1. Assume that \( x \) and \( y \) are two random variables having the joint pdf \[ f(x, y)=\left\{\begin{array}{ll}\frac{2 x+y}{4} & 0

Ask by Cole Lynch. in Nigeria

Feb 21,2025

We are given the joint probability density function (pdf) for x and y as   f(x, y) = (2x + y) / 4,   for 0 < x < 1 and 0 < y < 2,   f(x, y) = 0,        otherwise. We want to find   P(x + y ≤ 1) = ∬[region where x + y ≤ 1] f(x, y) dx dy. Step 1. Determine the region of integration Since x and y are defined in the rectangle 0 < x < 1 and 0 < y < 2, the condition x + y ≤ 1 limits the integration to the portion of that rectangle where x + y ≤ 1. For a given x, the inequality becomes:   y ≤ 1 - x. Notice that when x is in (0, 1), 1 - x is always less than or equal to 1. Therefore, since y can go up to 2 but the condition demands y ≤ 1 - x, the limits for y will be from 0 to 1 - x. Thus, the region of integration is:   x from 0 to 1,   y from 0 to 1 - x. Step 2. Set Up the Integral The probability is given by   P(x + y ≤ 1) = ∫[x = 0]¹ ∫[y = 0]^(1-x) (2x + y) / 4 dy dx. Step 3. Evaluate the Inner Integral (with respect to y) First, factor out the constant 1/4:   P(x + y ≤ 1) = (1/4) ∫[0]¹ { ∫[0]^(1-x) (2x + y) dy } dx. Now, compute the inner integral for a fixed x:   ∫[0]^(1-x) (2x + y) dy = ∫[0]^(1-x) 2x dy + ∫[0]^(1-x) y dy. Since x is treated as a constant with respect to y: • ∫[0]^(1-x) 2x dy = 2x * (1 - x). • ∫[0]^(1-x) y dy = [y²/2] evaluated from 0 to (1 - x) = (1 - x)² / 2. Thus, the inner integral becomes:   2x(1 - x) + (1 - x)² / 2. Step 4. Plug Back into the Outer Integral Now, substitute back and write the integral with respect to x:   P(x + y ≤ 1) = (1/4) ∫[0]¹ { 2x(1 - x) + (1 - x)² / 2 } dx. Step 5. Simplify the Integrand First, expand and simplify term-by-term.   2x(1 - x) = 2x - 2x².   (1 - x)² / 2 = (1 - 2x + x²) / 2 = 1/2 - x + x²/2. Combine the two results:   2x - 2x² + 1/2 - x + x²/2 = (2x - x) + ( - 2x² + x²/2) + 1/2              = x + 1/2 - (2x² - x²/2)              = x + 1/2 - ( (4x²/2) - (x²/2) )              = x + 1/2 - (3x²/2). Thus, the integrand is:   x + 1/2 - (3/2)x². Step 6. Perform the Outer Integration We now integrate from x = 0 to 1:   ∫[0]¹ { x + 1/2 - (3/2)x² } dx. Compute each term: 1. ∫[0]¹ x dx = [x²/2] from 0 to 1 = 1/2. 2. ∫[0]¹ (1/2) dx = 1/2 * (1 - 0) = 1/2. 3. ∫[0]¹ (3/2)x² dx = (3/2) * [x³/3] from 0 to 1 = (3/2) * (1/3) = 1/2. Now combine:   1/2 + 1/2 - 1/2 = 1/2. Thus, the outer integral equals 1/2. Step 7. Multiply by the constant outside Finally:   P(x + y ≤ 1) = (1/4) * (1/2) = 1/8. Answer: The probability P(x + y ≤ 1) is 1/8.