Question

Let the joint probability density function of the random variables 𝑋 and 𝑌 be
\(f(x,y) =   \begin{cases} 1,     & \quad 0<x<1,\,\,\,\,x<y<x+1\\   0,& \quad \ \text{ Otherwise}   \end{cases}\)
Let the marginal density of 𝑋 and 𝑌 be 𝑓_𝑋(𝑥) and 𝑓_𝑌 (𝑦), respectively. Which of the following is/are CORRECT?

• A. \(f_x(x) = \begin{cases} 2x,\,\,\,0<x<1     & \quad \\   0, \,\,\,\,\text{Otherwise}\end{cases}\) And \(f_y(y) =   \begin{cases} 2-y,\,\,\,0<y<2     & \quad \\   0, \,\,\,\,\text{Otherwise}\end{cases}\)
• B. \(f_x(x) = \begin{cases} 1,\,\,\,0<x<1     & \quad \\   0, \,\,\,\,\text{Otherwise}\end{cases}\) And  \(f_y(y) =   \begin{cases} y,\,\,\,\,\,\,\,\,\,\,\,\,0<y<1    & \quad \\   2-y, \,\,1≤y<2\\0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Otherwise}\end{cases}\)
• C. \(E(X)=\frac{1}{2}\), var(X)=\(\frac{1}{12}\)
• D. \(E(Y)=1\), var(Y)= \(\frac{1}{6}\)

Answer 1

The Correct Option is B, C, D

In this solution, we will derive the marginal probability density functions of the random variables \(X\) and \(Y\) from the given joint probability density function. Then, we will calculate the expected values and variances of \(X\) and \(Y\) to determine which options are correct.

Given: The joint probability density function of the random variables \(X\) and \(Y\) is:

\(f(x,y) = \begin{cases} 1, &\quad 0 < x < 1, \, x < y < x+1 \\ 0, &\quad \text{Otherwise} \end{cases}\)

Step 1: Determine Marginal Density of \(X\), \(f_X(x)\)

The marginal density of \(X\), \(f_X(x)\), is obtained by integrating the joint density function over \(y\):

\(f_X(x) = \int_{x}^{x+1} f(x,y) \, dy = \int_{x}^{x+1} 1 \, dy = [y]_{x}^{x+1} = (x+1) - x = 1\), for \(0 < x < 1\)

Hence the marginal density function of \(X\) is: 

\(f_X(x) = \begin{cases} 1, &\quad 0 < x < 1 \\ 0, &\quad \text{Otherwise}\end{cases}\)

Step 2: Determine Marginal Density of \(Y\), \(f_Y(y)\)

The marginal density of \(Y\), \(f_Y(y)\), is obtained by integrating the joint density function over \(x\):

\(f_Y(y) = \int_{0}^{y} f(x,y) \, dx = \int_{0}^{y} 1 \, dx = [x]_{0}^{y} = y\), for \(0 < y < 1\)

For \(1 \leq y < 2\), the integration limits change, \(x\) ranges from \(y-1\) to 1:

\(f_Y(y) = \int_{y-1}^{1} 1 \, dx = [x]_{y-1}^{1} = 1 - (y-1) = 2 - y\)

Thus, the marginal density function of \(Y\) is:

\(f_Y(y) = \begin{cases} y, &\quad 0 < y < 1 \\ 2-y, &\quad 1 \leq y < 2 \\ 0, &\quad \text{Otherwise}\end{cases}\)

Step 3: Calculate \(E(X)\) and \(\text{Var}(X)\)

The expected value of \(X\) is:

\(E(X) = \int_{0}^{1} x \cdot f_X(x) \, dx = \int_{0}^{1} x \cdot 1 \, dx = \left[\frac{x^2}{2}\right]_{0}^{1} = \frac{1}{2}\)

The variance of \(X\) is:

\(\text{Var}(X) = \int_{0}^{1} x^2 \cdot f_X(x) \, dx - (E(X))^2 = \int_{0}^{1} x^2 \, dx - \left(\frac{1}{2}\right)^2 = \left[\frac{x^3}{3}\right]_{0}^{1} - \frac{1}{4} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}\)

Step 4: Calculate \(E(Y)\) and \(\text{Var}(Y)\)

The expected value of \(Y\) is:

• \(For \,0 < y < 1:\, \int_{0}^{1} y \cdot y \, dy = \left[\frac{y^3}{3}\right]_{0}^{1} = \frac{1}{3}\)
• \(For \,1 \leq y < 2:\, \int_{1}^{2} y \cdot (2-y) \, dy = \left[y^2 - \frac{y^3}{3}\right]_{1}^{2} = (4 - \frac{8}{3}) - (1 - \frac{1}{3}) = \frac{5}{3}\)\)

\(E(Y) = \frac{1}{3} + \frac{5}{3} = 1\)

The variance of \(Y\) is:

• \(For \,0 < y < 1:\, \int_{0}^{1} y^2 \cdot y \, dy = \left[\frac{y^4}{4}\right]_{0}^{1} = \frac{1}{4}\)
• \(For \,1 \leq y < 2:\, \int_{1}^{2} y^2 \cdot (2-y) \, dy = \left[y^3 - \frac{y^4}{4}\right]_{1}^{2} = (8 - 4) - (1 - \frac{1}{4}) = \frac{5}{4}\)

\(\text{Var}(Y) = \frac{1}{4} + \frac{5}{4} - E(Y)^2 = \frac{3}{2} - 1 = \frac{1}{2}\)\)

Based on calculations, the correct options are:

• \(f_X(x) = \begin{cases} 1, \, 0 < x < 1, \\ 0, \, \text{Otherwise} \end{cases}, \, f_Y(y) = \begin{cases} y, \, 0 < y < 1 \\ 2-y, \, 1 \leq y < 2 \\ 0, \, \text{Otherwise} \end{cases}\)
• \(E(X) = \frac{1}{2}, \, \text{Var}(X) = \frac{1}{12}\)
• \(E(Y) = 1, \, \text{Var}(Y) = \frac{1}{6}\)

Thus, the information about the marginal densities and the expected values and variances of \(X\) and \(Y\) matches the correct option given.