Bill and Joe each put 10 into separate accounts at time t = 0, where t is measured in years. Bill’s account earns interest at a constant annual effective interest rate of K/25, K > 0.

Joe’s account earns interest at a force of interest, [math]\delta_t = \frac{1}{K + 0.25t}[/math]

At the end of four years, the amount in each account is X.

Calculate X.

• 20.7
• 21.7
• 22.7
• 23.7
• 24.7

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

Solution: A

Equating the accumulated values after 4 years provides an equation in K

[[math]] \begin{align*} 10(1+\frac{K}{25})^{4} &= 10\exp \left(\int_{0}^{*4}\frac{1}{K+0.25t}d t\right) \\ 4\ln(1+0.04K) &= \int_{0}^{4}\frac{1}{K+0.25t}d t=4\ln(K)+0.25t)\big|_{0}^{4} \\ &=4\ln(\operatorname{K}+0.25t)\big|_{0}^{4} \\ &=4\ln(K+1)-4\ln(K) = 4\ln\frac{K+1}{K} \\ & 1+0.04K = \frac{K+1}{K}\\ & 0.04K^2 = 1 \\ &K = 5 \end{align*} [[/math]]

Therefore [math]X = 10(1+5/25)^4=20.74.[/math]

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