# Fun With Series

In hindsight, one of the good things about teaching where I did was that I had time to think about other stuff during school. This was especially handy when I was studying for the actuarial exams (I passed the first two on first attempt) and needed to practice some of the little things I haven’t done in a while and that never comes up in the course of what little I was able to teach. So just for grins, here’s a little derivation practice.

Yes, $i$ is interest rate. It looks funny not being part of a complex number, but that’s how it goes sometimes
For instance, deriving the formula for annuities: $\annuai{n}{i}$

Since an annuity is just the sum of the present values of a series of payments, it’s pretty easy to derive. The key is to remember the equation for the present value $PV$ of some amount of money at a later date $FV$, given the periodic interest rate (per period) $i$ expected between now and later, and the number of periods $t$ between now and later. Valuation equations all derive from this most basic premise.

\begin{align} PV=\frac{1}{(1+i)^t} \cdot FV\label{a1} \end{align}

A useful shortcut. Why? Rearrangements.
\begin{align} v &=\frac{1}{1+i} \label{v1}\\ (1+i)\cdot v &=1 \notag\\ v + vi &= 1 \notag\\ vi &=1-v \label{v2} \end{align}
Consider the worst way to describe the present value of an annuity: “Figure a payment of, say, $50 at the end of each year for the next 10 years. Take the first payment and find its present value. Then the second, and add it to your first answer…” Ouch. Well, that’s what math is for. How about we write it in math instead? $PV_{annuity}=PV(first payment)+PV(second payment)+ \dotsb + PV(last payment) \notag$ Slightly closer, but not clean yet. We keep adding up the same sort of thing over and over again. Perhaps there’s a way to write that a little more concisely. $PV_{annuity}=\sum_{t=1}^{10}PV_{t} \notag$ The jump is a little big, but now it is starting to look like math!1 Substitute our definition of$PV$from$\eqref{a1}$, then use our definition of$v$at$\eqref{v1}, factor out the constant, and we get something useful. And as a side benefit, we also see that the amount of the payment really is irrelevant. \begin{align*} PV_{annuity}&=\sum_{t=1}^{10}50\cdot \frac{1}{(1+i)^t} \\ PV_{annuity}&=\sum_{t=1}^{10} 50 \cdot v^t \\ PV_{annuity}&=50 \cdot \sum_{t=1}^{10} v^t \\ \end{align*} The purpose of this isn’t the deriving of a formula, it’s about how we think about things and make them easier for ourselves. Thinking about it in English is difficult because the language is less than conducive to compacting a complex thought into an easily manipulated form. Crossing the bridge towards mathematical thinking is the single biggest hurdle my students face, and one they typically stumble on. Why? <rant>To paraphrase (and make suitable for a family publication): “Because math is a class to blow off, not a tool to be used to make life easier”2</rant> I take every opportunity I can find to point out that formalized math is there to let us both be lazier and also accomplish more at the same time. Back to the point… \begin{align} \annuai{n}{i}&=\sum_{t=1}^n v^t \label{q1} \\ \annuai{n}{i}&=v^1+v^2+v^3+\dotsb+v^{n-1}+v^n \label{q2} \\ v\cdot \annuai{n}{i}&=v\cdot v^1+v\cdot v^2+v\cdot v^3+\dotsb+v\cdot v^{n-1}+v\cdot v^n \label{q3} \\ v\cdot \annuai{n}{i}&=v^2+v^3+v^4+\dotsb+v^{n}+v^{n+1} \label{q4} \\ \annuai{n}{i} - v\cdot \annuai{n}{i}&=v^1 - v^{n+1} \label{q5} \\ \annuai{n}{i}(1-v)&=v^1 - v^{n+1} \label{q6} \\ \annuai{n}{i}\cdot i \cdot v&=v\cdot (1-v^n) \label{q7} \\ \annuai{n}{i}&=\frac{1-v^n}{i} \label{q8} \\ \end{align} In general, there are often steps that are either necessary to show or not. One such example here is\eqref{q3}$, which could easily be omitted. But if what you mean to do is multiply every term by$v$, then perhaps writing it out will help reduce mistakes. Another place is at$\eqref{q5}$, which could easily have another equation in front of it explicitly showing the subtraction. Getting MathJax up and running, and putting together the other pieces for this blog, have made this post take WAAAY longer than it should have. Doing a little math on paper is a good