The function f(x, y) is defined as k * (x^2 + 2*x*y), where k is a normalization constant to ensure that the function f(x, y) is a valid probability density function (pdf). The limits of x are (0, 1) and for y are (0, 2).
a. To make f(x, y) be a valid pdf, the double integral of f(x, y) over its entire range (0 ≤ x ≤ 1, 0 ≤ y ≤ 2) must equal 1.
Therefore, we have the equation:
∬f(x, y) dx dy = 1
Substitute f(x, y) = k * (x^2 + 2*x*y):
∬ [k * (x^2 + 2*x*y)] dx dy = 1
Take the integral part by part:
∫ from 0 to 2 [∫ from 0 to 1 [k * (x^2 + 2*x*y)] dx ] dy = 1
After performing the double integration, equate it to 1. Solve the resulting equation for k. This will give you the constant k that ensures f(x, y) is a valid pdf.
b. To find P(Y < 1 | X = 0.5), we first need to condition on the value X = 0.5. This results in the conditional pdf f(y|0.5) where x = 0.5 in f(x, y).
Now, to find P(Y < 1 | X = 0.5), we integrate this conditional pdf from 0 to 1. This single integral of the conditional pdf over the range (0, 1) will give you the probability P(Y < 1 | X = 0.5).