Question

Tworandom variables, X and Y, have a joint probability density function given by f(x,y) =kxy 0≤x≤1,0≤y≤1 =0 elsewhere

​a) Determine the value of k that makes this a valid probability density function.

Answer 1

To determine the value of k that makes the given function a valid probability density function, we integrate the function over its range and set the result equal to 1. This gives us the equation k/4 = 1, from which we find that k = 4.

Step-by-step explanation:

To determine the value of k that makes the given function a valid probability density function, we need to ensure that the integral over the entire range equals 1:

∫01 ∫01 kxy dy dx = 1

Integrating with respect to y first, we have:

k∫01 [x/2 * y2]

Simplifying the expression:

k(x/2) = kx/2

Integrating with respect to x next:

(k/2)∫01 x dx

Simplifying the expression:

(k/2)(1/2) = (k/4)

For the function to be a valid probability density function, the integral over the entire range must equal 1:

k/4 = 1

Solving for k:

k = 4