Question #213460
  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


1
Expert's answer
2021-07-12T13:26:55-0400

1.


f(y)={1/6for y=2,8,18,32,50,720elsewhere f(y) = \begin{cases} 1/6 &\text{for } y=2,8,18,32,50,72 \\ 0 &\text{elsewhere } \end{cases}


f(z)={1/6for z=2,1,0,1,2,30elsewhere f(z) = \begin{cases} 1/6 &\text{for } z=-2,-1,0,1,2,3 \\ 0 &\text{elsewhere } \end{cases}


2.


f(y)={y/18for y=1,2,5,100elsewhere f(y) = \begin{cases} y/18 &\text{for } y=1,2,5,10 \\ 0 &\text{elsewhere } \end{cases}


3.


x01234y21022\begin{matrix} x & 0 & 1 & 2 & 3 & 4 \\ y & 2 & 1 & 0 & 2 & 2 \end{matrix}


y012p(y)2/104/104/10\begin{matrix} y & 0 & 1 & 2 \\ p(y) & 2/10 & 4/10 & 4/10 \end{matrix}


f(y)={1/5for y=02/5for y=12/5for y=20elsewhere 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)={1/2xfor x=1,2,3,4,...0elsewhere f(x) = \begin{cases} 1/2^x &\text{for } x=1,2,3,4, ... \\ 0 &\text{elsewhere } \end{cases}

12+18+132+...=i=112n+1=12\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{32}+...=\displaystyle\sum_{i=1}^{\infin}\dfrac{1}{2^{n+1}}=\dfrac{1}{2}


f(y)={1/2for y=01/2for y=10elsewhere f(y) = \begin{cases} 1/2 &\text{for } y=0 \\ 1/2 &\text{for } y=1 \\ 0 &\text{elsewhere } \end{cases}

5.


f(x)={536(56)yfor y=1,0,1,2,3,4,...0elsewhere 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}



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