Two random variables are such that `sigma_(X+Y)gtsigma_(X-Y)`, then r (X, Y) is positive.

Answer

To solve the problem, we need to analyze the relationship between the two random variables \(X\) and \(Y\) given the condition that \(\sigma_{X+Y} > \sigma_{X-Y}\). We want to determine if this implies that the correlation coefficient \(r(X, Y)\) is positive.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \(A = X + Y\) and \(B = X - Y\).
- We are given that \(\sigma_A > \sigma_B\), where \(\sigma_A\) and \(\sigma_B\) are the standard deviations of \(A\) and \(B\), respectively.

2. **Expressing Variances**:
- The variances can be expressed as:
\[ \sigma^2_A = \sigma^2_{X+Y} \quad \text{and} \quad \sigma^2_B = \sigma^2_{X-Y} \]
- The relationship can be stated as:
\[ \sigma^2_{X+Y} > \sigma^2_{X-Y} \]

3. **Using the Formulas for Variance**:
- The variance of the sum and difference of two random variables can be expressed as:
\[ \sigma^2_{X+Y} = \sigma^2_X + \sigma^2_Y + 2 \text{Cov}(X, Y) \]
\[ \sigma^2_{X-Y} = \sigma^2_X + \sigma^2_Y - 2 \text{Cov}(X, Y) \]

4. **Setting Up the Inequality**:
- From the above expressions, we can set up the inequality:
\[ \sigma^2_X + \sigma^2_Y + 2 \text{Cov}(X, Y) > \sigma^2_X + \sigma^2_Y - 2 \text{Cov}(X, Y) \]

5. **Simplifying the Inequality**:
- By simplifying the inequality, we get:
\[ 2 \text{Cov}(X, Y) > -2 \text{Cov}(X, Y) \]
- This simplifies to:
\[ 4 \text{Cov}(X, Y) > 0 \]
- Thus, we find that:
\[ \text{Cov}(X, Y) > 0 \]

6. **Relating Covariance to Correlation**:
- The correlation coefficient \(r(X, Y)\) is defined as:
\[ r(X, Y) = \frac{\text{Cov}(X, Y)}{\sigma_X \sigma_Y} \]
- Since \(\sigma_X\) and \(\sigma_Y\) are both positive (as they are standard deviations), the sign of \(r(X, Y)\) is determined by the sign of \(\text{Cov}(X, Y)\).

7. **Conclusion**:
- Since we have established that \(\text{Cov}(X, Y) > 0\), it follows that:
\[ r(X, Y) > 0 \]
- Therefore, the statement that if \(\sigma_{X+Y} > \sigma_{X-Y}\), then \(r(X, Y)\) is positive is **true**.