Question

Let X represent the amount of energy a city uses (in megawatt-hours) in the Kanto region. Let Y represent the amount of mismanaged plastic waste (in metric tons) a city produces in the Kanto region. The following joint distribution relates the probability of a city using a certain amount of energy and creating a certain amount of plastic waste. Remember to list the relevant domains. fxy(x,y) = 2 sy s 2x < 10 2. Find P[X<2UX > 4). TO RECEIVE FULL CREDIT, YOU MUST FIND THE PROBABILITY THROUGH ITEGRATION. Hint: draw the area you are integrating over. You can keep as a fraction or round to one decimal place. 3. Find fy(y) 4. Find E[Y]. Keep as a simplified fraction or round to four decimal places. 5. Find fxy(x). 6. Find Myyy). 7. Find E[X]. Hint: you already found fx(x) in a previous problem. Keep as a simplified fraction or round to four decimal places. 8. Find E[XY]. Round to an integer. 9. Find Cov(X,Y). Keep as a simplified fraction or round to four decimal places. 10. Are X and Y independent?

Answer 1

Part 2: The probability of X≤2 or X≥4 is 0.5.

Part 3: The value of marginal probability of y is
`f_y(y)=(10-y)/(32)` for
`2\leq y\leq 10`

Part 4:The value of E(y) is 4.6667.

Part 5:The value of
`f_(xy)(x)` is
`(2)/(10-y)` for
`2\leq y\leq 2x\leq 10`

Part 6:The value of
`M_(x,y)(y)` is
`(y+10)/(4)`

Part 7:The value of E(x) is 3.6667.

Part 8:The value of E(x,y) is 36.

Part 9:The value of Cov(x,y) is 18.8886.

Part 10:X and Y are not independent variables as
`f_(xy)(x,y)\\eq f_x(x).f_y(y)\\`

Explanation:

As the complete question is here, however some of the values are not readable, thus the question is found online and is attached herewith.

From the given data, the joint distribution is given as

`f(x,y)=(1)/(16)` for
`2\leq y\leq 2x\leq 10`

Now the distribution of x is given as

`f_x(x)=\int\limits^(\infty)_(-\infty) {f(x,y)} \, dy`

Here the limits for y are
`2\leq y\leq 2x` So the equation becomes

`f_x(x)=\int\limits^(\infty)_(-\infty) {f(x,y)} \, dy\\f_x(x)=\int\limits^(2x)_(2) (1)/(16) \, dy\\f_x(x)=(1)/(16) (2x-2)\\f_x(x)=(x-1)/(8) \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 1\leq x\leq 5`

Part 2:

The probability is given as

`P(X\leq 2 U X\geq 4)=\int\limits^2_1 {f_x(x)} \, dx +\int\limits^5_4 {f_x(x)} \, dx\\P(X\leq 2 U X\geq 4)=\int\limits^2_1 {(x-1)/(8)} \, dx +\int\limits^5_4 {(x-1)/(8)} \, dx\\P(X\leq 2 U X\geq 4)=(1)/(16)+(7)/(16)\\P(X\leq 2 U X\geq 4)=0.5`

So the probability of X≤2 or X≥4 is 0.5.

Part 3:

The distribution of y is given as

`f_y(y)=\int\limits^(\infty)_(-\infty) {f(x,y)} \, dx`

Here the limits for x are
`y/2\leq x\leq 5` So the equation becomes

`f_y(y)=\int\limits^(\infty)_(-\infty) {f(x,y)} \, dx\\f_y(y)=\int\limits^(5)_(y/2) (1)/(16) \, dx\\f_y(y)=(1)/(16) (5-(y)/(2))\\f_y(y)=(10-y)/(32) \,\,\,\,\,\,\,\,\,\,\,\, for \,\,\,\,\,\,\,\,\,\ 2\leq y\leq 10`

So the value of marginal probability of y is
`f_y(y)=(10-y)/(32)` for
`2\leq y\leq 10`

Part 4

The value is given as

`E(y)=\int\limits^(10)_2 {yf_y(y)} \, dy\\E(y)=\int\limits^(10)_2 {y(10-y)/(32)} \, dy\\E(y)=(1)/(32)\int\limits^(10)_2 {10y-y^2} \, dy\\E(y)=4.6667`

So the value of E(y) is 4.6667.

Part 5

This is given as

`f_(xy)(x)=(f_(xy)(x,y))/(f_y(y))\\f_(xy)(x)=((1)/(16))/((10-y)/(32))\\f_(xy)(x)=(2)/(10-y)`

So the value of
`f_(xy)(x)` is
`(2)/(10-y)` for
`2\leq y\leq 2x\leq 10`

Part 6

The value is given as

`\geq M_(x,y)(y)=E(f_(xy)(x))=\int\limits^5_(y/2) {x f_(xy)(x)} \, dx \\M_(x,y)(y)=\int\limits^5_(y/2) {x (2)/(10-y)} \, dx \\M_(x,y)(y)=(2)/(10-y)\left[(x^2)/(2)\right]^5_{(y)/(2)}\\M_(x,y)(y)=(2)/(10-y)\left((25)/(2)-(y^2)/(8)\right)\\M_(x,y)(y)=(y+10)/(4)`

So the value of
`M_(x,y)(y)` is
`(y+10)/(4)`

Part 7

The value is given as

`E(x)=\int\limits^(5)_1 {xf_x(x)} \, dx\\E(x)=\int\limits^(5)_1 {x(x-1)/(8)} \, dx\\E(x)=(1)/(8)\left((124)/(3)-12\right)\\E(x)=(11)/(3) =3.6667`

So the value of E(x) is 3.6667.

Part 8

The value is given as

`E(x,y)=\int\limits^(5)_1 \int\limits^(10)_2 {xyf_(x,y)(x,y)} \,dy\, dx\\E(x,y)=\int\limits^(5)_1 \int\limits^(10)_2 {xy(1)/(16)} \,dy\, dx\\E(x,y)=\int\limits^(5)_1 (x)/(16)\left[(y^2)/(2)\right]^(10)_2\, dx\\E(x,y)=\int\limits^(5)_1 3x\, dx\\\\E(x,y)=3\left[(x^2)/(2)\right]^5_1\\E(x,y)=36`

So the value of E(x,y) is 36

Part 9

The value is given as

`Cov(X,Y)=E(x,y)-E(x)E(y)\\Cov(X,Y)=36-(3.6667)(4.6667)\\Cov(X,Y)=18.8886\\`

So the value of Cov(x,y) is 18.8886

Part 10

The variables X and Y are considered independent when

`f_(xy)(x,y)=f_x(x).f_y(y)\\`

Here

`f_x(x).f_y(y)=(x-1)/(8)(10-y)/(32) \\`

And

`f_(xy)(x,y)=(1)/(16)`

As these two values are not equal, this indicates that X and Y are not independent variables.