Question 1172515
Let's break down this problem step by step.

**Given:**

* Joint density function:
    * F(x, y) = 8xy if 0 ≤ y ≤ x ≤ 1
    * F(x, y) = 0 elsewhere

**a) Compute P(X + Y < 1)**

1.  **Sketch the Region:**
    * The region defined by 0 ≤ y ≤ x ≤ 1 is a triangle in the first quadrant with vertices (0, 0), (1, 0), and (1, 1).
    * The condition X + Y < 1 implies Y < 1 - X. This is the region below the line Y = 1 - X.
    * We need to find the intersection of these two regions.
    * The intersection is a triangle with vertices (0, 0), (1, 0), and (1/2, 1/2).

2.  **Set up the Integral:**
    * P(X + Y < 1) = ∫∫ F(x, y) dy dx
    * The limits of integration are:
        * 0 ≤ x ≤ 1/2
        * 0 ≤ y ≤ 1 - x

3.  **Evaluate the Integral:**
    * P(X + Y < 1) = ∫(from 0 to 1/2) ∫(from 0 to 1-x) 8xy dy dx
    * P(X + Y < 1) = ∫(from 0 to 1/2) [4xy^2](from 0 to 1-x) dx
    * P(X + Y < 1) = ∫(from 0 to 1/2) 4x(1 - x)^2 dx
    * P(X + Y < 1) = ∫(from 0 to 1/2) 4x(1 - 2x + x^2) dx
    * P(X + Y < 1) = ∫(from 0 to 1/2) (4x - 8x^2 + 4x^3) dx
    * P(X + Y < 1) = [2x^2 - (8/3)x^3 + x^4](from 0 to 1/2)
    * P(X + Y < 1) = 2(1/4) - (8/3)(1/8) + (1/16)
    * P(X + Y < 1) = 1/2 - 1/3 + 1/16
    * P(X + Y < 1) = (24 - 16 + 3)/48 = 11/48

**b) Find the Marginal Densities fX(x) and fY(y)**

1.  **Marginal Density of X (fX(x)):**
    * fX(x) = ∫(from 0 to x) 8xy dy
    * fX(x) = [4xy^2](from 0 to x)
    * fX(x) = 4x^3, 0 ≤ x ≤ 1

2.  **Marginal Density of Y (fY(y)):**
    * fY(y) = ∫(from y to 1) 8xy dx
    * fY(y) = [4x^2y](from y to 1)
    * fY(y) = 4y - 4y^3, 0 ≤ y ≤ 1

**c) Find the Conditional Density fY|X(y|1/2)**

1.  **Conditional Density Formula:**
    * fY|X(y|x) = F(x, y) / fX(x)

2.  **Substitute x = 1/2:**
    * fY|X(y|1/2) = 8(1/2)y / 4(1/2)^3
    * fY|X(y|1/2) = 4y/(1/2)
    * fY|X(y|1/2) = 8y.
    * Since 0 <= y <= x, and x = 1/2, then 0<=y<=1/2.

3.  **Result:**
    * fY|X(y|1/2) = 8y, 0 ≤ y ≤ 1/2

**d) Find the Conditional Expectation E[Y|X=1/2]**

1.  **Conditional Expectation Formula:**
    * E[Y|X=1/2] = ∫ y * fY|X(y|1/2) dy

2.  **Substitute and Evaluate:**
    * E[Y|X=1/2] = ∫(from 0 to 1/2) y * 8y dy
    * E[Y|X=1/2] = ∫(from 0 to 1/2) 8y^2 dy
    * E[Y|X=1/2] = [(8/3)y^3](from 0 to 1/2)
    * E[Y|X=1/2] = (8/3) * (1/8)
    * E[Y|X=1/2] = 1/3

**Summary**

* a) P(X + Y < 1) = 11/48
* b) fX(x) = 4x^3, 0 ≤ x ≤ 1; fY(y) = 4y - 4y^3, 0 ≤ y ≤ 1
* c) fY|X(y|1/2) = 8y, 0 ≤ y ≤ 1/2
* d) E[Y|X=1/2] = 1/3