Discount Rate

[Albert Chan]

(04-11-2022 01:18 AM)Albert Chan Wrote:  The formula [OP] is good when guess rate is under-estimated.
This explained why it has i0 = 2*(n-p)/(n*(n+1)) instead of 2*(n-p)/(p*(n+1))

To be more precise, OP formula preferred rate magnitude under-estimated.

Plotting errors of both rate estimate, error(n version) ≈ -2 * error(p version)
In other words, this is much better rate estimate.

\(\displaystyle I_0 = \frac{2\;(N-P)}{N+1}
\left( \frac{{1 \over N} + {2 \over P}}{3} \right)
\)

Examples:

I₀(N= 36, P= 30) = .010210      // true rate = .010207
I₀(N=-36, P=-50) = .018074     // true rate = .017958

Nice. But, for OP formula, we wanted guess rate under-estimated.
Perhaps give up some accuracy for safety in convergence, and make ratio 1:1 ?
While we are at it, why not assume N+1 ≈ N, to simplify further ?

\(\displaystyle I_0 = \frac{2\;(N-P)}{N}
\left( \frac{{1 \over N} + {1 \over P}}{2} \right)
= \frac{1}{P} - \frac{P}{N^2}
\)

This may be how I₀ = 1/P - P/N² comes from

Update: Perhaps formula designed also to match asymptote, when C is big.
When compounding factor C is big, literally all payments goes to paying interest.

C = N/P = I*N/(1-(1+I)^-N) ≈ N*I      → I = 1/P

Rate formula matched this behavior: I₀ = 1/P - P/N^2 = (1 - 1/C^2)/P ≈ 1/P

For better estimate, we can use continuous compouding for I, derived earlier.

(04-10-2022 04:03 AM)Albert Chan Wrote:  Let y = -C*exp(-C), solve for I

I ≈ 1/P + W(y)/N

When C = N/P is big, y is small, we have W(y) ≈ y

I ≈ C/N + (-C*exp(-C))/N = (1-exp(-N/P)) / P