<p>Other people have already hinted at the correct answer. Here is a complete justification:</p>

<p>Note that <span class="math-container">$V(x,y) = g(U(x,y))$</span> where <span class="math-container">$g$</span> is the function defined by <span class="math-container">$$g(u) = \dfrac{1}{\beta} \cdot \log(u).$$</span> Assuming that only positve levels of consumption are allowed (otherwise <span class="math-container">$V$</span> is not well defined), <span class="math-container">$g$</span> is continuously differentiable on the image of <span class="math-container">$U$</span> and satisfies
<span class="math-container">$$
g'(u) = \dfrac{1}{\beta u} &gt; 0
$$</span>
Therefore, <span class="math-container">$g$</span> is a monotone transformation. <span class="math-container">$\blacksquare$</span></p>

CC BY-SA 4.0
2020-06-01T18:30:05.407Z
2020-06-01T18:30:05.407Z
brunosalcedo
<p>Take the first derivative; if that is always positive for all positive values of your variable then the transformation is positive monotone (i.e maintains ordering)</p>

<p>Since you have two variables, you need to check how the transformation affects both by checking two first order derivatives.</p>

<p>Edit: if you assume the function U is always positive you only need to check the first derivative of the transformation with respect to U</p>

CC BY-SA 4.0
2020-06-01T15:33:29.770Z
Why should this depend on the number of variables? There's one transformation, $F(U)=\frac{\ln U}{\beta}$, so there's one derivative to check.
VARulle
CC BY-SA 4.0
2020-06-01T16:09:03.277Z
There are two transformations, the first is when you take the natural logarithm, the second is when you divide by $\beta$ you need to check that both of these do not change the original ordering of utilities given $x_1$ and $x_2$. Dividing by $\beta$ is a nonissue as long as $\beta$ is positive, it’s just multiplication by a constant. So this boils down to checking the log transformation, which is by definition a positive monotone transformation but to formally check it you need to show that it’s increasing in the values of the original utility function, made up of $x_1$ and $x_2$
tvbc
CC BY-SA 4.0
2020-06-01T16:23:52.790Z
You can always view it as one transformation, it won’t really make a difference, however, unless you assume U is always positive you can run into a situation where beta = 1/ln(x2) and x1 is constant...that would not be a positive monotone transformation
tvbc
CC BY-SA 4.0
2020-06-01T14:22:19.903Z
2020-06-01T16:17:30.537Z
2020-06-01T16:17:30.537Z
tvbc
<p>Consider the utility function</p>

<p><span class="math-container">$$
U(x_1,x_2) = x_1^\alpha x_2^\beta
$$</span>
for <span class="math-container">$0 &lt; \alpha, \beta &lt; 1$</span>. How do I then show that 
<span class="math-container">$$
V(x_1,x_2) = F(U(x_1,x_2)) = \frac{\alpha}{\beta} \ln(x_1) + \ln(x_2)
$$</span>
is a positive, monotone transformation of <span class="math-container">$U(x_1,x_2)$</span>. I thought about showing that MRS for both functions is the same. Is this approach alright? I also thought about taking <span class="math-container">$\ln(x)$</span> on <span class="math-container">$U(x_1,x_2)$</span> and see what happens:</p>

<p><span class="math-container">$$\ln(x_1^\alpha x_2^\beta) = a \ln(x_1) + \beta \ln(x_2)$$</span>
but I am not sure here if I am allowed to divide with <span class="math-container">$\frac{1}{\beta}$</span> now to get the desired result?</p>

<p>Thanks for your help in advance. </p>

CC BY-SA 4.0
2020-06-01T11:17:04.057Z
Well, is multiplication with $1/\beta$ a positive, monotone transformation?
VARulle
CC BY-SA 4.0
2020-06-01T11:46:19.053Z
I think so, yes. I don't see how it changes anything? Otherwise I am not sure how to get the desired result.
Mathias
<microeconomics>
How to prove that a function is a positive montone transformation
Mathias
# How to prove that a function is a positive montone transformation

(Asked by: username_0 on 2020-06-01)

Consider the utility function

$$
U(x\_1,x\_2) = x\_1^\alpha x\_2^\beta
$$
for $0 < \alpha, \beta < 1$. How do I then show that
$$
V(x\_1,x\_2) = F(U(x\_1,x\_2)) = \frac{\alpha}{\beta} \ln(x\_1) + \ln(x\_2)
$$
is a positive, monotone transformation of $U(x\_1,x\_2)$. I thought about showing that MRS for both functions is the same. Is this approach alright? I also thought about taking $\ln(x)$ on $U(x\_1,x\_2)$ and see what happens:

$$\ln(x\_1^\alpha x\_2^\beta) = a \ln(x\_1) + \beta \ln(x\_2)$$
but I am not sure here if I am allowed to divide with $\frac{1}{\beta}$ now to get the desired result?

Thanks for your help in advance.

**Tags:** microeconomics
> **Comments:**
> **username_1**: Well, is multiplication with $1/\beta$ a positive, monotone transformation?  
> **username_0**: I think so, yes. I don't see how it changes anything? Otherwise I am not sure how to get the desired result.  

---

## 2 answers

### Answer by username_2 on 2020-06-01. Score: 2

Other people have already hinted at the correct answer. Here is a complete justification:

Note that $V(x,y) = g(U(x,y))$ where $g$ is the function defined by $$g(u) = \dfrac{1}{\beta} \cdot \log(u).$$ Assuming that only positve levels of consumption are allowed (otherwise $V$ is not well defined), $g$ is continuously differentiable on the image of $U$ and satisfies
$$
g'(u) = \dfrac{1}{\beta u} > 0
$$
Therefore, $g$ is a monotone transformation. $\blacksquare$


---

### Answer by username_3 on 2020-06-01. Score: 0

Take the first derivative; if that is always positive for all positive values of your variable then the transformation is positive monotone (i.e maintains ordering)

Since you have two variables, you need to check how the transformation affects both by checking two first order derivatives.

Edit: if you assume the function U is always positive you only need to check the first derivative of the transformation with respect to U

> **Comments:**
> **username_1**: Why should this depend on the number of variables? There's one transformation, $F(U)=\frac{\ln U}{\beta}$, so there's one derivative to check.  
> **username_3**: There are two transformations, the first is when you take the natural logarithm, the second is when you divide by $\beta$ you need to check that both of these do not change the original ordering of utilities given $x_1$ and $x_2$. Dividing by $\beta$ is a nonissue as long as $\beta$ is positive, it’s just multiplication by a constant. So this boils down to checking the log transformation, which is by definition a positive monotone transformation but to formally check it you need to show that it’s increasing in the values of the original utility function, made up of $x_1$ and $x_2$  
> **username_3**: You can always view it as one transformation, it won’t really make a difference, however, unless you assume U is always positive you can run into a situation where beta = 1/ln(x2) and x1 is constant...that would not be a positive monotone transformation  

---

