<p>The value of $h(g(f(u(x,y))$ is not necessarily positive but the transformation $u \rightarrow h \circ g \circ f \circ u$ is indeed a positive monotone transformation.</p>

<p>To see this, take any $(x,y,z,w)$. We have the following chain of equivalences:
\begin{align}
u(x,y) \geq u(z,w) &amp; \Leftrightarrow f(u(x,y)) \geq f(u(z,w)) \text{ since } f \text{ is increasing} \\
&amp; \Leftrightarrow g(f(u(x,y)) \geq g(f(u(z,w)) \text{ since } g \text{ is increasing} \\
&amp; \Leftrightarrow h(g(f(u(x,y))) \geq h(g(f(u(z,w))) \text{ since } h \text{ is increasing}
\end{align}
Thus, the transformation $u \rightarrow h \circ g \circ f \circ u$ preserves the ordering, i.e. it is a positive monotone transformation.</p>

<p>Another way to see it is to assume that all functions are differentiable and notice that
\begin{equation*}
\frac{d (h \circ g \circ f )(x)}{dx}= f'(x) g'(f(x)) h'(g(f(x))
\end{equation*}
which is positive since $f'&gt;0,g'&gt;0,h'&gt;0$.</p>

CC BY-SA 3.0
2017-04-04T06:06:59.313Z
2017-04-04T06:06:59.313Z
Oliv
<p>Suppose there is an economic agent with the utility function $u(x,y)$. A second agent has the utility function $h(g(f(u(x,y))))$.</p>

<p>Am I correct in thinking that if $f'(x)&gt;0$, $g'(x)&gt;0$, and $h'(x)&gt;0$, then  $h(g(f(u(x,y))))$ must also be positive and therefore  $h(g(f(u(x,y))))$ is a positive monotonic transformation of $u(x,y)$? </p>

<p>If not, how come?</p>

<microeconomics><mathematical-economics><utility>
Positive Monotonic Transformations and Nested Functions
Elizabeth 
# Positive Monotonic Transformations and Nested Functions

(Asked by: username_0 on 2017-04-04)

Suppose there is an economic agent with the utility function $u(x,y)$. A second agent has the utility function $h(g(f(u(x,y))))$.

Am I correct in thinking that if $f'(x)>0$, $g'(x)>0$, and $h'(x)>0$, then $h(g(f(u(x,y))))$ must also be positive and therefore $h(g(f(u(x,y))))$ is a positive monotonic transformation of $u(x,y)$?

If not, how come?

**Tags:** microeconomics, mathematical-economics, utility

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## 1 answers

### Answer by username_1 on 2017-04-04. Score: 4 | Accepted answer

The value of $h(g(f(u(x,y))$ is not necessarily positive but the transformation $u \rightarrow h \circ g \circ f \circ u$ is indeed a positive monotone transformation.

To see this, take any $(x,y,z,w)$. We have the following chain of equivalences:
\begin{align}
u(x,y) \geq u(z,w) & \Leftrightarrow f(u(x,y)) \geq f(u(z,w)) \text{ since } f \text{ is increasing} \\
& \Leftrightarrow g(f(u(x,y)) \geq g(f(u(z,w)) \text{ since } g \text{ is increasing} \\
& \Leftrightarrow h(g(f(u(x,y))) \geq h(g(f(u(z,w))) \text{ since } h \text{ is increasing}
\end{align}
Thus, the transformation $u \rightarrow h \circ g \circ f \circ u$ preserves the ordering, i.e. it is a positive monotone transformation.

Another way to see it is to assume that all functions are differentiable and notice that
\begin{equation\*}
\frac{d (h \circ g \circ f )(x)}{dx}= f'(x) g'(f(x)) h'(g(f(x))
\end{equation\*}
which is positive since $f'>0,g'>0,h'>0$.


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