Given Present Value of two annuities-due, find annuities-immediate for time period 4n

Question

For convenience I will denote annuities-due as: a"[n] and annuities-immediate as: a'[n] and let C=level payments or contribution.

Given: a"[n]=12 and a"[2n]=21

Problem: a'[4n] = ?

So if I read the problem correctly, it is ultimately asking for the value of annuities-immediate after time period of 4n.

Now, I understand that

$$ Ca"[n] = C(1+i)a'[n] = C(1+i) \frac{1}{i} (1-(1+i)^{-n}) $$

However, the question doesn't express the existence of any C although, since they give the value as 12 for a"[n], which is greater than 1, I know that C is implied. So my first step was assume n=1 which ultimately cancels out every variable except for C giving us, C= 12.

$$ a"[n=1] = C(1+i) \frac{1}{i} (1 - (1+i)^{-n}) $$

$$ = C * \frac{1+i}{i} * \frac{(1+i)-1}{1+i} $$

$$ C = 12 $$

Next, I plug C=12 into a"[2n] to which ultimately I get the value of i=1/3 or .33333

Now lastly, we are asked to find a'[4n] rather than a"[4n] that is, the present value of annuities-immediate and not annuities-due. Since $$ a"[4n] = C (1+i) \frac{1}{i} *(1 - (1+i)^{-4}) $$ which is approximately, 32.8125

and a'[4n] is simply: $$ a'[4n] = \frac{a"[4n]}{1+i}$$ or $$ C [ \frac{(1 - (1+i)^{-4})}{i}] $$ $$ = 12* \frac{ (1 - (1.3333)-4)}{.33333} $$

which is approximately, 24.609 and it gives the same value when dividing a"[4n] by (1+i).

So therefore, my answer was 24.609 however, my answer key says 32.8125 which is the value of the annuities-due a"[4n].

I do not understand why when the question asks for the present value of the annuities-immediate, the answer key gives the value of annuities-due. I even checked online for answers and it still regards a'[4n] as a"[4n]. Can someone please explain? As I do not think this is a typo.

Answer 1

Recall that $$\require{enclose} a_{\enclose{actuarial}{n}} = \frac{1 - v^n}{i}, \quad \ddot a_{\enclose{actuarial}{n}} = (1+i) a_{\enclose{actuarial}{n}},$$ where $v = 1/(1+i)$ is the periodic present value discount factor. So consider $$\frac{\ddot a_{\enclose{actuarial}{2n}}}{\ddot a_{\enclose{actuarial}{n}}} = \frac{1 - v^{2n}}{1-v^n} = \frac{(1 - v^n)(1 + v^n)}{1 - v^n} = 1 + v^n. $$ Hence $$v^n = \frac{21}{12} - 1 = \frac{3}{4},$$ and now substituting back, we obtain $$12 = \ddot a_{\enclose{actuarial}{n}} = (1+i) \frac{1 - v^n}{i} = \frac{1+i}{i} \left(1 - \frac{3}{4}\right).$$ Solving for $i$, we get $$i = \frac{1}{47}.$$ Now we have all the pieces we need: $$a_{\enclose{actuarial}{4n}} = \frac{1 - v^{4n}}{i} = \frac{1 - (3/4)^4}{1/47} = \frac{8225}{256} \approx 32.1289.$$ If the question had a typo (as the answer key would suggest), the intended quantity to evaluate would be $$\ddot a_{\enclose{actuarial}{4n}} = (1+i) \frac{8225}{256} = \frac{525}{16} = 32.8125.$$ Moreover, it is worth noting that this could also be computed without solving for $i$: $$\ddot a_{\enclose{actuarial}{4n}} = (1+i)\frac{1 - v^{4n}}{i} = (1+i) \frac{1 - v^{2n}}{i} (1 + v^{2n}) = \ddot a_{\enclose{actuarial}{2n}} (1 + v^{2n}) = 21 \left(1 + \left(\tfrac{3}{4}\right)^2\right) = \frac{525}{16}.$$

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The reason why your solution does not work is because you made an assumption about $n$ that is not correct. There is no $C$ in the given conditions. $a_{\enclose{actuarial}{n}}$ is simply an annuity-immediate of $1$ for $n$ periods with periodic interest rate $i$. It is like saying you are paid $1$ dollar a month for $n$ months where the monthly interest rate is $i$. As such, the given conditions result in a system of two equations in exactly two unknowns, $i$ and $n$. We established that $i = 1/47$. You can then proceed to solve for $n$; the result is $$n = \frac{\log \frac{4}{3}}{\log \frac{48}{47}} \approx 13.6644.$$ This is not an integer so it is a bit strange to conceptualize in terms of a discrete cash flow, but the formula works; moreover, these values for $i$ and $n$ are the unique values that satisfy the given conditions.