Suppose I take out a loan.
It gains interest at rate \(r\), compounded continuously.
I repay the loan by making constant, continuous payments until time \(T\).
How does the repaid share of my loan vary over time?
And how does it depend on \(r\) and \(T\)?
Let \(P_0\) be the initial value of my loan: the “principal.”
Then my continuous payments \(C\) must satisfy
$$\begin{align} P_0 &= \int_0^TCe^{-r\tau}\,\mathrm{d}\tau \\ &= \frac{C}{r}\left(1-e^{-rT}\right) \end{align}$$
and so the value of my remaining payments at time \(t\in[0,T]\) equals
$$\begin{align} P_t &\equiv \int_t^TCe^{-r(\tau-t)}\,\mathrm{d}\tau \\ &= \frac{C}{r}\left(1-e^{-r(T-t)}\right) \\ &= P_0\left(\frac{e^{-rt}-e^{-rT}}{1-e^{-rT}}\right)e^{rt}. \end{align}$$
If I don’t make any payments before time \(t\) then the principal grows to \(P_0e^{rt}\).
Therefore, the value of my repayments up to time \(t\) equals the difference \((P_0e^{rt}-P_t)\).
Now let \(x\equiv t/T\in[0,1]\) be share of payments I’ve made up to time \(t\).
The chart below plots the corresponding share
$$\frac{P_0e^{rt}-P_t}{P_0e^{rt}}\bigg\rvert_{t=xT}=\frac{1-e^{-xrT}}{1-e^{-rT}}$$
of the loan that I’ve repaid.
This share grows with \(x\) at a decreasing rate.
Intuitively, my repayment “slows down” because the interest on the principal and payments grows larger than the payments themselves.
This slowing effect is stronger when the interest rate \(r\) is larger and time horizon \(T\) is longer.
