Let's solve each part step by step. --- ## 1. Find the value of \( k \) The joint pdf must integrate to 1 over the region: \[ \int_^1 \int_^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculate the integral: \[ \int_^1 \int_^1 (x^2 + y^2) \, dx \, dy = \int_^1 \left[ \int_^1 x^2 dx + \int_^1 y^2 dx \right] dy \] \[ = \int_^1 \left[ \frac1}{3} + y^2 \cdot (1 - ) \right] dy = \int_^1 \left( \frac{1}{3} + y^2 \right) dy \] \[ = \int_^1 \frac{1}{3} dy + \int_^1 y^2 dy = \frac{1}{3} \cdot 1 + \frac{1}{3} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] So, \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Find \( P(X + Y \leq 1) \) We want: \[ P(X + Y \leq 1) = \int_^1 \int_^{1-y} k(x^2 + y^2) dx \, dy \] Substitute \( k = \frac{3}{2} \): \[ = \int_^1 \int_^{1-y} \frac{3}{2}(x^2 + y^2) dx \, dy \] \[ = \frac{3}{2} \int_^1 \left[ \int_^{1-y} x^2 dx + \int_^{1-y} y^2 dx \right] dy \] \[ = \frac{3}{2} \int_^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] \[ = \frac{3}{2} \int_^1 \left[ \frac{(1-y)^3}{3} + y^2 - y^3 \right] dy \] Now, expand and integrate: - \(\int_^1 (1-y)^3 dy = \int_^1 (1 - 3y + 3y^2 - y^3) dy = [y - \frac{3}{2}y^2 + y^3 - \frac{1}{4}y^4]_^1 = 1 - \frac{3}{2} + 1 - \frac{1}{4} = \frac{1}{4}\) - \(\int_^1 y^2 dy = \frac{1}{3}\) - \(\int_^1 y^3 dy = \frac{1}{4}\) So, \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \left[ \frac{2}{12} \right] = \frac{3}{2} \cdot \frac{1}{6} = \frac{1}{4} \] --- ## 3. Find \( E[X] \) and \( E[Y] \) \[ E[X] = \int_^1 \int_^1 x f(x,y) dx\,dy = \int_^1 \int_^1 x \cdot \frac{3}{2}(x^2 + y^2) dx\,dy \] \[ = \frac{3}{2} \int_^1 \int_^1 (x^3 + x y^2) dx\,dy \] \[ = \frac{3}{2} \int_^1 \left[ \int_^1 x^3 dx + \int_^1 x dx \cdot y^2 \right] dy \] \[ = \frac{3}{2} \int_^1 \left[ \frac{1}{4} + \frac{1}{2} y^2 \right] dy = \frac{3}{2} \left[ \frac{1}{4} \cdot 1 + \frac{1}{2} \int_^1 y^2 dy \right] = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{2} \cdot \frac{1}{3} \right] = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry, \( E[Y] \) will be the same: \[ E[Y] = \frac{5}{8} \] --- ## **Summary of Answers** 1. **\( k = \frac{3}{2} \)** 2. **\( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **\( E[X] = E[Y] = \frac{5}{8} \)**

# Given Information Let the joint probability density function (pdf) of random variables \( X \) and \( Y \) be defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant to be determined. # Objectives 1. Find the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Determine \( E[X] \) and \( E[Y] \). --- ## 1. Find the value of \( k \) To ensure the joint pdf integrates to 1, we compute: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculate the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy \] This can be split into two parts: \[ = \int_0^1 \left[ \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dx \right] dy \] Calculating each inner integral: \[ \int_0^1 x^2 \, dx = \frac{1}{3}, \quad \text{and} \quad \int_0^1 y^2 \, dx = y^2 \cdot 1 = y^2 \] Thus, \[ = \int_0^1 \left( \frac{1}{3} + y^2 \right) \, dy = \frac{1}{3} + \int_0^1 y^2 \, dy = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Setting up the equation: \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Find \( P(X + Y < 1) \) We need to calculate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Splitting the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Putting it together: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Integrating: - \(\int_0^1 (1-y)^3 \, dy = \frac{1}{4}\) - \(\int_0^1 y^2 \, dy = \frac{1}{3}\) - \(\int_0^1 y^3 \, dy = \frac{1}{4}\) Final calculation: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{2}{12} \right] = \frac{1}{4} \] --- ## 3. Find \( E[X] \) and \( E[Y] \) For \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] Calculating: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{1}{2} y^2 \right] dy \] Final computation: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **\( k = \frac{3}{2} \)** 2. **\( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **\( E[X] = E[Y] = \frac{5}{8} \)**

