Annuity formula proof $\frac{a_{\overline{n}|}}{a_{\overline{k}|}}$

Question

I have the actuarial exam FM in 2 days and there is one more thing that I would like to understand.

I cam across a problem having to do with identities and this is the following.

A perpetuity paying $50 on the last day of each year was purchased on January 1, 1928.

On January 1, 1978, the perpetuity was exchanged for a 15-year annuity-due with semi-annual payments of amount X.

The interest rate is 6 percent, convertible monthly.

Find X.

When there are $n$ payments of $1$ with $k$ conversion periods between each payments at the end of each $n$ conversion period, I can see why the present value of those payments will be

$$\frac{a_{\overline{n}|}}{s_{\overline{k}|}}$$

I think the argument is that if a payment of $1$ is to be made in one payment period at the end, the equivalent payments that we can pretend that there is is $1 \over {s_{\overline{k}|}}$. Thus calculating the present value is straight forward.

However, if the payments are made at the beginning I am thinking that each payment of $1$ is equivalent to $1 \over {a_{\overline{k}|}}$ but the numerator should be an annuity-due rather than an annuity-immediate which is expressed as

$$\frac{a_{\overline{n}|}}{a_{\overline{k}|}}$$

Either I am not even deriving the first formula or there is something missing in my argument... can someone help me out?

Answer 1

Let be $i^{(12)}=6\%$ the nominal interest convertible monthly, so that the effective annual interest rate is $i=\left(1+\frac{i^{(12)}}{12}\right)^{12}=6.17\%$. The value of the perpetuity at the time of the exchange is $$ 50 a_{\overline{\infty}\!|i}=\frac{50}{i}=\frac{50}{0.0617}= 810.66 \tag 1 $$

The semiannual interest rate is $i^{(2)}=6.08\%$ found by $$ \left(1+\frac{i^{(2)}}{2}\right)^{2}=\left(1+\frac{i^{(12)}}{12}\right)^{12}\Longrightarrow i^{(2)}=2\left[\left(1+\frac{i^{(12)}}{12}\right)^{6}-1\right]=6.08\% $$ For the 15-year annuity-due with semi-annual payments we have 30 payments at the beginning of each period. The present value is $$ X\,\ddot a_{\overline{30\,}\!|i^{(2)}}=X\,\frac{1-v^n}{1-v}=X\, \times 14.4839\tag 2 $$ where $v=\frac{1}{1+i^{(2)}}$.

Putting $(1)=(2)$, we find

$$ 50 a_{\overline{\infty}\!|i}=X\,\ddot a_{\overline{30\,}\!|i^{(2)}}\Longrightarrow \boxed{X= 55.97 } $$