First, I will clarify some of the notation: $$ x_i \in S_i,\; i\in \{1,2,\dots, n\} \quad x\in S, \quad S\equiv \prod_{i=1}^nS_i \text{ (direct product set)} $$ So basically, we have $x \in S$ which is, if $x_i \in \mathbb{R}$, $x= (x_1,x_2,\dots , x_n)$.

Next, we have the optimization problem: $$\max_{x\in \prod_{i=1}^n S_i} \sum_{i=1}^n\left [H_i(x_i) - \sum_{k=1}^m \lambda^k C_i^k(x_i)\right ]$$ where $C_i^k$ is some real valued function.

Since $S\equiv \prod_{i=1}^n S_i$, the choices $x_i$ may be made independently. Why is this?

I think it is because the $x_i$ don't depend on each other (intuitively, they obviously don't. I don't know how to say this rigorously), and I think that maximizing over a sum of a function of different independent variables -- where the independent variables are indexed by what we are summing over -- is the same as summing the max of the function of each independent variable (I am trying to figure out why this is exactly, but I intuitively it seems true).

Thank you.

Edit: If necessary I can try to be more specific, but I think there is enough here/I have included everything relevant?

Also, example is taken from "Generalized Lagrange Multiplier Method" by Hugh Everett III

  • $\begingroup$ like $S_1 \times ... \times S_n$? $\endgroup$ – BCLC Jan 31 '16 at 5:24
  • $\begingroup$ It is cartesian product if $x_i \in \mathbb{R}$, but if $x_i \not \in \mathbb{R}$ I don't think it will necessarily be cartesian product. My understanding isn't much on the subject. $\endgroup$ – majmun Jan 31 '16 at 5:26
  • $\begingroup$ What do you mean by "independent"? $\endgroup$ – user228113 Jan 31 '16 at 6:15
  • $\begingroup$ @G.Sassatelli I meant to say independent variables. But I guess that doesn't make much sense either, so I rolled the question back to a previous form. Sorry $\endgroup$ – majmun Jan 31 '16 at 6:34

I'm not sure, but if I were to guess I would say that it is because $x\in S \equiv \prod_{i=1}^n S_i$ implies that $x= (f(1),f(2),f(3),\dots ,f(n))$ which in turn implies $x_1 = f(1), x_2 = f(2) \dots x_n = f(n)$ so we have the sum of functions (or the composition of functions) of different independent variables.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.