Question

The random variable X measures the concentration of ethanol in a chemical solution, and the random variable Y measures the acidity of the solution. They have a joint probability density function f (x, y) = A (20 - x - 2y), 0 lessthanorequalto x lessthanorequalto 5, 0 lessthanorequalto y lessthanorequalto 5 and f (x, y) = 0 elsewhere. (a) What is the value of A? (b) What is P (1 lessthanorequalto X lessthanorequalto 2, 2 lessthanorequalto Y lessthanorequalto 3)? (c) Construct the marginal probability density functions for X and Y. (d) Are the ethanol concentration and the acidity independent? (e) What are the expectation and the variance of the ethanol concentration? (f) What the expectation and the variance of the acidity? (g) If the ethanol concentration is 3, what is the conditional probability density function of the acidity? (h) What is the covariance between the ethanol concentration and the acidity? (i) What is the correlation between the ethanol concentration and the acidity?

Answer 1

My explanation is too long so, I had to limit my characters

Find answers within explanation.

Step-by-step explanation:

Given

f (x, y) = A (20 - x - 2y), 0 ≤ x ≤ 5, 0 ≤ y ≤ 5 and f (x, y) = 0 elsewhere.

(a) To solve for A,the joint probability density function must satisfy the following condition

∫∫f(x,y) = 1

So, we have

∫∫ A (20 - x - 2y) dydx.{0,5}{0,5} = 1

First, we integrate with respect to y

∫[∫A (20 - x - 2y){0,5}dy]dx{0,5} = 1

A∫[∫ (20 - x - 2y){0,5}dy]dx{0,5} = 1

A∫[(20y - xy - y²){0,5}] dx {0,5} = 1

A∫[(20(5) - x(5) - (5)²)] dx{0,5} = 1

A∫[(100 - 5x - 25)] dx {0,5} = 1

A∫[(75- 5x)] dx {0,5} = 1

Then we differentiate with respect to x

A[(75x- 5x²/2)] {0,5} = 1

A[(75(5)- 5(5)²/2)] = 1

A(375 - 125/2)= 1

625A/2 = 1

625A = 2

A = 2/625

b. Here we have

∫∫ A (20 - x - 2y) dydx.{2,3}{1,2} where A = 2/625

First, we integrate with respect to y

∫[∫A (20 - x - 2y){2,3}dy]dx{1,2}

A∫[∫ (20 - x - 2y){2,3}dy]dx{1,2}

A∫[(20y - xy - y²){2,3}] dx {1,2}

A∫[(20(3) - x(3) - (3)²) - (20(2) - x(2) - (2)²] dx{1,2}

A∫[(60 - 3x - 9) - (40 - 2x - 4)] dx {1,2}

A∫[(20- x - 5)] dx {1,2}

A∫[(15 - x)] dx {1,2}

Then we differentiate with respect to x

A[(15x- x²/2)] {1,2}

A[(15(2)- (2)²/2) - (15(1) - 1²/2]

A(28 - 29/2)

A(27/2) ------ Substitute 2/625 for A

2/625 * 27/2

27/625

So, P (1 ≤ X ≤ 2, 2 ≤ Y ≤ 3) = 27/625

c. Calculating the marginal probability density function for X;

This is given by

fx(x) = ∫ f(x,y) dy

Where f(x,y) = f (x, y) = A (20 - x - 2y), 0 ≤ y ≤ 5 and A = 2/625

So, we have

fx(x) = ∫ A (20 - x - 2y) dy {0,5}

A ∫(20 - x - 2y) dy {0,5}

Integrate with respect to y

A (20y - xy - y²) {0,5}

A(20(5) - x(5) - 5²)

A(100 - 5x - 25)

A(75-5x)

A * 5(15-x)

5A(15-x)

5 * 2/625 * (15 - x)

2/125 * (15 - x)

(30 - 2x)/125

So, fx(x) = (30 - 2x)/125

Calculating the marginal probability density function for Y;

This is given by

fy(y) = ∫ f(x,y) dx

Where f(x,y) = f (x, y) = A (20 - x - 2y), 0 ≤ x ≤ 5 and A = 2/625

So, we have

fy(y) = ∫ A (20 - x - 2y) dx {0,5}

A ∫(20 - x - 2y) dx {0,5}

Integrate with respect to x

A (20x - x²/2 - 2xy) {0,5}

A(20(5) - 5²/2 - 2*5y)

A(100 - 25/2 - 10y)

A(175/2 - 10y)

A * (175 - 20y)/2

2/625 * (175 - 20y)/2

(175 - 20y)/625

(35 - 4y)/125

So, fy(y) = (35 - 4y)/125

d. If the product of the marginal distribution of variables X and Y emails the joint probability density function, then they are independent.

