I have utility function given by:

$U(x_1, x_2) = \begin{cases} x_1+x_2 & \text{if $x_1+x_2<6$} \\ 6 & \text{if $6\leq x_1+x_2\leq10$} \\ x_1 + x_2 & \text{if $x_1+x_2>10$} \end{cases}$

If I transform this to this:

$F(U(x_1, x_2)) = \begin{cases} U(x_1, x_2)+4 & \text{if $U(x_1, x_2)<6$} \\ 10 & \text{if $U(x_1, x_2)=6$} \\ U(x_1, x_2) & \text{if $U(x_1, x_2)>10$} \end{cases}$

Is it strictly increasing function transformation?

Thanks guys

  • $\begingroup$ Are you sure your definition makes sense? If I were you, I would first F alone, without reference to U. $\endgroup$ – Arthur Oct 16 '19 at 23:14
  • $\begingroup$ @Arthur Why would that be helpful? Even from this formulation, one can see the problem. $\endgroup$ – Giskard Oct 17 '19 at 6:32
  • $\begingroup$ Thanks, guys, I got it later its a non-decreasing function but not a strictly increasing one. @Arthur I wrote the function like that because it is more apparent rather than writing it without referring to U, anyways thanks for trying $\endgroup$ – CheeseBurger Oct 17 '19 at 19:57

This is a strictly increasing monotonic transformation simply because the preferences in the previous function are the same in the transformed function as well.

  • 1
    $\begingroup$ This answer is incorrect. $\endgroup$ – Giskard Oct 17 '19 at 6:34

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