Annuities

For most of us, we aren’t able to put a large sum of money in the bank today. Instead, we save for the future by depositing a smaller amount of money from each paycheck into the bank. This idea is called a savings annuity. Most retirement plans like 401k plans or IRA plans are examples of savings annuities. An annuity can be described recursively in a fairly simple way. Recall that basic compound interest follows from the relationship [latex-display]{{P}_{m}}=\left(1+\frac{r}{k}\right){{P}_{m-1}}[/latex-display] For a savings annuity, we simply need to add a deposit, d, to the account with each compounding period: [latex-display]{{P}_{m}}=\left(1+\frac{r}{k}\right){{P}_{m-1}}+d[/latex-display] Taking this equation from recursive form to explicit form is a bit trickier than with compound interest. It will be easiest to see by working with an example rather than working in general. Suppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. In this example: r = 0.06 (6%) k = 12 (12 compounds/deposits per year) d = $100 (our deposit per month) Writing out the recursive equation gives [latex-display]{{P}_{m}}=\left(1+\frac{0.06}{12}\right){{P}_{m-1}}+100=\left(1.005\right){{P}_{m-1}}+100[/latex-display] Assuming we start with an empty account, we can begin using this relationship: Continuing this pattern, after m deposits, we’d have saved: In other words, after m months, the first deposit will have earned compound interest for m-1 months. The second deposit will have earned interest for m­-2 months. Last months deposit would have earned only one month worth of interest. The most recent deposit will have earned no interest yet. This equation leaves a lot to be desired, though – it doesn’t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005: [latex-display]1.005{{P}_{m}}=1.005\left(100{{\left(1.005\right)}^{m-1}}+100{{\left(1.005\right)}^{m-2}}+\cdots+100(1.005)+100\right)[/latex-display] Distributing on the right side of the equation gives [latex-display]1.005{{P}_{m}}=100{{\left(1.005\right)}^{m}}+100{{\left(1.005\right)}^{m-1}}+\cdots+100{{(1.005)}^{2}}+100(1.005)[/latex-display] Now we’ll line this up with like terms from our original equation, and subtract each side [latex-display]\begin{align}&\begin{matrix}1.005{{P}_{m}}&=&100{{\left(1.005\right)}^{m}}+&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&{}\\{{P}_{m}}&=&{}&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&+100\\\end{matrix}\\&\\\end{align}[/latex-display] Almost all the terms cancel on the right hand side when we subtract, leaving [latex-display]1.005{{P}_{m}}-{{P}_{m}}=100{{\left(1.005\right)}^{m}}-100[/latex-display] Solving for Pm [latex-display]\begin{align}&0.005{{P}_{m}}=100\left({{\left(1.005\right)}^{m}}-1\right)\\&{{P}_{m}}=\frac{100\left({{\left(1.005\right)}^{m}}-1\right)}{0.005}\\\end{align}[/latex-display] Replacing m months with 12N, where N is measured in years, gives [latex-display]{{P}_{N}}=\frac{100\left({{\left(1.005\right)}^{12N}}-1\right)}{0.005}[/latex-display] Recall 0.005 was r/k and 100 was the deposit d. 12 was k, the number of deposit each year. Generalizing this result, we get the saving annuity formula.

Example 7

Example 8

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