Suppose there is an item that deteriorates with use, its condition expressed as a number between $0$ (completely broken condition) and $1$ (perfect condition), for which I want to assign a monetary value. To make the units nicer, let's think of the condition also as a monetary value; i.e., someone has guaranteed to pay me $T$ dollars if I want to sell the item at any time, where $T$ is its condition. $M(T)$ is an implicit multiplicative markup on $T$ (i.e., $M(T) = 2$ would mean I value the item at twice the guaranteed resale value), and my overall value is $V(T) = M(T) * T$.

The first intuition (which I'm treating as an axiom) is that $V(T)$ should be monotonically increasing in $T$. The second is that $M(T)$ should be monotonically decreasing in $T$.

I can also estimate my $V$ at a handful of points in $T$. So far, I have tried Lagrange Polynomial Interpolation on such a handful of points to output a candidate $V(T)$, but when I divide this by $T$ to produce the corresponding $M(T)$, the result violates the second axiom; namely, polynomials produce non-decreasing regions in $M(T)$ around the LPI input points.

Are there ways to construct a $V(T)$, in elementary but perhaps non-polynomial functions, so that the two axioms are satisfied, and it also passes through a handful of input points (the points can be trusted to themselves satisfy the axioms)? Outside the regions corresponding to $0 \leq T \leq 1$, of course, the behavior of $V(T)$ and $M(T)$ do not matter. Is there a standard way to express valuations for these sorts of items?

  • $\begingroup$ How many is "a handful" of points in $T$? Does your $V(T)$ have to exactly fit the estimated points or is an approximate fit enough? $\endgroup$
    – VARulle
    Commented Mar 9, 2023 at 14:56
  • $\begingroup$ @VARulle An approximate fit is enough; the points are just subjectively determined by asking "about how much would I be willing to pay for the item at this particular condition?" I was using 5 points with my Lagrange Polynomial Interpolation, but I guess up to around 20 points would be manageable. $\endgroup$
    – user10478
    Commented Mar 9, 2023 at 15:22
  • $\begingroup$ O.k., but then instead of exact interpolation why not just use a simple regression model? Even linear regression could do, depending on how your data points look like. $\endgroup$
    – VARulle
    Commented Mar 10, 2023 at 9:29
  • $\begingroup$ @VARulle Linear regression would be incompatible with the two axioms, wouldn't it? To be clear, I'm not looking to express merely a trend in my data points. I want a function I could hand to someone and say, "this is what I'm willing to pay for any $T$." It's just that I wouldn't be very sensitive to, i.e., a $2-3\%$ error. $\endgroup$
    – user10478
    Commented Mar 10, 2023 at 19:24
  • $\begingroup$ Linear regression for V. If you get a positive intercept, then dividing by T gives you a decreasing M. $\endgroup$
    – VARulle
    Commented Mar 12, 2023 at 13:12


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