The lines of regression of y on x and x on y are respectively `y=x and 4x-y-3=0` and the second moment about the origin for x is 2, variance of y is

Options

A. 5
B. 4
C. 9
D. none of these

Answer

To solve the problem, we need to find the variance of \( y \) given the regression lines and the second moment about the origin for \( x \).

### Step-by-Step Solution:

1. **Identify the Regression Lines:**
- The regression line of \( y \) on \( x \) is given as \( y = x \).
- The regression line of \( x \) on \( y \) can be rearranged from \( 4x - y - 3 = 0 \) to \( y = 4x - 3 \).

2. **Determine the Regression Coefficients:**
- From the regression line \( y = x \), we can identify the regression coefficient \( b_{yx} \) (the slope of \( y \) on \( x \)):
\[ b_{yx} = 1 \]
- From the regression line \( y = 4x - 3 \), we can identify the regression coefficient \( b_{xy} \) (the slope of \( x \) on \( y \)):
\[ b_{xy} = \frac{1}{4} \]

3. **Calculate the Correlation Coefficient \( r \):**
- The relationship between the regression coefficients and the correlation coefficient is given by:
\[ r = \sqrt{b_{yx} \cdot b_{xy}} \]
- Substituting the values we found:
\[ r = \sqrt{1 \cdot \frac{1}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

4. **Use the Second Moment about the Origin:**
- The second moment about the origin for \( x \) is given as \( E(X^2) = 2 \).
- The variance of \( x \) can be calculated as:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 \]
- Since we do not have \( E(X) \), we will denote it as \( \mu_x \).

5. **Relate Variance of \( y \) to Variance of \( x \):**
- The formula relating the variances and correlation is:
\[ \text{Var}(Y) = r^2 \cdot \frac{\text{Var}(X)}{b_{xy}^2} \]
- We know \( r = \frac{1}{2} \) and \( b_{xy} = \frac{1}{4} \):
\[ r^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
- The variance of \( x \) can be calculated as:
\[ \text{Var}(X) = E(X^2) - (E(X))^2 = 2 - \mu_x^2 \]

6. **Substituting Values:**
- Now substituting into the variance formula for \( y \):
\[ \text{Var}(Y) = \frac{1}{4} \cdot \frac{2 - \mu_x^2}{\left(\frac{1}{4}\right)^2} = \frac{1}{4} \cdot \frac{2 - \mu_x^2}{\frac{1}{16}} = \frac{1}{4} \cdot 16(2 - \mu_x^2) = 4(2 - \mu_x^2) \]

7. **Finding Variance of \( y \):**
- Since we need the variance of \( y \) and we have \( \text{Var}(Y) = 4(2 - \mu_x^2) \), we need to find \( \mu_x^2 \) to finalize the variance of \( y \). However, without loss of generality, we can assume \( \mu_x = 0 \) for simplicity (as it does not affect the variance):
\[ \text{Var}(Y) = 4(2 - 0) = 8 \]

8. **Conclusion:**
- The variance of \( y \) is \( 8 \), which is not one of the provided options (5, 4, 9, none). Therefore, the answer is **none**.