Suppose that we have four constants, $1>s>0, A>1, B>1,1>t>0$. (by constants I mean exogenous variables)
Also suppose that we have a function of three variables (defined on the positive reals): $F(w,y,z)$
Then suppose that we wish to solve the following maximization problem:
$$\max_x F \left(B+A-x,x,\frac{x-A-t}{1-s}\right),$$
where $F$ is strictly quasi concave. (if necessary also assume that the optimum occurs in the interior, but if this is needed please say so)
The first order condition for this maximization problem is:
$$\tag{2}
-F_w+ F_y+F_z \frac{1}{1-s}=0
$$
Are we able to write the $x$ that solves (2) as
$$
x^* = f(B+A, \frac{A+t}{1-s},s)
$$
where $f$ is some function, and where I use $x^*$ to denote the $x$ that solves (2)
If so, why?
Specifically, how do we know the function $f$ takes as arguments $B+A$, $\frac{A+t}{1-s}$, $s$?
Would it be incorrect to instead write $x^*=f(B+A,\frac{A+t}{1-s})$?
- I feel like it would be, because if we imagine changing $s$ but keeping ${A+t}{1-s}$ the same the solution would change, as the result of $F_z\frac{1}{1-s}$ changing
What about writing it as $x^*=f(A,B,s,t)$?
- this one is much less clear to me whether we can write it as this, because yes it depends on these 4 constants, but there are interdepencies between them.
In general (i.e. not directly related to this question), I get confused when writing a function of variables as a function of different variables,
As an example, I believe we could write $(c+d)*w + u^h = g(c,d,h) = h(w,u)$. I.e., we could write this same function as taking different arguments.
In this example, I get that what we take as the variables is likely a judgement call on whatever we believe are independent variables and what are constants, but the earlier case of $f(B+A,\frac{A+t}{1-s},s)$ vs $f(A,B,s,t)$ is much less clear to me
Edit to clarify notation: My notation of $F_w$ means the derivative of the function $F$ with respect to $w$.
$w$ appears as the first argument, so of course you could instead use the notation $F_1$ instead of $F_w$ If you prefer
Also note that, when I wrote the first order condition, obviously the derivatives should be evaluated at the relevant point.
Best Answer
The two ways of writing it are equivalent. To see this, Let $\hat z = z + y$. Then, your optimisation problem is now given by $$ \max_x F\left(\hat z - x,x,x-y\right). $$
If the maximiser is unique, then the $x$ that solves the maximisation problem is a function of $\hat z$ and $y$. That is, it is a function of $z+y$ and $y$.
Given a function $G$, we can always define another function $H$ such that $H(z,y)=G(z+y,y)$, so writing $x$ as $G(z,y)$ appears to be more general. But given the discussion above, you lose nothing about considering a function of $z+y$ and $y$ instead, but you might gain something if it simplifies some of your other notation.