Equivalence between convex combination and multiplicative form

72 Views Asked by At

I am interested in understanding the relationship between convex combination and a multiplicative form.

Let $x,y > 0$. Let the convex combination be $f(\gamma) = \gamma x + (1-\gamma) y$ where $\gamma \in [0,1]$. Let the "multiplicative combination" be $g(\theta) = x^\theta y^{1-\theta}$ where $\theta \in [0,1]$.

Intuitively, both of these combinations vary between $x$ and $y$ as $\gamma$ and $\theta$ change. Clearly when $\gamma = \theta = 0$, $f(0) = g(0) = y$ and when $\gamma=\theta= 1$, $f(1) =g(1)=x$.

I am wondering if these two formulations are essentially equal and if there is a way to write $\gamma$ in terms of $\theta$?

Thank you!

1

There are 1 best solutions below

2
On

The "multiplicative combination" is just taking the convex combination of $\log x$ and $\log y$ and then exponentiating back: $$\log g(\theta) = \theta \log(x) + (1-\theta) \log(y).$$

Although both approaches parameterize the interval between $x$ and $y$, they are not "essentially equal." Take for instance $x=2$ and $y=1$. Then $f(\gamma)=1 + \gamma$ but $g(\theta) = 2^\theta$, even though $f(0)=g(0)=1$ and $f(1)=g(1)=2$