I've seen this answer and I understand the methodology, but I am wondering why my original solution using a different method did not work.
This is the sample problem in my study guide for Exam FM:
Olga buys a 5-year increasing annuity for X. Olga will receive 2 at the end of the first month, 4 at the end of the second month, and for each month thereafter the payment increases by 2. The nominal interest rate is 9% convertible quarterly. Calculate X.
$X$ is clearly the present value of this increasing annuity.
My study guide lists a formula for $(I^{(m)}a)^{(m)}_{\bar{n}|}$ as $$\frac{\ddot{a}^{(m)}_{\bar{n}|}-nv^n}{i^{(m)}}$$ This is supposed to be the present value of payments $\frac{1}{m^2}, \frac{2}{m^2}, \frac{3}{m^2},…,\frac{mn}{m^2}$ made at times $\frac{1}{m}, \frac{2}{m}, \frac{3}{m}, n$ respectively.
I initially thought the present value would be given by $288(I^{(12)}a)^{(12)}_{\bar{5}|}$ – this would give payments every month of $\frac{288}{144}=2, \frac{288*2}{144}=4, …$. Applying the formula, however, gives me an answer that is far too large.
$$i \approx 9.3083\%$$
$$i^{(12)}\approx8.9333\%$$
$$\ddot{a}^{(12)}_{\bar{5}|} = \ddot{a}_{\bar{60}|0.7444\%} \approx 48.6083$$
$$288(I^{(12)}a)^{(12)}_{\bar{5}|} = 288\frac{48.6083-5(1.093083^{-5})}{0.08933} = 288(508.2575)=146378.16$$
The actual answer is $\approx 2729$.
Is this payment stream not accurately described using $(I^{(m)}a)^{(m)}_{\bar{n}|}$? If so why not? Admittedly, this is an overly-complex way to solve the problem compared to the linked answer but I'm hoping to better understand this type of increasing annuity.
Best Answer
$$\require{enclose}$$ Your confusion is precisely why I do not recommend memorizing all of the formulas for annuities. All you really need is to understand and memorize the basic formulas for the present and accumulated values of level annuities-immediate, -due; the increasing (and optionally decreasing) annuities; and the level annuities paid $m$ times per year. Geometrically increasing/decreasing annuities are simply level annuities with a modified interest rate. Increasing/decreasing annuities paid on a non-annual basis are equivalent to ones that are paid annually, again by converting the interest rate the effective $m^{\rm th}$ly rate. Over-reliance on formulas is the downfall of many students preparing for the actuarial exams.
Let us recall the following:
$$\begin{align} i^{(m)} &= m((1+i)^{1/m} - 1) \tag{1} \\ v &= (1+i)^{-1} \tag{2} \\ a_{\enclose{actuarial}{n}i} &= \frac{1 - v^n}{i} \tag{3} \\ \ddot a_{\enclose{actuarial}{n}i} &= (1+i)a_{\enclose{actuarial}{n}i} \tag{4} \\ a_{\enclose{actuarial}{n}i}^{(m)} &= \frac{1}{m} a_{\enclose{actuarial}{mn}j} = \frac{1 - v^n}{i^{(m)}}, \quad j = i^{(m)}/m \tag{5} \\ (Ia)_{\enclose{actuarial}{n}i} &= \frac{\ddot a_{\enclose{actuarial}{n}i} - nv^n}{i}. \tag{6} \end{align}$$ To understand what the scheulde of payments for the symbols $(Ia)_{\enclose{actuarial}{n}}^{(m)}$ and $(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}}$ look like, we have the table $$\begin{array}{c|c|c|c|c|c} \text{time} & a_{\enclose{actuarial}{n}} & a_{\enclose{actuarial}{n}}^{(m)} & (Ia)_{\enclose{actuarial}{n}} & (Ia)_{\enclose{actuarial}{n}}^{(m)} & (I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}} \\ \hline 0 & 0 & 0 & 0 & 0 & 0\\ 1/m & 0 & 1/m & 0 & 1/m & 1/m^2 \\ 2/m & 0 & 1/m & 0 & 1/m & 2/m^2 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 1 & 1 & 1/m & 1 & 1/m & 1/m = m/m^2\\ 1 + 1/m & 0 & 1/m & 0 & 2/m & 1/m + 1/m^2 = (m+1)/m^2 \\ 1 + 2/m & 0 & 1/m & 0 & 2/m & 1/m + 2/m^2 = (m+2)/m^2 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 2 & 1 & 1/m & 2 & 2/m & 2/m = 2m/m^2 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ (n-1) + 1/m & 0 & 1/m & 0 & n/m & (n-1)/m + 1/m^2 = (mn-m+1)/m^2 \\ (n-1) + 2/m & 0 & 1/m & 0 & n/m & (n-1)/m + 2/m^2 = (mn-m+2)/m^2 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ n & 1 & 1/m & n & n/m & n/m = mn/m^2 \end{array}$$
Then it is clear that $(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}}$ is simply an increasing annuity of $1/m^2$ over a term of length $mn$ with periodic rate $j$; i.e., $$\begin{align}(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}} &= \frac{1}{m^2} (Ia)_{\enclose{actuarial}{mn}j} \\ &= \frac{\ddot a_{\enclose{actuarial}{mn}j} - mn v^n}{m^2 j} \\ &= \frac{\frac{1}{m} \ddot a_{\enclose{actuarial}{mn}j} - n v^n}{i^{(m)}} \\ &= \frac{\ddot a_{\enclose{actuarial}{n}}^{(m)} - nv^n}{i^{(m)}}. \tag{7} \end{align}$$
This proves the formula in your question.
Now, the next part pertains to how formula $(7)$ is applied to the specific problem. In this case, we have $n = 5$ years, $m = 12$ payments per year (i.e., monthly). We also have that the first payment is $2$ at the end of the first month and increases by $2$ each month. Since the first payment of the symbol $(I^{(m)}a)^{(m)}_{\enclose{actuarial}{n}}$ is $1/m^2$, it follows that we must multiply this by $2(12^2) = 288$, as you correctly deduced. Finally, we are given that $i^{(4)} = 0.09$, which means $$i = \left(1 + \frac{i^{(4)}}{4}\right)^4 - 1 \approx 0.0930833$$ and $v = (1+i)^{-1} \approx 0.914843$. We also have $$i^{(12)} \approx 0.0893333.$$ So far, we are in agreement. However,
$$\ddot a_{\enclose{actuarial}{5}}^{(12)} = (1+i)^{1/12} \frac{1 - v^5}{i^{(12)}} \approx (1.0930833)^{1/12} \frac{1 - (1.0930833)^{-5}}{0.0893333} \approx 4.05064401427.$$ Your calculation is incorrect because you must divide your result by $12$, as the monthly payments are each $1/12^{\rm th}$ of the equivalent annual payment over $60$ periods.
Finally, we have $$288(I^{(12)}a)^{(12)}_{\enclose{actuarial}{5}} \approx 288 \frac{4.05064401427 - 5(0.914843)^5}{0.0893333} \approx 2729.21$$ as claimed.
But as I warned at the very beginning, this is all entirely unnecessary. There is too much needless memorization and too much opportunity to make an error. The simplest approach is to scale the time period accordingly: the effective periodic interest rate is $j = i^{(12)}/12 \approx 0.00744444$, hence the present value is simply $$2(Ia)_{\enclose{actuarial}{60}j} = 2\frac{\ddot a_{\enclose{actuarial}{60}j} - 60v_j^{60}}{j},$$ where all we need to do is compute $v_j = (1+j)^{-1} \approx 0.992611$ and $$\ddot a_{\enclose{actuarial}{60}j} = 48.6077.$$ We get the same result with less complexity and fewer formulas to memorize.