Revision as of 19:24, 19 November 2023 by Admin (Created page with "'''Solution: B''' Let X be the price of Bond X. Then, for the two bonds: <math display = "block"> \begin{array}{l}{{X=10,000(0.03)a_{\overline{{{2n}}}|0.035}+c(1.035)^{-2n}}}\\ {{X-969.52=10,000(0.025)a_{\overline{{{2n}}}|0.035}+(c+50)(1.035)^{-2n}}}\end{array} </math> Subtracting the second equation from the first gives <math display = "block"> \begin{array}{l}{{969.52=50a_{\overline{{{2n}}}|0.035}-50(1.035)^{-2n}}}\\ {{969.52=\frac{50}{0.035}[1-(1.035)^{-2n}]-50(1...")
Exercise
ABy Admin
Nov 19'23
Answer
Solution: B
Let X be the price of Bond X. Then, for the two bonds:
[[math]]
\begin{array}{l}{{X=10,000(0.03)a_{\overline{{{2n}}}|0.035}+c(1.035)^{-2n}}}\\ {{X-969.52=10,000(0.025)a_{\overline{{{2n}}}|0.035}+(c+50)(1.035)^{-2n}}}\end{array}
[[/math]]
Subtracting the second equation from the first gives
[[math]]
\begin{array}{l}{{969.52=50a_{\overline{{{2n}}}|0.035}-50(1.035)^{-2n}}}\\ {{969.52=\frac{50}{0.035}[1-(1.035)^{-2n}]-50(1.035)^{-2n}}}\\
{{1.035^{-2n}=459.05/1478.57=0.310469}}\\ {{m=-(0.5)\ln(0.310469)/\vert\ln(1.035)=17.}}
\end{array}
[[/math]]
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