Given a joint PDF, f subscript X Y end subscript (x comma y )equals c x y comma space 0 less than y less than x less than 4, (1) (5 pts) Determine the constant c value such that the above joint PDF is valid. (2) (6 pts) Find P (X greater than 2 comma space Y less than 1 )(3) (9 pts) Determine the marginal PDF of X given Y

(1) Looks like the joint density is

In order for this to be a proper density function, integrating it over its support should evaluate to 1. The support is a triangle with vertices at (0, 0), (4, 0), and (4, 4) (see attached shaded region), so the integral is

(2) The region in which X > 2 and Y < 1 corresponds to a 2x1 rectangle (see second attached shaded region), so the desired probability is

(3) Are you supposed to find the marginal density of X, or the conditional density of X given Y?

In the first case, you simply integrate the joint density with respect to y:

In the second case, we instead first find the marginal density of Y:

Then use the marginal density to compute the conditional density of X given Y:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°