Answer: The simple return is always less than the continuously compounded return.

## Explanation **Correct Answer: B** - The simple return is always less than the continuously compounded return. **Mathematical Relationship:** - Simple return: \( R_{simple} = \frac{P_t - P_{t-1}}{P_{t-1}} \) - Continuously compounded return (log return): \( r_{cc} = \ln(1 + R_{simple}) \) **Key Insight:** For any positive simple return \( R_{simple} > 0 \), the continuously compounded return \( r_{cc} \) is always less than \( R_{simple} \) because: \[ r_{cc} = \ln(1 + R_{simple}) < R_{simple} \] This follows from the inequality \( \ln(1+x) < x \) for \( x > 0 \). **Analysis of Other Options:** **A. Incorrect** - This is the opposite of the true relationship. The continuously compounded return is always less than the simple return, not greater. **C. Incorrect** - The approximation error actually increases as the simple return increases. For small returns, \( \ln(1+R) \approx R \), but as R grows larger, the difference becomes more significant. **D. Incorrect** - Simple returns have a lower bound of -100% (total loss), while continuously compounded returns can go below -100% (approaching negative infinity as the simple return approaches -100%).

Author: Nikitesh Somanthe