Chapter 7.3, Problem 1E

To determine

Obtain the probability distribution of the random variable $Y=2X-1$.

Answer to Problem 1E

Therefore, the probability distribution of Y is provided below:

$g\left(y\right)=\{\begin{array}{l}\frac{1}{3},y=1,3,5\\ 0,\text{ elsewhere}\end{array}$

Explanation of Solution

In this context, let X be the random variable and the probability distribution of X is provided below:

$f\left(x\right)=\{\begin{array}{l}\frac{1}{3},x=1,2,3\\ 0,\text{ elsewhere}\end{array}$

Since the random variable X takes the values from 1 to 3, the random variable Y will take the following values.

Here, $Y=2X-1$

For x=1

$\begin{array}{c}y=2x-1\\ =2\left(1\right)-1\\ =1\end{array}$

In similar ways, the remaining values are obtained.

X	1	2	3
$Y=2X-1$	1	3	5

Theorem 7.1 states that X is a discrete random variable with probability distribution f(x). It is assumed that $Y=u\left(X\right)$ defines one-to-one transformation between the values of X and Y. Therefore, the equation $y=u\left(x\right)$ is uniquely solved for x in terms of y. That is, $x=w\left(y\right)$. Then, the probability distribution of Y is will be as given below:

$\begin{array}{c}g\left(y\right)=P\left(Y=y\right)\\ =P\left(X=w\left(y\right)\right)\\ =f\left(w\left(y\right)\right)\end{array}$

Express x as a function of y using Theorem 7.1.

$\begin{array}{c}y=2x-1\\ y+1=2x\\ x=\frac{y+1}{2}\end{array}$

Therefore, the probability distribution of Y is provided below:

$\begin{array}{c}g\left(y\right)=f\left(x\right)\\ =f\left(\frac{y+1}{2}\right)\\ =\{\begin{array}{l}\frac{1}{3},y=1,3,5\\ 0,\text{ elsewhere}\end{array}\end{array}$