Mathematically, f(x,y) = fx(x) * fy(y) for all values of x and y

Let x∈(0,5) and y∈(0,5)

Then

f(x,y) ≠ fx(x) * fy(y)

So, x and y are not independent

e. Here, we're asked to find E(x) and Var(x)

Calculating E(x)

E(x) = ∫xfx(x) dx

Where fx(x) = (30 - 2x)/125 for 0 ≤ x ≤ 5

So, E(x) = ∫x (30 - 2x)/125 dx {0,5}

1/125 ∫ x(30-2x) dx {0,5}

1/125∫30x - 2x² dx {0,5}

1/125 (15x² - 2x³/3) {0,5}

1/125(15(5)² - 2(5)³/3)

1/125(375-250/3)

1/125(875)

7/3

So, E(x) = 7/3

Var(x) = E(x²) - (E(x))²

Calculating E(x²)

E(x²) = ∫x²fx(x) dx

Where fx(x) = (30 - 2x)/125 for 0 ≤ x ≤ 5

So, E(x²) = ∫x² (30 - 2x)/125 dx {0,5}

1/125 ∫ x ²(30-2x) dx {0,5}

1/125∫30x² - 2x³ dx {0,5}

1/125 (10x³ - ½x⁴) {0,5}

1/125(10(5)³ - ½(5)⁴)

1/125(1250 - 625/2)

1/125(1875/2)

E(x²) = 15/2

So,Var(x) = E(x²) - (E(x))² becomes

Var(x) = 15/2 - (7/3)²

Var(x) = 15/2 - 49/9

Var(x) = (135 - 98)/9

Var(x) = 37/18

f. Here, we're asked to find E(y) and Var(y)

Calculating E(y)

E(y) = ∫yfy(y) dy

Where fy(y) = (35 - 4y)/125 for 0 ≤ y ≤ 5

So, E(y) = ∫y (35 - 4y)/125 dy {0,5}

1/125 ∫ y(35 - 4y) dy {0,5}

1/125∫35y - 4y² dy {0,5}

1/125 (35y²/2 - 4y³/3) {0,5}

1/125(35(5)²/2 - 4(5)³/3)

1/125(875/2 - 500/3)

7/2 - 4/3

(21 - 8)/6

So, E(y) = 13/6

Var(y) = E(y²) - (E(y))²

Calculating E(x²)

E(y²) = ∫y²fy(y) dy

Where fy(y) = (35 - 4y)/125 for 0 ≤ y ≤ 5

So, E(y²) = ∫y² (35 - 4y)/125 dy {0,5}

1/125 ∫ y²(35 - 4y) dy {0,5}

1/125∫35y² - 4y³ dy {0,5}

1/125 (35y³/3 - y⁴) {0,5}

1/125(35(5)³/3 - (5)⁴)

1/125(4375/3 - 625)

35/3 - 5

(35 - 15)/3

E(y²) = 20/3

So,Var(y) = E(y²) - (E(y))² becomes

Var(y) = 20/3 - (13/6)²

Var(y) = 71/36

g. Here, we're asked to solve for

fy|x = x(y).

This can be solved using the following

fy|x = x(y) = f(x,y)/fx(x)

So, fy|x = x(y) = f(x,y)/fx(x)

fy|x = 3(y) = f(3,y)/fx(3)

Let y∈(0,5); so, we have

fy|x = x(y) = A(20-3-2y)/(30-(2*3)/125)

fy|x = x(y) = 125A(17-2y)/24

Substitute 2/625 for A

fy|x = x(y) = (17-2y)/60

h. Formula for Covariance is

Cov(X,Y) = E(XY) - E(X)E(Y)

Calculating E(XY)

E(XY) = ∫∫xy f(x,y) dy dx

∫∫ xy * A(20-x-2y) dy dx {0,5}{0,5}

A∫∫ xy * (20-x-2y) dy dx {0,5}{0,5}

A∫∫ 20xy - x²y -2xy² dy dx {0,5}{0,5}

First, we integrate with respect to y

A∫10xy² - x²y²/2 - 2xy³/3 {0,5} dx {0,5}

A∫10x(5²) - x²(5²)/2 - 2x(5³)/3 dx {0,5}

A∫250x - 25x²/2 - 250x/3 dx {0,5}

A∫500x/3 - 25x²/2 dx {0,5}

Then we integrate with respect to x

A(500x²/6 - 25x³/6) {0,5}

A(500(5)²/6 - 25(5)³/6)

A(12500/6 - 3125/6)

A(9375/6)

Substitute 2/625 for A

2/625 * 9375/6

E(XY) = 5

So, Cov(X,Y) = 5 - 7/3*13/6

Cov(X,Y) = -1/18

i. Correlation is calculated as follows;

Cor(x,y) = Cov(x,y)/√(Var(y)*(Var(x)

Cor(x,y) = (-1/18)/√(71/36 *37/18)

Cor(x,y) = -0.0276