# Given Information Let the joint probability density function (pdf) of random variables \( X \) and \( Y \) be defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant to be determined. # Objectives 1. Find the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Determine \( E[X] \) and \( E[Y] \). --- ## 1. Find the value of \( k \) To ensure the joint pdf integrates to 1, we compute: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy \] This can be split into two parts: \[ = \int_0^1 \left[ \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dx \right] dy \] Calculating each inner integral: \[ \int_0^1 x^2 \, dx = \frac{1}{3}, \quad \text{and} \quad \int_0^1 y^2 \, dx = y^2 \cdot 1 = y^2 \] Thus, \[ = \int_0^1 \left( \frac{1}{3} + y^2 \right) \, dy = \frac{1}{3} + \int_0^1 y^2 \, dy = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Setting up the equation: \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Find \( P(X + Y < 1) \) We need to calculate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Splitting the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Putting it together: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Integrating: - \(\int_0^1 (1-y)^3 \, dy = \frac{1}{4}\) - \(\int_0^1 y^2 \, dy = \frac{1}{3}\) - \(\int_0^1 y^3 \, dy = \frac{1}{4}\) Final calculation: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{2}{12} \right] = \frac{1}{4} \] --- ## 3. Find \( E[X] \) and \( E[Y] \) For \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] Calculating: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{1}{2} y^2 \right] dy \] Final computation: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **\( k = \frac{3}{2} \)** 2. **\( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **\( E[X] = E[Y] = \frac{5}{8} \)**

# Given Information Let the joint probability density function (pdf) of random variables \( X \) and \( Y \) be defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant to be determined. # Objectives 1. Find the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Determine \( E[X] \) and \( E[Y] \). --- ## 1. Find the value of \( k \) To ensure the joint pdf integrates to 1, we compute: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral involves evaluating: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy \] This integral can be separated into two parts: \[ = \int_0^1 \left[ \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dx = y^2 \cdot 1 = y^2 \] Putting this together: \[ = \int_0^1 \left( \frac{1}{3} + y^2 \right) \, dy = \frac{1}{3} + \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Setting up the equation for \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Find \( P(X + Y < 1) \) To find this probability, we compute: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] We now split the inner integral into two parts: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives us: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now integrating each term separately: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] - **Third term**: \[ \int_0^1 y^3 \, dy = \frac{1}{4} \] Putting all integrals together, we have: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{2}{12} \right] = \frac{1}{4} \] --- ## 3. Find \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] Calculating this gives: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives us: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Now we integrate: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry, since \( f(x,y) \) is symmetric in \( x \) and \( y \): \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values \( E[X] = E[Y] = \frac{5}{8} \)** This structured approach ensures clarity and thoroughness in deriving the results for the joint pdf and its properties, demonstrating a solid understanding of the concepts involved in probability theory and expectation calculations.

# Given Information Let the joint probability density function (pdf) of random variables \( X \) and \( Y \) be defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that we need to determine. # Objectives 1. Find the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Determine \( E[X] \) and \( E[Y] \). --- ## 1. Find the value of \( k \) To ensure that the joint pdf integrates to 1, we compute: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy \] This can be separated into two parts: \[ = \int_0^1 \left[ \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dx = y^2 \cdot 1 = y^2 \] Combining these results gives: \[ = \int_0^1 \left( \frac{1}{3} + y^2 \right) \, dy = \frac{1}{3} + \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Setting up the equation for \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Find \( P(X + Y < 1) \) To find this probability, compute: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Splitting the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we integrate each term separately: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Combining these results leads to: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{2}{12} \right] = \frac{1}{4} \] --- ## 3. Find \( E[X] \) and \( E[Y] \) To compute \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] Calculating this gives: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Now we integrate: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry, since \( f(x,y) \) is symmetric in \( x \) and \( y \): \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values \( E[X] = E[Y] = \frac{5}{8} \)** This structured approach confirms the calculations and illustrates the properties of the joint pdf and expected values, ensuring clarity and accuracy in understanding the behavior of the random variables involved.

