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I'm currently studying for SOA Exam FM using Harold Cherry's study manual (11th edition). In practice exam 5, there is a question (#22) that asks the following:

Determine the modified duration of a growing perpetuity, such that the first payment is 1 dollar 1 year from now, and increases each year by 1 dollar. Interest rate is 8%.

Typically you'd just use the formula for Macaulay Duration and multiply it by the discounting factor $v = (1+i)^{-1}$, no problem:

$vMacD = v\frac{v + 4v^2 + 9v^3 + ...}{v + 2v^2 + 3v^3 + ...}$

But in this case when you apply that formula, you get the following in the numerator:

$v + 4v^2 + 9v^3 + ...$

I am wondering if there is a neat way to evaluate this infinite sum. I searched around and could not really find anything. In the case of the denominator, which is just $v + 2v^2 + 3v^3 + ...$, there is a formula derived previously which is easy to use.

The solution in the back suggests evaluating it as $Mod D = \frac{-P'}{P}$, where $P = \frac{1}{i} + \frac{1}{i^2}$, which is fine (and I would definitely rather do this approach when I sit for the exam) but I am wondering if there is a way to evaluate this infinite sum otherwise. I don't think the same strategy used in the derivation for $v + 2v^2 + 3v^3 + ... = \frac{1}{i} + \frac{1}{i^2}$ could be used, because it takes advantage of the fact that the increase from term to term is only 1.

Any help would be appreciated. This is not a homework question, I've just got a bit of an itch now that I've seen this and can't figure it out.

jason
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  • Yes, that part is fine and simplifies to $1/i + 1/i^2$. I'm wondering about the other part. – jason Aug 02 '21 at 16:52
  • Start with $\frac1{1-x}=\sum_{n=0}^\infty x^n$ Differentiate both sides; multiply both sides by $x$; differentiate again and multiply by $x$ again. – saulspatz Aug 02 '21 at 16:54

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