What is the intuitive explanation of weighting factor $\alpha$ and $1-\alpha$ in the equations such as score, optimization, smoothing etc, that takes the form below:
$$ f(\alpha) = \alpha \cdot A + (1-\alpha)B, \text{ where } 0 < \alpha < 1$$
Examples:
Edit: Why do we use $\alpha$ and $1-\alpha$ instead of two independent values $\alpha$ and $\beta$?
On
When $\alpha = 0$, we get $f(0) = 0 \cdot A + (1 - 0)B = B$. Thus, at $\alpha = 0$, the output is "purely" $B$.
When $\alpha = 1$, we get $f(1) = 1 \cdot A + (1 - 1)B = A$. Thus, at $\alpha = 1$, the output is "purely" $A$.
For values in between, you get a mixture of $A$ and $B$ that can be viewed as summing to 100%. For example, when $\alpha = 0.2$, we get $f(0.2) = 0.2A + 0.8B$, which can be viewed as a mixture that is 20% type $A$ and 80% type $B$.
Another useful interpretation is that it is the equation of the line through the points $(0, B)$ and $(1,A)$. This is more readily seen by rewriting the function as $f(\alpha) = B + (A-B) \alpha$. This is the more familiar description of the line with $y$-intercept $(0,B)$ and slope $A-B$.
We want our "weight coefficients", i.e., the numbers that tell us what percent of the whole this factor is, to sum up to $1$. For example, if we have $A$ and $B$, and we say $g = .3A + .7B$, we are saying the "makeup" of $g$ is $30 \%$ $A$ and $70 \%$ $B$, and the percents should add up to $100 \%$ since together they make up $1$ unit of $g$. So our "weight coefficients", $.3$ and $.7$, should add up to $1$.
That's why, when there are $2$ coefficients, we say one is $\alpha$ (which is a number between $0$ and $1$) and the other is $1 - \alpha$. If there are $3$ coefficients, we could say one is $\alpha$, one is $\beta$, (both are between $0$ and $1$) and the third is $1 - \alpha - \beta$. Or we could say the three are $a, b, c$ with the condition that $0 \leq a, b ,c \leq 1$ and $ a+ b + c = 1$.