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

CC BY-SA 4.0
2019-10-17T06:34:17.990Z
This answer is incorrect.
Giskard
CC BY-SA 4.0
2019-10-13T18:25:19.470Z
2019-10-13T18:25:19.470Z
ghghghghghg
<p>I have utility function given by:</p>

<p><span class="math-container">$U(x_1, x_2) =
  \begin{cases}
                                   x_1+x_2 &amp; \text{if $x_1+x_2&lt;6$} \\
                                   6 &amp; \text{if $6\leq x_1+x_2\leq10$} \\
                                   x_1 + x_2 &amp; \text{if $x_1+x_2&gt;10$}
  \end{cases}$</span></p>

<p>If I transform this to this:</p>

<p><span class="math-container">$F(U(x_1, x_2)) =
  \begin{cases}
                                   U(x_1, x_2)+4 &amp; \text{if $U(x_1, x_2)&lt;6$} \\
                                   10 &amp; \text{if $U(x_1, x_2)=6$} \\
                                   U(x_1, x_2) &amp; \text{if $U(x_1, x_2)&gt;10$}
  \end{cases}$</span></p>

<p>Is it strictly increasing function transformation?</p>

<p>Thanks guys</p>

CC BY-SA 4.0
2019-10-16T23:14:24.943Z
Are you sure your definition makes sense? If I were you, I would first F alone, without reference to U.
Arthur
CC BY-SA 4.0
2019-10-17T06:32:05.950Z
@Arthur Why would that be helpful? Even from this formulation, one can see the problem.
Giskard
CC BY-SA 4.0
2019-10-17T19:57:29.750Z
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
CheeseBurger
<utility>
Strictly increasing function transformation
CheeseBurger
# Strictly increasing function transformation

(Asked by: username_0 on 2019-10-13)

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

**Tags:** utility
> **Comments:**
> **username_1**: Are you sure your definition makes sense? If I were you, I would first F alone, without reference to U.  
> **username_2**: @Arthur Why would that be helpful? Even from this formulation, one can see the problem.  
> **username_0**: 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  

---

## 1 answers

### Answer by username_3 on 2019-10-13. Score: -2

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

> **Comments:**
> **username_2**: This answer is incorrect.  

---

