Annuity/Perpetuity problem

financesolution-verification

(Repost but with a proposed solution) Mary-Jane, a $20$-year-old woman, purchases a perpetuity-due as a pension benefit that begins when she is $60$ years old. The benefit pays an annual total of $48000$ with level payments at the start of every month. To purchase this benefit, she has to make $40$ annual premium payments, the first of which will be at the time of the policy purchase i.e. her $20$th birthday. Every payment after the first will increase by $6$%. Assuming a flat term structure and annual effective interest rate of $8$% for all valuations, what should be the amount of the initial payment?

My attempt:

Notice that the first $40$ payments (time $0$ to time $39$) form an annuity due whereas the payments from payment $41$ (time 40 onwards) form a perpetuity due. Suppose Mary-Jane pays $P$ as her first payment.

$t = 0 \rightarrow P$ (just turned 20)

$t = 1 \rightarrow P(1.06)$

$t = 2 \rightarrow P(1.06)^2$
.
.

$t = 39 \rightarrow P(1.06)^39$ (just turned 59)

$t = 40 \leftarrow 48000$ (just turned 60)

$t = 41 \leftarrow 48000$
.
.

The present value of the pension plan is $48000 \hat{a}_{\infty, 0.08}$, where $\hat{a} = \frac{1}{d}$ and $d$ is the discount rate. Hence $PV = 48000 \cdot \frac{1}{d} = \frac{48000}{0.0741} = 648000$.

This should be equal to the future value of the premiums at $t = 40$. This is given by:

$FV = P \cdot (1+i)(\frac{(1+i)^n – (1+g)^n}{i – g})$, where $i$ is interest and $g$ is the growth rate. Hence $FV = P \cdot (1.08)(\frac{(1.08)^{40} – (1.06)^{40}}{0.02}) = 617.6954P$

This gives $P = \frac{648000}{617.6954} = 1049.0608$.

Is this correct? I did quite a bit of research but I am still unsure as to whether this is the right way to do such a problem. Any assistance is much appreciated.

Best Answer

$\require{enclose}$ When the perpetuity begins to pay out when the annuitant turns $60$ years of age, the schedule of payments is monthly, not annually. I quote:

The benefit pays an annual total of 48000 with level payments at the start of every month.

The $48000$ is therefore divided into $12$ equal payments of $4000$.

Consequently, the equation of value at the time that she turns $60$ is: $$P(1+i)^{40} + (1.06)P(1+i)^{39} + \cdots + (1.06)^{39}P(1+i) = 4000 + 4000v^{1/12} + 4000v^{2/12} + \cdots,$$ where $v = 1/(1+i)$ is the annual present value discount factor. The first thing to note is that I set the valuation time at age $60$, rather than at age $20$, because this makes the calculation a little cleaner. On the left-hand side is the accumulated value of a geometrically increasing annuity-due with annual payments. On the right-hand side is the present value of a perpetuity-due with monthly payments.

The geometric annuity-due is expressible as a level payment annuity-due with modified interest rate; e.g. factoring out $(1.06)^{40} P$, we get $$P(1+i)^{40} + (1.06)P(1+i)^{39} + \cdots + (1.06)^{39}P(1+i) = (1.06)^{40} P \left( \left(\frac{1+i}{1.06}\right)^{40} + \left(\frac{1+i}{1.06}\right)^{39} + \cdots + \frac{1+i}{1.06}\right).$$ So if $i = 0.08$, the factor in parentheses is a level-annuity-due with effective annual interest rate $$j = \frac{1.08}{1.06} - 1 \approx 0.01887,$$ and the level payment is adjusted to be $(1.06)^{40}P$. Thus this side of the equation of value is $$(1.06)^{40} P \ddot s_{\enclose{actuarial}{40}j} = (1.06)^{40} P (1+j) \frac{(1+j)^{40} - 1}{j}.$$

Meanwhile, the right-hand side is a perpetuity-due with effective monthly rate $$k = (1+i)^{1/12} - 1 \approx 0.00643403,$$ which you can verify yields an effective annual rate of $i$, since $(1+k)^{12} = 1+i$. So the equivalent monthly present value discount factor is $$v^{1/12} = \frac{1}{1+k} \approx 0.993607.$$ Then the right-hand side has present value

$$4000 \ddot a_{\enclose{actuarial}{\infty} k} = 4000 \left(1 + \frac{1}{k}\right).$$

Hence the required level payment is $$P = \frac{4000 \left(1 + \frac{1}{k}\right)j}{(1.06)^{40} (1+j)((1+j)^{40} - 1)} \approx \frac{4000 (156.424)(0.01887)}{(10.2857)(1.01887)(1.11211)} \approx 1012.9496.$$

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