Final answer:
The function f(x, y) is non-negative because it can be expressed as a sum of squares and a non-negative constant. To find k for f(x, y) to be a joint probability density function, one must integrate f over the region where it's defined and set this equal to 1, the total probability.
Step-by-step explanation:
Non-Negativity of the Function
To show that f(x, y) is non-negative, we need to analyze the quadratic form inside the brackets. The function f(x, y) = k[x² + y² - xy - y + 3] is given to be non-negative when 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1. We simplifly the expression by completing the square in terms of x and y. Focusing on the square terms and the interaction term -xy, we notice that (x - ½y)² = x² - xy + ¼y². Therefore, we can rewrite the expression as:
f(x, y) = k[(x - ½y)² + ¾(y² - ½y²) - y + 3] = k[(x - ½y)²+ ¾½y² - y + 3].
Simplifying further, we get:
f(x, y) = k[(x - ½y)² + ¾½y² + (3 - y)].
Since the sum of squares is always non-negative, and (3 - y) is non-negative for y ≤ 1, f(x, y) ≥ 0.
Finding the Constant k
To find the value of k such that f(x, y) is a joint probability density function, we must ensure that the integral of f(x, y) over the region 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 is 1, since for continuous probability distributions, PROBABILITY = AREA. This means:
∫∫ f(x, y) dxdy = k ∫∫ [x² + y² - xy - y + 3] dxdy = 1, where the integration is taken over the square region [0,1]x[0,1].
Upon evaluating this double integral, we can solve for k and thus ensure the function is a valid probability density function.