Answer:
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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
