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

Ask by Hanson Rose. in Nigeria

Feb 21,2025

We are given that x and y have the joint probability density function   f(x, y) = (2x + y)/4   for 0 < x < 1 and 0 < y < 2, and f(x, y) = 0 otherwise. We want to find P(x + y ≤ 1). Step 1. Identify the Region of Integration The inequality x + y ≤ 1, together with the boundaries 0 < x < 1 and 0 < y < 2, defines the region of integration. Notice that for a fixed x in [0, 1], y must satisfy   0 < y ≤ 1 − x. Thus, the integration limits can be set as:   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+y ≤ 1] f(x, y) dy dx. Substitute the limits:   P(x + y ≤ 1) = ∫₀¹ [∫₀^(1−x) (2x + y)/4 dy] dx. Step 3. Compute the Inner Integral (with respect to y) First, factor out the constant 1/4 from the inner integral:   = 1/4 ∫₀^(1−x) (2x + y) dy. Separate the integral:   = 1/4 [2x∫₀^(1−x) dy + ∫₀^(1−x) y dy]. Now, compute each part:   • ∫₀^(1−x) dy = (1 − x),   • ∫₀^(1−x) y dy = ½(1 − x)². Thus, the inner integral becomes:   1/4 [2x(1 − x) + ½(1 − x)²]. Step 4. Compute the Outer Integral (with respect to x) Now, the probability is   P(x + y ≤ 1) = 1/4 ∫₀¹ [2x(1 − x) + ½(1 − x)²] dx. It is often easier to combine the terms. Factor out (1 − x) from the expression inside the integral:   2x(1 − x) + ½(1 − x)² = (1 − x)[2x + ½(1 − x)]. For simplicity, we can leave it as is and integrate term by term. However, let’s first combine the terms inside the brackets on a common denominator:   = 1/4 ∫₀¹ [2x(1 − x) + ½(1 − x)²] dx     = 1/4 ∫₀¹ [(4x(1 − x) + (1 − x)²)/2] dx         [since 2x = (4x)/2]     = 1/8 ∫₀¹ [(1 − x)(4x + (1 − x))] dx. Simplify 4x + (1 − x) = 3x + 1, so the integrand becomes:   = 1/8 ∫₀¹ (1 − x)(3x + 1) dx. Expand the integrand:   (1 − x)(3x + 1) = 3x + 1 − 3x² − x = 1 + 2x − 3x². Thus, the probability is   P(x + y ≤ 1) = 1/8 ∫₀¹ (1 + 2x − 3x²) dx. Step 5. Evaluate the Integral Now, integrate term by term:   ∫₀¹ 1 dx = [x]₀¹ = 1,   ∫₀¹ 2x dx = [x²]₀¹ = 1,   ∫₀¹ 3x² dx = [x³]₀¹ = 1. So, the combined integral is:   1 + 1 − 1 = 1. Therefore,   P(x + y ≤ 1) = 1/8 × 1 = 1/8. Final Answer:   P(x + y ≤ 1) = 1/8.