mind-clearing exercise sometimes, but this was nearly the opposite. Given our basic definition$\act{a}{n}{I}=\frac{1-v^n}{i}$, there are a few other useful varieties of basic annuities worth having fun with. Consider an annuity that pays 1 at the beginning of each month, instead of at the end. This is referred to as an “annuity due” and uses the symbol$\actd{a}{n}{i}. Each payment has one fewer compounding period, so \begin{align*} \actd{a}{n}{i} &= v^0 + v^1 + \dotsb + v^{n-2} + v^{n-1} \\ \actd{a}{n}{i} &= \frac{v^1}{v} + \frac{v^2}{v} + \dotsb + \frac{v^{n-1}}{v} + \frac{v^{n}}{v} \\ \actd{a}{n}{i} &= \frac{\act{a}{n}{i}}{v} \\ \actd{a}{n}{i} &= \frac{1-v^n}{iv} \\ \end{align*} Consider an annuity that pays 1 at the end of the first month, 2 the second, three the third, and so on fornmonths. The usual name for this is an increasing annuity. \begin{align*} \act{(Ia)}{n}{i} &= PV(1) + PV(2) + \dotsb + PV(n-1) + PV(n) \\ \act{(Ia)}{n}{i} &= \sum_{t=1}^n PV(t) \\ \act{(Ia)}{n}{i} &= \sum_{t=1}^n tv^t \\ \act{(Ia)}{n}{i} &= 1v^1 + 2v^2 + \dotsb + (n-1)v^{n-1} + nv^n \\ v\left(\act{(Ia)}{n}{i}\right) &= 1v^2 + 2v^3 + \dotsb + (n-1)v^n + nv^{n+1} \\ (1-v)\left(\act{(Ia)}{n}{i}\right) &= v^1 + v^2 + \dotsb + v^{n-1} + v^n - v^{n+1} \\ (iv)\left(\act{(Ia)}{n}{i}\right) &= \act{a}{n}{i} - v^{n+1} \\ \act{(Ia)}{n}{i} &= \frac{\act{a}{n}{i} - v^{n+1}}{iv} \\ \act{(Ia)}{n}{i} &= \frac{\actd{a}{n}{i} - v^n}{i} \\ \end{align*} And then there is the decreasing annuity, which starts by payingn$at the end of the first month,$(n-1)$the second, and so on down to 1 the$nth month. \begin{align*} \act{(Da)}{n}{i} &= PV(n) + PV(n-1) + \dotsb + PV(2) + PV(1) \\ \act{(Da)}{n}{i} &= \sum_{t=1}{n} PV(n-t+1) \\ \act{(Da)}{n}{i} &= \sum_{t=1}{n} (n-t+1)v^t \\ \act{(Da)}{n}{i} &= (n)v^1 + (n-1)v^2 + \dotsb + 2v^{n-1} + 1v^n \\ v\left(\act{(Da)}{n}{i}\right) &= (n)v^2 + (n-1)v^3 + \dotsb + 2v^{n} + 1v^{n+1} \\ (1-v)\left(\act{(Da)}{n}{i}\right) &= nv^1 - 1v^2 - v^3 - \dotsb - v^n - v^{n+1} \\ (iv)\left(\act{(Da)}{n}{i}\right) &= nv - v\left(\act{a}{n}{i}\right) \\ \act{(Da)}{n}{i} &= \frac{nv - v\left(\act{a}{n}{i}\right)}{vi} \\ \act{(Da)}{n}{i} &= \frac{n - \left(\act{a}{n}{i}\right)}{i} \\ \end{align*} Yes, this is what I did for relaxation when I was stressed at school. There’s another one that is entertaining (but I’m not sure I see the point) in which the annuity pays 1 the first month, 2 the second, up ton$in the$n^{\text{th}}$month, and then decreases back to 1 in the$(2n-1)^{\text{th}}$month. I have heard it called a “rainbow annuity” (even though it is a triangle, not an arc) and I don’t know the symbol for it so I’ll just call it$\act{RA}{n}{i}$. (Mind you,$nrefers to the peak payment, not the number of payments) \begin{align*} \act{RA}{n}{i} &= v^1 + 2v^2 + \dotsb + (n-1)v^{n-1} + nv^n + (n-1)v^{n+1} + \dotsb + 2v^{2n-2} + v^{2n-1} \\ v\left(\act{RA}{n}{i}\right) &= v^2 + 2v^3 + \dotsb + (n-1)v^{n} + nv^{n+1} + (n-1)v^{n+2} + \dotsb + 2v^{2n-1} + v^{2n} \\ (1-v)\left(\act{RA}{n}{i}\right) &= v + v^2 + \dotsb + v^n + (-1)v^{n+1} + (-1)v^{n+2} + \dotsb + (-1)v^{2n-1} + (-1) v^{2n} \\ (iv)\left(\act{RA}{n}{i}\right) &= \act{a}{n}{i} + (-1)v^{n+1} + (-1)v^{n+2} + \dotsb + (-1)v^{2n-1} + (-1)v^{2n} \\ (iv)\left(\act{RA}{n}{i}\right) &= \act{a}{n}{i} - v^n\left(v^1 + v^2 + \dotsb + v^{n-1} + v^{n}\right) \\ (iv)\left(\act{RA}{n}{i}\right) &= \act{a}{n}{i} - v^n\left(\act{a}{n}{i}\right) \\ (iv)\left(\act{RA}{n}{i}\right) &= (1-v^n)\left(\act{a}{n}{i}\right) \\ \act{RA}{n}{i} &= \frac{1-v^n}{iv} \act{a}{n}{i} \\ \act{RA}{n}{i} &= \frac{1}{v} \frac{1-v^n}{iv} \act{a}{n}{i} \\ \act{RA}{n}{i} &= \frac{1}{v} \act{a}{n}{i} \act{a}{n}{i} \\ \act{RA}{n}{i} &= \frac{\left(\act{a}{n}{i}\right)^2}{v} \\ \end{align*} One of the neat bits that shows up in each and every one of these is the ongoing pattern of “take the first line, multiply byv$, and subtract it off” which collapses the sequence into something more manageable, followed by representing the left side with a factor of$(1-v)$which turns into$iv\$, conveniently canceling out something on the right. One tool, many variations. Isn’t that what math is all about?

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