Under exponential interpolation, if he $t _ { i }$ -tear yield is $y \left( t _ { i } \right)$ , $i = 1,2$ , then the interpolated yield for $t$ lying between $t _ { 1 }$ and $t _ { 2 }$ is given by $y ( t ) = y \left( t _ { 1 } \right) e ^ { x \cdot \left( t - t _ { 1 } \right) }$ where $x$ is a parameter. Suppose the yield at one year is 4% and the yield at two years is 5%. Then, the closest yield at one and a half years, using exponential interpolation in time $t$ , is
A) 4.47%
B) 4.50%
C) 4.53%
D) 4.55%

Question

Quizplus · Accepted Answer

To find the interpolated yield at 1.5 years, we first calculate the parameter $x$ using the formula $x = \frac{\ln(y(t_2)/y(t_1))}{t_2 - t_1}$. Substituting the given yields, $x = \frac{\ln(5/4)}{2-1} = \ln(1.25)$. Then, the yield at 1.5 years is $y(1.5) = 4 \times e^{\ln(1.25) \times (1.5-1)}$. Calculating this gives approximately 4.47%.

Under Exponential Interpolation, If He tit _ { i }ti​ -Tear Yield Is

Under Exponential Interpolation, If He $t _ { i }$ -Tear Yield Is