The present value of cash flows at an effective interest rate of i is given by the following function:

[[math]] P\left(i\right)=100+500\left(1+i\right)^{-3}-1000\left(1+i\right)^{-4},{}~{\mathrm{for}}\,i\gt0. [[/math]]

Determine which of the following expressions represents the modified duration of these cash flows.

• [[math]]\frac{1500(1+i)^{-4}-4000(1+i)^{-5}}{100+500(1+i)^{-3}-1000(1+i)^{-4}}[[/math]]
• [[math]]\frac{1500(1+i)^{-3}-4000(1+i)^{-4}}{100+500(1+i)^{-3}-1000(1+i)^{-4}}[[/math]]
• [[math]]\frac{-1500(1+i)^{-4}+4000(1+i)^{-5}}{100+500(1+i)^{-3}-1000(1+i)^{-4}}[[/math]]
• [[math]]\frac{-1500(1+i)^{-3}+4000(1+i)^{-4}}{100+500(1+i)^{-3}-1000(1+i)^{-4}}[[/math]]
• [[math]]\frac{-6000(1+i)^{-5}+20,000(1+i)^{-6}}{100+500(1+i)^{-3}-1000(1+i)^{-4}}[[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: A

[[math]] -\frac{P^{\prime}\left(i\right)}{P(i)}=\frac{1500\left(1+i\right)^{-4}-4000\left(1+i\right)^{-5}}{100+500\left(1+i\right)^{-4}-1000\left(1+i\right)^{-4}} [[/math]]

Copyright 2023 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.