parameter estimation for propotional equations of three variables

Question

I am modeling a system that should estimate a parameter $\beta \in [0,1]$ which is directly proportional to two other variables $P \in [0,1]$ and $NV \in \{0,1,2,\cdots,N\}$, and inversely proportional to a third variable $NW \in \{0,1,2,\cdots,N\}$.

My simple model is based on the proportional equation: $\beta \propto \frac{P\ \cdot\ NV}{NW}$ and then, $\beta = \frac{K\ \cdot\ P\ \cdot\ NV}{NW}$ where $K$ is a constant.

My concern here:

• is this model a good one? is there a better alternative?
• how can I estimate the value of the constant $K$ such that the value of $\beta$ remains $\in [0,1] $

Answer 1

You want proportionality to two variables and inverse proportionality to a third. Given these constraints, the model equation is the ONLY choice. That is, this IS THE DEFINITION of the words "direct and inverse proportion". However, if you mean "proportional" in a non rigorous sense (this goes up, that goes up), the relationship need not be linearly proportional (this doubles, that doubles). For example, $\beta= K\cdot P^2 \cdot NV/NW$ meets that non-rigorous definition ($P$ doubles, $\beta$ quadruples). You could take other nonlinear functions of the variables as well. For example $\beta= K\cdot P \cdot NV/\log(NW)$. This is an option when $NW$ spans several orders of magnitude and you expect that the effect depends on the order of magnitude, not linearly on the value. You can even take weighted sums of terms with various forms, like

$$ \beta= K_1\cdot P \cdot \frac{NV}{NW}+K_2\cdot P^2 \cdot \frac{\log(NV)}{NW} $$

Without knowing more about what you are actually doing, and what data this model should apply to, it is impossible to say if the model you propose is "good".

IF you want the model you propose, figuring out $K$ is possible. Since all your variables are positive, then for any $K>0$, $\beta$ will always be greater than $0$. So we just have to make sure that $\beta$ is not too large (not larger than $1$). Neglecting the $K$, the largest finite value we can get is when $P=1$, $NV=N$, and $NW=1$, giving $\beta_\text{max}=N$. If we want that value to be $1$ at max, then we simply divide by this value, giving $K=1/N$ and a final equation of $$ \beta= \frac{P \cdot NV}{N\cdot NW} $$

Answer 2

If $\beta$ ranges from $-\infty$ to $\infty$, then $e^\beta/(1+e^\beta)$ will range from 0 to 1. That is a nifty transformation that is usually used the other way around in logistic regression. But you can also use it in reverse here.

edit: Maybe I should clarify a little bit. If your $NV$ can be arbitrarily small and your numerator can be negative, then your proposed function will grow without bounds in both directions. So $\beta$ can not be contained in $(0,1)$. So try fitting $\log(\beta/(1-\beta)) = K \cdot P \cdot NV / NW$ instead and recover $\beta$ by transformation.