Answer to Question #213460 in Statistics and Probability for nicki

Question

1. Suppose the pmf of a random variable x is given by f(x)={1/6,for x=1,2,3,4,5,6

{0,elsewhere

Obtain the pmf of y=2x² and z=x-3

2.Let the pmf of a random variable X be given by f(x)={(x²+1)/18 for x=0,1,2,3

{0,elsewhere

determine the pmf of Y=x²+1

3.Suppose the pmf of a random variable x is given by f(x)={x/10 for x=0,1,2,3,4

{0,elsewhere

Obtain the pmf of y=|x-2|

4.Let the pmf of a random variable x be given by f(x)={(1/2)^x for x=1,2,3

{0,elsewhere

determine the pmf

y={1 if x is odd

{0 if x is even

5.Suppose the random variable x has the pmf is given by f(x)={1/6(5/6)^x for x=0,1,2,3,....

{0,elsewhere

Obtain the pmf of y=x-1

Answer 1

1.

$$f(y) = \begin{cases} 1/6 &\text{for } y=2,8,18,32,50,72 \\ 0 &\text{elsewhere } \end{cases}$$

$$f(z) = \begin{cases} 1/6 &\text{for } z=-2,-1,0,1,2,3 \\ 0 &\text{elsewhere } \end{cases}$$

2.

$$f(y) = \begin{cases} y/18 &\text{for } y=1,2,5,10 \\ 0 &\text{elsewhere } \end{cases}$$

3.

$$\begin{matrix} x & 0 & 1 & 2 & 3 & 4 \\ y & 2 & 1 & 0 & 2 & 2 \end{matrix}$$

$$\begin{matrix} y & 0 & 1 & 2 \\ p(y) & 2/10 & 4/10 & 4/10 \end{matrix}$$

$$f(y) = \begin{cases} 1/5 &\text{for } y=0 \\ 2/5 &\text{for } y=1 \\ 2/5 &\text{for } y=2 \\ 0 &\text{elsewhere } \end{cases}$$

4.

$$f(x) = \begin{cases} 1/2^x &\text{for } x=1,2,3,4, ... \\ 0 &\text{elsewhere } \end{cases}$$

$$\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+...=\displaystyle\sum_{i=1}^{\infin}\dfrac{1}{2^{n+1}}=\dfrac{1}{2}$$

$$f(y) = \begin{cases} 1/2 &\text{for } y=0 \\ 1/2 &\text{for } y=1 \\ 0 &\text{elsewhere } \end{cases}$$

5.

$$f(x) = \begin{cases} \dfrac{5}{36}\bigg(\dfrac{5}{6}\bigg)^y &\text{for } y=-1,0,1,2,3,4, ... \\ 0 &\text{elsewhere } \end{cases}$$