If `U=(X-a)/(h) and V=(Y-b)/(k)` then `b_(UV)=b_(XY)`

Answer

To solve the problem, we need to prove whether the statement \( b_{UV} = b_{XY} \) is true or false given the transformations \( U = \frac{X - a}{h} \) and \( V = \frac{Y - b}{k} \).

### Step-by-Step Solution:

1. **Understanding the Transformations**:
- We have \( U = \frac{X - a}{h} \) and \( V = \frac{Y - b}{k} \).
- Here, \( a \) and \( b \) are constants that shift the origin, and \( h \) and \( k \) are constants that scale the variables.

2. **Correlation Coefficient Property**:
- The correlation coefficient \( r \) is invariant under linear transformations of the form \( U = aX + b \) and \( V = cY + d \). This means that the correlation between \( U \) and \( V \) remains the same as that between \( X \) and \( Y \).
- Therefore, we can say that \( r_{UV} = r_{XY} \).

3. **Calculating the Standard Deviations**:
- The standard deviation of \( U \) can be calculated as:
\[ \sigma_U = \frac{\sigma_X}{h} \]
- The standard deviation of \( V \) can be calculated as:
\[ \sigma_V = \frac{\sigma_Y}{k} \]

4. **Calculating the Covariance**:
- The covariance \( b_{UV} \) can be expressed in terms of the correlation coefficient:
\[ b_{UV} = r_{UV} \cdot \frac{\sigma_U}{\sigma_V} \]
- Substituting the values we have:
\[ b_{UV} = r_{XY} \cdot \frac{\frac{\sigma_X}{h}}{\frac{\sigma_Y}{k}} = r_{XY} \cdot \frac{k \sigma_X}{h \sigma_Y} \]

5. **Comparing with \( b_{XY} \)**:
- The covariance \( b_{XY} \) is given by:
\[ b_{XY} = r_{XY} \cdot \frac{\sigma_X}{\sigma_Y} \]
- Now we can compare \( b_{UV} \) and \( b_{XY} \):
\[ b_{UV} = r_{XY} \cdot \frac{k \sigma_X}{h \sigma_Y} \quad \text{and} \quad b_{XY} = r_{XY} \cdot \frac{\sigma_X}{\sigma_Y} \]
- From this, we can see that:
\[ b_{UV} = \frac{k}{h} \cdot b_{XY} \]

6. **Conclusion**:
- Since \( b_{UV} \) is not equal to \( b_{XY} \) unless \( k = h \), we conclude that the statement \( b_{UV} = b_{XY} \) is **false**.