weighted average with exponential weighting I want to create weighted average, where weights depend on value of number. If I want exponential weights is this regular?
$average = \log_e(\frac{\sum_{i=1}^n e^{v_i}}{n})$
Isn't it just average of weights?
If it works, what if I want weighting quadratic or linear?
EDIT: weighted average mean is defined as:
$average_{weighted}=\frac{\sum_{i=1}^n w_i \cdot x_i}{\sum_{i=1}^n w_i}$
If I want weights increased exponentialy this should work:
$average= \frac{\sum_{i=1}^n e^{x_i} \cdot x_i}{\sum_{i=1}^n e^{x_i}}$
Am I right?
 A: Any continuous one-to-one function $f(x)$ can form the basis for a so-called quasi-arithmetic mean (aka Kolmogorov mean) by defining the average $<x>$ of the numbers $x_1, x_2, .. x_n$ as
$$n f(<x>) = f(x_1) + f(x_2) + .. + f(x_n) $$
For $f(x) = x$ the arithmetic mean is recovered, for $f(x) = 1/x$ we get the harmonic mean, and for generic $f(x) = x^p$ the power mean. For $f(x) = ln(x)$ the geometric mean is recovered.
You are proposing $f(x) = e^x$ which I will refer to as the anti-geometric mean (not sure if it has a name already). This mean lacks the desirable property of being homogeneous. In other words, if the anti-geometric mean of $x_1, x_2, .. x_n$ equals $<x>$, it doesn't follow that the anti- geometric mean of $c x_1, c x_2, .. c x_n$ equals $c <x>$. 
The power means and geometric means are all homogeneous. That's what sets them aside from the other quasi-arithmetic means, and what renders them widely useful.

Edit - following the recent edits made in the question, it seems that what is asked is: can a meaningful generalized weighted mean $<x>$ be defined in which the weights depend on the values included in the mean. More specifically: 
$$(w(x_1) + w(x_2) + .. + w(x_n)) <x> = x_1 w(x_1) + x_2 w(x_2) + .. + x_n w(x_n) $$
The answer is "yes". An example is the Lehmer mean, a special case of the above weighted average with weights $w(x) = x^p$. The choice $w(x) = e^x$ is also valid and would lead to a maximum value dominated mean. 
