Can we clarify this "accumulated money flow" application of integration?

Question

I read about this model/application in Calculus with Applications, 11th Edition by Lial, Greenwell, and Ritchey (example), where if you have a function $f(t)$ that models some revenue stream, the rate at which you're accruing money over time, and some "interest rate" $r$, then $\newcommand{\dt}{\,\mathrm{d}t}\newcommand{\ex}{\mathrm{e}}$

• the number $\int_0^T f(t) \dt$ is your total money flow, the money you've made by time $t = T$,
• the number $\int_0^T f(t)\ex^{-rt} \dt$ is the present value of your money flow, how much that revenue you get from $f(t)$ by $t = T$ is presently worth,
• and the number $\ex^{rT}\int_0^T f(t)\ex^{-rt} \dt$ is the accumulated amount of money flow, the final amount of money flow including interest after the money comes in from $f(t)$.

This model doesn't make sense to me, especially the fact that there is only one rate $r$ involved here. For a positive rate $r$, $\int_0^T f(t)\ex^{-rt} \dt < \int_0^T f(t)\dt$, so I'd think $r$ has to be tied to the rate of inflation since a fixed amount of money loses its value over time (a job paying \$1000 after each month for twelve months, is worth less that \$12000 right now). With this understanding of the present value of the money flow, the money hasn't been invested yet. Only when you let $P_0 = \int_0^T f(t)\ex^{-rt} \dt$ and "invest your money flow" at an interest rate of $R$ do you have $\ex^{RT}\int_0^T f(t)\ex^{-rt} \dt$ money after time $T$. Can someone validate my understanding here? Or point out my misunderstanding?

Answer 1

The rate $r$ is the interest you receive for money in the bank. There is nothing in the problem about buying things, so we don't care about inflation. The interest rate is quoted as percent per year and assumed to be compounded instantly. The first integral simply adds up all the money you receive. The second one takes money that you receive at time $t$ and discounts it by multiplying by $e^{-rt}$. That gives the amount you would have to have now to have that amount at time $t$. The value of the integral is the amount one could deposit in the bank now and just be able to pay you the whole revenue stream. If you deposit that amount now, after time $T$ it will be multiplied by $e^{rT}$, which is the factor out in front of the integral. If we move that constant into the integral, it becomes $$\int_0^T f(t)e^{r(T-t)}dt$$ The explanation for this is if you earn $f(t)dt$ at time $t$ it draws interest for a period $T-t$ and is then worth $f(t)e^{r(T-t)}dt$

Answer 2

There are two distinct issues here that you're conflating. One is the issue of changes in price levels - that's what you're referring to as inflation, although deflation can also occur, where price levels fall and the value of \$1 rises. The other issue is that money in the future is worth less than money in the present. This is the time value of money, and the general argument is that, when given a choice, a rational person would prefer some amount of money today over the same amount in the future, after accounting for changes in price levels.

Inflation and time value of money are distinct issues, and they are independent. Since $f$ is measured in nominal (dollar) terms, it automatically incorporates inflation. If prices go up, so does the number of dollars of revenue, and vice versa. So, in these formulas, $r$ just represents discounting for the time value of money, and we don't need to adjust for inflation.

Even if we didn't include inflation expectations in $f$, we'd still end up with the same formula. Let $\pi$ represent the change in the price level and $i$ represent the nominal interest rate. Then, our present value formula would be:

$$ \int_0^T f(t)e^{-it}e^{\pi t} dt $$ $$ = \int_0^T f(t)e^{-(i-\pi)t} dt $$ $$ = \int_0^T f(t)e^{-rt} dt , $$

where $r = i - \pi$ is the real interest rate and represents the post-inflation market price to borrow/lend \$1. This is the same relationship we get from the Fisher equation.

Similarly, your accumulated money flow will look like this:

$$ e^{iT}e^{-\pi T}\int_0^T f(t)e^{-rt}e^{\pi t} dt $$ $$ = e^{(i-\pi)T}\int_0^T f(t)e^{-rt} dt $$ $$ = e^{rT}\int_0^T f(t)e^{-rt} dt . $$