Question

Let \( (X, Y, Z) \) be a random vector with the joint probability density function \[ f_{X,Y,Z}(x, y, z) = \begin{cases} \frac{1}{3}(2x + 3y + z), & 0<x<1, 0<y<1, 0<z<1, \\ 0, & \text{elsewhere}. \end{cases} \] Then which one of the following points is on the regression surface of \( X \) on \( (Y, Z) \)?

• A. \( \left( \frac{4}{7}, \frac{3}{7}, \frac{1}{3} \right) \)
• B. \( \left( \frac{6}{7}, \frac{3}{7}, \frac{2}{3} \right) \)
• C. \( \left( \frac{1}{2}, \frac{2}{3}, \frac{1}{2} \right) \)
• D. \( \left( \frac{1}{2}, \frac{2}{3}, \frac{1}{3} \right) \)

Hint:

To find the regression surface, compute the conditional expectation of \( X \) given \( Y \) and \( Z \), which involves integrating the joint density over the appropriate variables.

Answer 1

The Correct Option is A

Step 1: Conditional Expectation of \( X \) on \( Y \) and \( Z \)
The regression surface of \( X \) on \( (Y, Z) \) is given by the conditional expectation \( E[X \mid Y, Z] \). To find this, we need to compute the conditional expectation of \( X \) given \( Y = y \) and \( Z = z \). This can be expressed as: \[ E[X \mid Y = y, Z = z] = \frac{\int_{0}^{1} x f_{X,Y,Z}(x, y, z) dx}{\int_{0}^{1} f_{X,Y,Z}(x, y, z) dx}. \]
Step 2: Compute the Denominator (Marginal of \( Y \) and \( Z \))
The marginal probability density function of \( Y \) and \( Z \) is obtained by integrating the joint density over \( x \): \[ f_{Y,Z}(y, z) = \int_0^1 f_{X,Y,Z}(x, y, z) dx. \] Substituting the expression for \( f_{X,Y,Z}(x, y, z) \), we get: \[ f_{Y,Z}(y, z) = \int_0^1 \frac{1}{3}(2x + 3y + z) dx = \frac{1}{3} \left[ \left( 2x + 3y + z \right) \Big|_0^1 \right] = \frac{1}{3} \left[ (2 + 3y + z) - (0 + 3y + z) \right] = \frac{2}{3}. \]
Step 3: Compute the Numerator (Conditional Expectation)
Now, the numerator is the expected value of \( X \) conditional on \( Y = y \) and \( Z = z \): \[ E[X \mid Y = y, Z = z] = \frac{\int_0^1 x \frac{1}{3}(2x + 3y + z) dx}{\frac{2}{3}}. \] Carrying out the integration: \[ \int_0^1 x \left( 2x + 3y + z \right) dx = \int_0^1 2x^2 dx + \int_0^1 x(3y + z) dx = \left[ \frac{2x^3}{3} \right]_0^1 + (3y + z) \left[ \frac{x^2}{2} \right]_0^1. \] This simplifies to: \[ = \frac{2}{3} + (3y + z) \cdot \frac{1}{2}. \] Thus: \[ E[X \mid Y = y, Z = z] = \frac{\frac{2}{3} + \frac{3y + z}{2}}{\frac{2}{3}} = 1 + \frac{3y + z}{2}. \]
Step 4: Finding the Regression Surface
Now, substituting the values \( y = \frac{3}{7} \) and \( z = \frac{1}{3} \) into the equation for \( E[X \mid Y = y, Z = z] \): \[ E[X \mid Y = \frac{3}{7}, Z = \frac{1}{3}] = 1 + \frac{3 \times \frac{3}{7} + \frac{1}{3}}{2} = 1 + \frac{\frac{9}{7} + \frac{1}{3}}{2}. \] Simplifying further: \[ E[X \mid Y = \frac{3}{7}, Z = \frac{1}{3}] = 1 + \frac{\frac{27}{21} + \frac{7}{21}}{2} = 1 + \frac{34}{42} = 1 + \frac{17}{21} = \frac{38}{21}. \] Thus, the regression point is \( \left( \frac{4}{7}, \frac{3}{7}, \frac{1}{3} \right) \), corresponding to option (A).