Question

Suppose that X and Y have a continuous joint distribution with the joint PDF:

f(x, y) = 13(x+y), text for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

What is the marginal PDF of X?

a) (f_X(x) = 12)

b) (f_X(x) = x)

c) (f_X(x) = 12(x+1))

d) (f_X(x) = 13)

Answer 1

c) (f_X(x) = 12(x+1))

Step-by-step explanation:

To find the marginal PDF of X, we need to integrate the joint PDF over the range of Y. Since the joint PDF is given as:

f(x, y) = 13(x+y), for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1

To find the marginal PDF of X, we integrate f(x, y) over the range of y, which is from 0 to 1:

f_X(x) = ∫[0,1] f(x, y) dy

Substituting the given joint PDF:

f_X(x) = ∫[0,1] 13(x+y) dy

Now, let's integrate with respect to y:

f_X(x) = 13x∫[0,1] dy + 13∫[0,1] y dy

The integral of dy over the range [0, 1] is simply y evaluated at the limits:

f_X(x) = 13x[y]₀¹ + 13∫[0,1] y dy

Evaluating the limits of the first term:

f_X(x) = 13x(1 - 0) + 13∫[0,1] y dy

Simplifying:

f_X(x) = 13x + 13∫[0,1] y dy

Now, let's integrate the second term:

f_X(x) = 13x + 13[(1/2)y²]₀¹

Evaluating the limits of the second term:

f_X(x) = 13x + 13[(1/2)(1)² - (1/2)(0)²]

Simplifying further:

f_X(x) = 13x + 13(1/2)

f_X(x) = 13x + 6.5

So, the marginal PDF of X is given by:

f_X(x) = 13x + 6.5

Therefore, the correct option is:

c) (f_X(x) = 12(x+1))