It is a common observation that if a loan is paid off in equal installments, and if the number of installments is large, then the early installments consist mostly of payment of interest, while, as the final installment nears, the repayment of principal increases dramatically.
I'm looking for a conceptual explanation of the observation that the present value of the repayment of principal in an amortization schedule is constant.
Details: A loan in the amount $P$ is to be repaid in $n$ monthly installments, starting at the end of the first month. The monthly interest rate is $r$. The monthly repayment $A$ can be computed as follows: $$ \begin{aligned} \text{present value of loan}&=\text{present value of all future payments}\\ P&=\sum_{i=1}^nA(1+r)^{-i}\\ &=A\frac{(1+r)^{-n-1}-(1+r)^{-1}}{(1+r)^{-1}-1}\\ &=\frac{A}{r}[1-(1+r)^{-n}], \end{aligned} $$ with the result that $$ A=\frac{Pr(1+r)^n}{(1+r)^n-1}. $$ Let $B_i$ be the balance on the loan at the end of $i$ months. So $B_0=P$, $B_1=(1+r)P-A$, and in general $$ B_{i+1}=(1+r)B_i-A. $$ The repayment of principal on the $i^\text{th}$ payment, $R_i$, is the reduction in the balance resulting from that payment, $R_i=B_{i-1}-B_i$. One can show by induction that $$ \begin{aligned} R_i&=(1+r)^{i-1}(A-rP)\\ &=(1+r)^{i-1}rP\frac{1}{(1+r)^n-1}, \end{aligned} $$ which increases geometrically with growth factor $1+r$. It follows immediately that the present value of the repayment of principle is constant: $$ \begin{aligned} \text{PV}(R_i)&=(1+r)^{-i}(1+r)^{i-1}rP\frac{1}{(1+r)^n-1}\\ &=\frac{rP}{1+r}\cdot\frac{1}{(1+r)^n-1}. \end{aligned} $$ I can't escape the feeling that, if only I had the right conceptual framework, I would have been able to predict this (not the precise value, but the fact that it is constant) without doing the calculation above. Is there a way of looking at things that makes this clear?
Here's one possible answer. I'd be interested in seeing a cleaner argument.
A key observation is that, at any point during the life of the loan, the outstanding balance can be treated as a new loan. So $$ \text{present value of loan}=\text{present value of all future payments} $$ must hold, not only at time $0$, but at all times, $0,1,2,\ldots,n$. As a consequence, $$ B_k(1+r)^{-k}=\sum_{i=k+1}^nA(1+r)^{-i} $$ and therefore $$ \begin{aligned} R_k=B_{k-1}-B_k&=\sum_{i=k}^nA(1+r)^{k-1-i}-\sum_{i=k+1}^nA(1+r)^{k-i}\\ &=A(1+r)^{k-1-n}, \end{aligned} $$ with the consequence that $(1+r)^{-k}R_k$ is constant.
In words: the only difference between the way the debt looks at time $k-1$ and the way it looks at time $k$ is that the debt at time $k-1$ requires $n-k+1$ payments, whereas the debt at time $k$ requires $n-k$ payments. The repayment of principal at time $k$ is therefore the value at time $k$ of a fictitious $n-k+1^\text{st}$ payment to be made at time $n+1$. As $k$ increases, the value at time $k$ of this fictitious payment at time $n+1$ increases, while its value at time $0$ (its present value) is constant.