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i. Show that \(d^{(p)}\), the nominal rate of discount payable \(p\) times a year, can be written as \(d^{(p)}=p\left(1-e^{-\delta / p}\right)\), where \(\delta\) is the force of interest.

ii. Starting from the equation derived in i., show that \(\lim _{p \rightarrow \infty} d^{(p)}=\delta\).
in Data Science & Statistics by Platinum (119,120 points) | 100 views

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&\left(1-\frac{d^{(p)}}{p}\right)^{p}=v=e^{-\delta} \Rightarrow\left(1-\frac{d^{(p)}}{p}\right)=e^{-\delta / p} \\
&\frac{d^{(p)}}{p}=1-e^{-\delta / p} \Rightarrow d^{(p)}=p\left(1-e^{-\delta i p}\right)
&\lim _{p \rightarrow \infty} d^{(p)}=\lim _{p \rightarrow \infty}\left[p\left(1-e^{-\delta / p}\right)\right]=\lim _{p \rightarrow \infty}\left[p\left(1-\left(1-\frac{\delta}{p}+\frac{\delta^{2}}{2 ! p^{2}}-\ldots\right)\right)\right] \\
&=\lim _{p \rightarrow \infty}\left[p\left(\frac{\delta}{p}+\frac{\delta^{2}}{2 ! p^{2}}-\ldots\right)\right]=\lim _{p \rightarrow \infty}\left[\delta-\frac{\delta^{2}}{2 ! p}+\frac{\delta^{3}}{3 ! p^{2}} \ldots\right]=\delta
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