# Given Information Consider the joint probability density function (pdf) of random variables \( X \) and \( Y \) defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant to be determined. # Objectives 1. Find the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Determine \( E[X] \) and \( E[Y] \). --- ## 1. Find the value of \( k \) The joint pdf must integrate to 1 over the allowed region, which can be expressed as: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy \] We can separate this integral into two parts: \[ = \int_0^1 \left[ \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dx = y^2 \cdot 1 = y^2 \] Combining these results: \[ = \int_0^1 \left( \frac{1}{3} + y^2 \right) \, dy = \frac{1}{3} + \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Setting the equation for \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Find \( P(X + Y < 1) \) To compute this probability, we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Splitting the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results leads to: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we integrate each term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Calculating the combined result: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{2}{12} \right] = \frac{1}{4} \] --- ## 3. Find \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] Calculating this gives: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these results: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating gives: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry, since \( f(x,y) \) is symmetric in \( x \) and \( y \): \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values \( E[X] = E[Y] = \frac{5}{8} \)** This comprehensive approach confirms the calculations and enhances understanding of the behavior of the joint pdf and the expected values of the random variables involved.

# Given Information We are given the joint probability density function (pdf) of random variables \( X \) and \( Y \) defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral, we can break it down: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dx \right) dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dx = y^2 \cdot 1 = y^2 \] Putting this together: \[ = \int_0^1 \left( \frac{1}{3} + y^2 \right) dy = \frac{1}{3} + \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To find the probability \( P(X + Y \leq 1) \), we need to evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] We can separate the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Combining these results: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{2}{12} \right] = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution provides a clear, step-by-step calculation of the joint pdf properties, ensuring a thorough understanding of the random variables involved.

# Given Information We are given the joint probability density function (pdf) of random variables \( X \) and \( Y \) defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that we need to determine. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral can be broken down as follows: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To find the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] We can separate the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Calculating the combined result: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] \] \[ = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution provides a thorough and structured analysis of the joint pdf properties, ensuring clarity and understanding of the concepts involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the region \( (0 < x < 1, 0 < y < 1) \), we set up the following equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To find the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution clearly demonstrates the calculations and properties of the joint pdf, ensuring a comprehensive understanding of the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that we need to determine. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral, we can separate it as follows: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Hence, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To find the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry, since the function \( f(x, y) \) is symmetric in \( x \) and \( y \): \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution provides a clear and detailed analysis of the joint pdf properties, ensuring a comprehensive understanding of the behavior of the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the following equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To find the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution presents a comprehensive approach to understanding the properties of the joint pdf and the expected values of the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that we need to determine. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To find the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution provides a detailed and clear analysis of the joint pdf properties, ensuring a thorough understanding of the behavior of the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is given by: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Find the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Find the Value of \( k \) To ensure that the joint pdf integrates to 1 over the valid region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Adding these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution effectively outlines the calculations and properties of the joint pdf, ensuring a comprehensive understanding of the behavior of the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that we need to determine. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral involves: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To find the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution systematically outlines the calculations and properties of the joint pdf, providing a comprehensive understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the valid region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution effectively outlines the calculations and properties of the joint pdf, ensuring a comprehensive understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that we need to determine. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure the joint pdf integrates to 1 over the defined region, we set up the following equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral, we can express it as: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution effectively outlines the calculations and properties of the joint pdf, ensuring a comprehensive understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution provides a comprehensive outline of the calculations and properties of the joint pdf, ensuring a thorough understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the following equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral involves: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution systematically outlines the calculations and properties of the joint pdf, providing a comprehensive understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution clearly outlines the calculations and properties of the joint pdf, providing a comprehensive understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution provides a comprehensive outline of the calculations and properties of the joint pdf, ensuring a thorough understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral involves: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution thoroughly outlines the calculations and properties of the joint pdf, ensuring a complete understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral gives: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution systematically outlines the calculations and properties of the joint pdf, providing a comprehensive understanding of the relationships between the random variables involved.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution provides a thorough examination of the joint pdf, ensuring a comprehensive understanding of the relationships between the random variables.

# Given Information The joint probability density function (pdf) of random variables \( X \) and \( Y \) is defined as: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that needs to be determined. # Objectives 1. Determine the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute \( E[X] \) and \( E[Y] \). --- ## 1. Determine the Value of \( k \) To ensure that the joint pdf integrates to 1 over the defined region, we set up the following equation: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] Calculating the integral gives: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. **For \( y^2 \)**: \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Combining these results gives: \[ = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] Thus, we find \( k \): \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) To calculate the probability \( P(X + Y \leq 1) \), we evaluate: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Separating the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. **For \( x^2 \)**: \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. **For \( y^2 \)**: \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these results gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now we compute each integral term: - **First term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Putting it together: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) To calculate \( E[X] \): \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Integrating yields: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution presents a clear examination of the calculations and properties of the joint pdf, enhancing understanding of the relationships between the random variables involved.

