A researcher finds that two continuous, random variables of interest, X and Y, have a joint

probability density function (pdf) given by:
f(x,y)={cxy 0<=x<=1,0<=y<=1,x+y=>1,
.........0 otherwise
where c is a constant.
(i) Find the value of c so that f(x,y) represents a pdf.
(iii) Calculate the conditional probability P{X > 0.25 | Y = 0.5}.
(iv) Calculate Cov(X,Y) and interpret the obtained value.
(v) Find E[X + 1/Y].

(ii) Calculate the marginal density functions of the random variables X and Y, and
E(X) and E(Y).

(i) To find the value of c so that f(x,y) represents a probability density function (pdf), we need to ensure that the total probability under the function integrates to 1 over the specified range.

To calculate the integral, we need to split it into two parts:

For 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 - x:
∫∫ f(x,y) dxdy = ∫∫ cxy dxdy

Performing the integration:

∫∫ cxy dxdy = c∫(0 to 1)∫(0 to 1-x)xy dydx
= c∫(0 to 1) [x/2(1-x)^2] dx
= (-c/3) * (2/3)

For the remaining region (x + y ≥ 1):
∫∫ f(x,y) dxdy = ∫∫ 0 dxdy = 0

To ensure that f(x,y) represents a pdf, the total probability must be 1. Therefore, we set the first part equal to 1:

(-c/3) * (2/3) = 1
c = -9/2

(ii) To calculate the marginal density functions of X and Y, we integrate the joint pdf over the respective variables:

For X:
f_X(x) = ∫(0 to 1-x) f(x,y) dy
= ∫(0 to 1-x) (-9/2)xy dy
= (-9/2) * (x/2)(1-x)^2
= (-9/4) * x(1-x)^2

For Y:
f_Y(y) = ∫(0 to 1-y) f(x,y) dx
= ∫(0 to 1-y) (-9/2)xy dx
= (-9/2) * (y/2)(1-y)^2
= (-9/4) * y(1-y)^2

To calculate the expected values (E(X) and E(Y)), we integrate the respective variables multiplied by their marginal density functions:

E(X) = ∫(0 to 1) x f_X(x) dx
= ∫(0 to 1) x (-9/4) x(1-x)^2 dx

E(Y) = ∫(0 to 1) y f_Y(y) dy
= ∫(0 to 1) y (-9/4) y(1-y)^2 dy

(iii) To calculate the conditional probability P{X > 0.25 | Y = 0.5}, we first find the conditional density function f(x|y), then integrate it:

f(x|y) = f(x,y) / f_Y(y)
= cxy / ((-9/4) * y(1-y)^2)
= (-2/9) * (x / (y(1-y)^2))

P{X > 0.25 | Y = 0.5} = ∫(0.25 to 1) (-2/9) * (x / (0.5(1-0.5)^2)) dx

(iv) The covariance Cov(X,Y) is calculated using the formulas:

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

To find E(XY), we integrate X multiplied by Y multiplied by the joint pdf:

E(XY) = ∫∫ xy f(x,y) dxdy
= ∫∫ cxy^2 dxdy

To interpret Cov(X,Y), we can say that it quantifies the degree of linear dependence or relationship between X and Y. A positive covariance indicates a positive relationship, a negative covariance indicates a negative relationship, and a covariance close to zero indicates little or no relationship.

(v) To find E[X + 1/Y], we calculate the expected value by integrating over the joint pdf:

E[X + 1/Y] = ∫∫ (x + 1/y) f(x,y) dxdy

To find the value of c so that f(x,y) represents a probability density function (pdf), we need to ensure that the total probability over the entire range of the variables is equal to 1.

To derive c, integrate the joint pdf f(x,y) over the entire range of x and y and set it equal to 1:

∫∫ f(x,y) dx dy = 1

∫∫ cxy dx dy = 1

First, integrate with respect to x:

∫ cxy dx = c/2 * x^2 * y

Next, integrate c/2 * x^2 * y with respect to y:

∫∫ c/2 * x^2 * y dy dx

Integrating with respect to y gives us:

c/4 * x^2 * y^2 + constant

To satisfy the condition that the pdf integrates to 1, we set the constant to 0:

c/4 * x^2 * y^2 = 1

Now, solve for c:

c = 4 / ∫∫ x^2 * y^2 dx dy

To calculate the conditional probability P{X > 0.25 | Y = 0.5}, we use the conditional probability formula:

P{X > 0.25 | Y = 0.5} = f(x|y) / f(y)

To find f(x|y), we need to calculate the conditional density function given by:

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

The marginal density function f(y) is found by integrating f(x,y) over the range of x:

f(y) = ∫ f(x,y) dx

In this case, because f(x,y) is equal to 0 when x+y < 1, we need to integrate only over the range where x ≥ 1-y.

Once we have f(x|y) and f(y), we can calculate the conditional probability P{X > 0.25 | Y = 0.5} by substituting these values into the conditional probability formula.

To calculate the covariance (Cov) between X and Y, we use the formula:

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

First, calculate E(XY) by finding the joint expectation:

E(XY) = ∫∫ xy * f(x,y) dx dy

Calculate E(X) and E(Y) by finding the marginal expectations:

E(X) = ∫ x * f(x) dx

E(Y) = ∫ y * f(y) dy

Finally, substitute these values into the covariance formula to obtain the Cov(X,Y) value.

To find the expectation E[X + 1/Y], we use the linearity of expectation:

E[X + 1/Y] = E(X) + E(1/Y)

Calculate E(X) as mentioned earlier, and then calculate E(1/Y) by finding the marginal expectation of 1/Y weighted by the marginal density function of Y.

By following these steps, you will be able to answer all the sub-questions (i) to (v) regarding the given joint probability density function.