Inflation over Several Years

When calculating inflation over several years you have to multiply values as prices are increasing from a new base each time. E.g., if in the first year inflation was 3% and in the second 5%, then the price level grew by $(1+0.03)\times (1+0.05)=1.0815$, so total inflation over the two years was $1.0815-1=0.0815$, or 8.15%. Similarly, average inflation is computed as a geometric mean: for this example $\sqrt{1.03\times 1.05}=1.0399$, an average of $1.0399-1=0.0399$, or 3.99% per year. The geometric mean is usually less than the arithmetic mean, being equal only when all values are the same. Multiyear inflation is larger than the sum of annual inflations. In general, if we have inflation over $n$ years as $i_1..i_n$, the total inflation will be:

If inflation is constant for all years, then this is simplified to ${(1+i)}^n-1$. Similarly, if $I$ is inflation over $n$ years, then average annual inflation is $\sqrt[n]{1+I}-1$. (The $n$-th root $\sqrt[n]{x}$ is the number that multiplied by itself $n$ times gives $x$; on a calculator or in a spreadsheet, raise to the power $1∕n$ — a form this chapter uses repeatedly.)

For example, if you have $100 today, and inflation is 2.5% annually, you’d need $128.01 in 10 years to have the same purchasing power. To strip inflation back off a dollar figure, divide by the same growth factor you would have multiplied by: $102.50 a year from now buys what $\$102.50∕1.025=\$100$ buys today. If your net worth is not growing faster than inflation, you are losing money. $100,000 left uninvested loses $\$100,000-\frac{\$100,000}{1.025}=\$2,439$ in the first year alone; in 10 years it would be worth just $\frac{\$100,000}{{(1+0.025)}^{10}}=\$78,120$ in today’s dollars.