# Given Information We have the joint probability density function (pdf) of the random variables \( X \) and \( Y \) defined as follows: \[ f(x,y) = k(x^2 + y^2), \quad 0 < x < 1, \quad 0 < y < 1 \] where \( k \) is a constant that must be determined to ensure that the pdf integrates to 1 over the valid region. # Objectives 1. Find the value of \( k \). 2. Calculate \( P(X + Y < 1) \). 3. Compute the expected values \( E[X] \) and \( E[Y] \). --- ## 1. Find the Value of \( k \) For a function to be a valid probability density function, it must satisfy the condition that the total probability integrates to 1 over its defined range. Thus, we need to solve the following integral: \[ \int_0^1 \int_0^1 k(x^2 + y^2) \, dx \, dy = 1 \] First, we calculate the double integral without \( k \): \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \int_0^1 \left( \int_0^1 x^2 \, dx + \int_0^1 y^2 \, dy \right) \] Calculating the inner integrals: 1. For \( x^2 \): \[ \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] 2. For \( y^2 \): \[ \int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1}{3} \] Adding these results together gives: \[ \int_0^1 \int_0^1 (x^2 + y^2) \, dx \, dy = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \] To find \( k \), we set up the equation: \[ k \cdot \frac{2}{3} = 1 \implies k = \frac{3}{2} \] --- ## 2. Calculate \( P(X + Y < 1) \) Next, we need to compute the probability \( P(X + Y < 1) \). This can be expressed as: \[ P(X + Y \leq 1) = \int_0^1 \int_0^{1-y} k(x^2 + y^2) \, dx \, dy \] Substituting \( k = \frac{3}{2} \): \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} (x^2 + y^2) \, dx \right] dy \] Now we can separate the inner integral: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^{1-y} x^2 \, dx + \int_0^{1-y} y^2 \, dx \right] dy \] Calculating the inner integrals: 1. For \( x^2 \): \[ \int_0^{1-y} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^{1-y} = \frac{(1-y)^3}{3} \] 2. For \( y^2 \): \[ \int_0^{1-y} y^2 \, dx = y^2(1-y) \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{(1-y)^3}{3} + y^2(1-y) \right] dy \] Now, we will compute each integral term separately: - **Calculating the first term**: \[ \int_0^1 (1-y)^3 \, dy = \left[ \frac{(1-y)^4}{4} \right]_0^1 = \frac{1}{4} \] - **Calculating the second term**: \[ \int_0^1 y^2 \, dy = \frac{1}{3} \] Thus, combining these integrals results in: \[ = \frac{3}{2} \left[ \frac{1}{3} \cdot \frac{1}{4} + \frac{1}{3} - \frac{1}{4} \right] = \frac{3}{2} \left[ \frac{1}{12} + \frac{4}{12} - \frac{3}{12} \right] = \frac{3}{2} \cdot \frac{2}{12} = \frac{1}{4} \] --- ## 3. Compute \( E[X] \) and \( E[Y] \) The expected value of \( X \) can be calculated as follows: \[ E[X] = \int_0^1 \int_0^1 x f(x,y) \, dx \, dy = \int_0^1 \int_0^1 x \cdot \frac{3}{2}(x^2 + y^2) \, dx \, dy \] This simplifies to: \[ = \frac{3}{2} \int_0^1 \left[ \int_0^1 (x^3 + xy^2) \, dx \right] dy \] Evaluating the inner integrals step by step: 1. **For \( x^3 \)**: \[ \int_0^1 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^1 = \frac{1}{4} \] 2. **For \( xy^2 \)**: \[ \int_0^1 xy^2 \, dx = y^2 \cdot \int_0^1 x \, dx = y^2 \cdot \left[ \frac{x^2}{2} \right]_0^1 = \frac{y^2}{2} \] Combining these gives: \[ = \frac{3}{2} \int_0^1 \left[ \frac{1}{4} + \frac{y^2}{2} \right] dy \] Now, we compute the integral: \[ = \frac{3}{2} \left[ \frac{1}{4} + \frac{1}{6} \right] = \frac{3}{2} \cdot \frac{5}{12} = \frac{15}{24} = \frac{5}{8} \] By symmetry of the joint pdf, we can conclude that: \[ E[Y] = \frac{5}{8} \] --- ## Summary of Answers 1. **Value of \( k = \frac{3}{2} \)** 2. **Probability \( P(X + Y \leq 1) = \frac{1}{4} \)** 3. **Expected values: \( E[X] = E[Y] = \frac{5}{8} \)** This solution systematically addresses the calculations and properties of the joint pdf, providing a comprehensive understanding of the relationships between the random variables involved, while maintaining clarity and precision throughout the derivation.