Revision as of 21:56, 20 November 2023 by Admin (Created page with "Let <math>Y</math> indicate the nominal value of the two-year bond, then: <math>9465=\frac{0.05 Y}{1.08}+\frac{1.05 Y}{1.08^2}</math>, so <math>Y=10,000</math>. Thus, the amount of liability at the end of the second year is 10,500 . Hence, the liability at the end of the first year is: <math display = "block"> \frac{10,500}{2}=5250 </math> So, the amount invested in the one-year bond is: <math display = "block"> \frac{5250-10,000(0.05)}{1.06}=4481 \text {. } </math> {...")
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Exercise


Nov 20'23

Answer

Let [math]Y[/math] indicate the nominal value of the two-year bond, then: [math]9465=\frac{0.05 Y}{1.08}+\frac{1.05 Y}{1.08^2}[/math], so [math]Y=10,000[/math].

Thus, the amount of liability at the end of the second year is 10,500 . Hence, the liability at the end of the first year is:

[[math]] \frac{10,500}{2}=5250 [[/math]]

So, the amount invested in the one-year bond is:

[[math]] \frac{5250-10,000(0.05)}{1.06}=4481 \text {. } [[/math]]

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

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