P}(-\mathrm{U}\le \mathrm{y}-1)$

$=\mathrm{P}(\mathrm{U}\le \mathrm{y})+\mathrm{P}(\mathrm{U}\le 1-\mathrm{y})$

$=\mathrm{P}(\mathrm{U}\le \mathrm{y})-\mathrm{P}(\mathrm{U}\le 1-\mathrm{y})$

$=\mathrm{y}-1+\mathrm{y}=2\mathrm{y}-1$

$\frac{1}{2}<\mathrm{y}<1$

Density function is,

${\mathrm{f}}_{\max (\mathrm{U},1-\mathrm{U})}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}{\mathrm{F}}_{\max (\mathrm{U},1-\mathrm{U})}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}(2\mathrm{y}-1)=2$

$\frac{1}{2}<\mathrm{y}<1$