Consider the following problem \begin{equation*} \begin{aligned} & \underset{x_1, x_2}{\text{maximize}} & & u(x_1, x_2) \\ & \text{subject to} & & p_1x_1 + p_2x_2 = y \end{aligned} \end{equation*} Suppose I found optimal values $(x_1^*, x_2^*)$ that maximize my function $u$.

Now consider problem

\begin{equation*} \begin{aligned} & \underset{x_1, x_2}{\text{maximize}} & & v(u(x_1, x_2)) \\ & \text{subject to} & & p_1x_1 + p_2x_2 = y \end{aligned} \end{equation*}

Does the optimal value change?

First order conditions (FOCs) for the first problem are \begin{equation*} \begin{aligned} & \frac{\partial u}{\partial x_1} - \lambda p_1 = 0 \\ & \frac{\partial u}{\partial x_2} - \lambda p_2 = 0 \end{aligned} \end{equation*}

For the second one

\begin{equation*} \begin{aligned} & \frac{\partial v}{\partial u}\frac{\partial u}{\partial x_1} - \lambda p_1 = 0 \\ & \frac{\partial v}{\partial u}\frac{\partial u}{\partial x_2} - \lambda p_2 = 0 \end{aligned} \end{equation*}

Not sure how to proceed from here. One way to find an example of $v$ that changes optimal value. I tried several candidates but get same optimal values. A link or description of relevant material will be helpful.

  • $\begingroup$ Set $v(x)=-x$. Then clearly a maximum for $u$ is a minimum for $v$ and conversely. But in general you shouldn't expect any relation between the two.. $\endgroup$
    – b00n heT
    Sep 9 '16 at 9:31
  • $\begingroup$ @b00n heT Thanks for your comment. What if I specify that $v$ is a positive monotonic transformation? Such as for example $v = x^2$ or $ v = e^x$? Then there will be a relation? $\endgroup$
    – tosik
    Sep 9 '16 at 9:34
  • $\begingroup$ The relation must be monotonic increasing. $\endgroup$
    – user65203
    Sep 9 '16 at 9:35
  • $\begingroup$ Of course there will be: if some function is monotone, then maximizing/minimizing the function is equivalent to maximizing/minimizing its argument. $\endgroup$
    – b00n heT
    Sep 9 '16 at 9:36

If $v$ is strictly growing on the domain where $u$ take its values, it doesn't change the optimums

If $v$ is strictly decreasing on the domain where $u$ take its values, it turns minimas into maximas, and maximas into minimas

Rq : This is often exploited to simplify problems

Following comment : I don't think this is worth calling theorem, and I don't think there is a special name for this property. The proof : (for one case)

Let us consider a set $S$, $f$ a function on $S$ such that the maximum of $f$ on $S$ exists, $\max_{x \in S} f(x) = x_0$, and $g$ a growing fuinction on $f(S)$.

$\forall x \in S, f(x) \leq f(x_0)$ so $\forall x \in S, g(f(x)) \leq g(f(x_0)) $

  • $\begingroup$ Thanks for your answer. It will be really helpful if you could give me a name or a link to a theorem that ensures that positive monotonic transformation doesn't change optimal values $\endgroup$
    – tosik
    Sep 9 '16 at 9:55
  • $\begingroup$ Ahh, so obviously, thanks $\endgroup$
    – tosik
    Sep 9 '16 at 12:01

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