Question:
Let the random vector \( (X, Y) \) have the joint probability density function
\[f(x, y) = \begin{cases} \frac{1}{x}, & \text{if } 0 < y < x < 1 \\0, & \text{otherwise} \end{cases}.\]
Then \( \text{Cov}(X, Y) \) is equal to:
Let the random vector \( (X, Y) \) have the joint probability density function
\[f(x, y) = \begin{cases} \frac{1}{x}, & \text{if } 0 < y < x < 1 \\0, & \text{otherwise} \end{cases}.\]
Then \( \text{Cov}(X, Y) \) is equal to:
\[f(x, y) = \begin{cases} \frac{1}{x}, & \text{if } 0 < y < x < 1 \\0, & \text{otherwise} \end{cases}.\]
Then \( \text{Cov}(X, Y) \) is equal to:
Updated On: Oct 1, 2024
- \( \frac{1}{6} \)
- \( \frac{1}{12} \)
- \( \frac{1}{18} \)
- \( \frac{1}{24} \)
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The Correct Option is D
Solution and Explanation
The correct option is (D): \( \frac{1}{24} \)
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