<p>Suppose $v=\alpha + \beta u$ for $\beta &gt; 0$ i.e. $v$ is a positive affine transformation of $u$, then
$$ \frac{v(g^1) - v(g^2)}{v(g^2) - v(g^3)} = \frac{\alpha + \beta u(g^1) - \alpha - \beta u(g^2)}{\alpha + \beta u(g^2) - \alpha - \beta u(g^3)}  = \frac{u(g^1) - u(g^2)}{u(g^2) - u(g^3)}$$</p>

<p>Converse is <strong>not</strong> true because $v=-u$ also satisfy</p>

<p>$$ \frac{v(g^1) - v(g^2)}{v(g^2) - v(g^3)}   = \frac{u(g^1) - u(g^2)}{u(g^2) - u(g^3)}$$</p>

<p>for all $g_1, g_2, g_3\in\mathcal{G}$.</p>

<p><strong>Added Later</strong></p>

<p>By the way, converse is also true. We are given this information that $u$ and $v$ both represent $\succsim$. So, $v=-u$ is not a valid example as they cannot both represent $\succsim$.</p>

<p>Here is the proof of the converse:</p>

<p>Fix any two lotteries $b$ and $w$ such that $u(b) &gt; u(w)$ and consequently, $v(b) &gt; v(w)$. For any lottery $g\in\mathcal{G}$, we know that the following is true</p>

<p>$$ \frac{v(g) - v(w)}{v(w) - v(b)}   = \frac{u(g) - u(w)}{u(w) - u(b)}$$</p>

<p>So, $$v(g) = v(w) + (v(w) - v(b))\frac{u(g) - u(w)}{u(w) - u(b)} $$
or equivalently,</p>

<p>\begin{eqnarray*} v(g) &amp; = &amp; v(w)  - \frac{v(w) - v(b)}{u(w) - u(b)}u(w)+ \frac{v(w) - v(b)}{u(w) - u(b)}u(g) \\ &amp; = &amp;  \left(\frac{u(w)v(b) - v(w)u(b)}{u(w) - u(b)}\right)+ \left(\frac{v(w) - v(b)}{u(w) - u(b)}\right)u(g) \end{eqnarray*} </p>

<p>Therefore, $v = \alpha + \beta u$ where $\displaystyle\alpha = \frac{u(w)v(b) - v(w)u(b)}{u(w) - u(b)}$ and $\displaystyle\beta = \frac{v(w) - v(b)}{u(w) - u(b)} &gt; 0$.</p>

CC BY-SA 3.0
2017-05-12T19:44:09.447Z
An MSc/PhD micro coursework is being outsourced to you.
Giskard
CC BY-SA 3.0
2017-05-12T15:21:59.797Z
2017-05-13T03:36:25.187Z
2017-05-13T03:36:25.187Z
Amit
<p>Well, the "if" part is certainly straightforward.  Make v = a+bu where b>0.  The a's cancel through subtraction in each numerator and denominator; once this is done, the b's cancel in numerator and denominator.  </p>

<p>The "only if" part is a little harder.</p>

CC BY-SA 3.0
2017-05-12T13:11:19.817Z
2017-05-12T13:11:19.817Z
David Croson
<p>Let $u$ and $v$ be utility functions (not necessarily VNM) representing $\succsim$ on $\mathcal{G}$. Show that $v$ is a positive affine transformation of $u$ if and only if for all gambles $g^1, g^2, g^3$ in $\mathcal{G}$, with no two indifferent, we have:</p>

<p>$\frac{u(g^1)-u(g^2)}{u(g^2)-u(g^3)} =\frac{v(g^1)-v(g^2)}{v(g^2)-v(g^3)}$</p>

<p>I have tried to figure it out, but I am not sure where to start. Anyone who got a hint for me?</p>

<microeconomics><uncertainty>
Help with a choice under uncertainty exercise
Andrea F.L.
# Help with a choice under uncertainty exercise

(Asked by: username_0 on 2017-05-12)

Let $u$ and $v$ be utility functions (not necessarily VNM) representing $\succsim$ on $\mathcal{G}$. Show that $v$ is a positive affine transformation of $u$ if and only if for all gambles $g^1, g^2, g^3$ in $\mathcal{G}$, with no two indifferent, we have:

$\frac{u(g^1)-u(g^2)}{u(g^2)-u(g^3)} =\frac{v(g^1)-v(g^2)}{v(g^2)-v(g^3)}$

I have tried to figure it out, but I am not sure where to start. Anyone who got a hint for me?

**Tags:** microeconomics, uncertainty


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## 2 answers

### Answer by username_1 on 2017-05-12. Score: 2

Suppose $v=\alpha + \beta u$ for $\beta > 0$ i.e. $v$ is a positive affine transformation of $u$, then
$$ \frac{v(g^1) - v(g^2)}{v(g^2) - v(g^3)} = \frac{\alpha + \beta u(g^1) - \alpha - \beta u(g^2)}{\alpha + \beta u(g^2) - \alpha - \beta u(g^3)} = \frac{u(g^1) - u(g^2)}{u(g^2) - u(g^3)}$$

Converse is **not** true because $v=-u$ also satisfy

$$ \frac{v(g^1) - v(g^2)}{v(g^2) - v(g^3)} = \frac{u(g^1) - u(g^2)}{u(g^2) - u(g^3)}$$

for all $g\_1, g\_2, g\_3\in\mathcal{G}$.

**Added Later**

By the way, converse is also true. We are given this information that $u$ and $v$ both represent $\succsim$. So, $v=-u$ is not a valid example as they cannot both represent $\succsim$.

Here is the proof of the converse:

Fix any two lotteries $b$ and $w$ such that $u(b) > u(w)$ and consequently, $v(b) > v(w)$. For any lottery $g\in\mathcal{G}$, we know that the following is true

$$ \frac{v(g) - v(w)}{v(w) - v(b)} = \frac{u(g) - u(w)}{u(w) - u(b)}$$

So, $$v(g) = v(w) + (v(w) - v(b))\frac{u(g) - u(w)}{u(w) - u(b)} $$
or equivalently,

\begin{eqnarray\*} v(g) & = & v(w) - \frac{v(w) - v(b)}{u(w) - u(b)}u(w)+ \frac{v(w) - v(b)}{u(w) - u(b)}u(g) \\ & = & \left(\frac{u(w)v(b) - v(w)u(b)}{u(w) - u(b)}\right)+ \left(\frac{v(w) - v(b)}{u(w) - u(b)}\right)u(g) \end{eqnarray\*}

Therefore, $v = \alpha + \beta u$ where $\displaystyle\alpha = \frac{u(w)v(b) - v(w)u(b)}{u(w) - u(b)}$ and $\displaystyle\beta = \frac{v(w) - v(b)}{u(w) - u(b)} > 0$.

> **Comments:**
> **username_2**: An MSc/PhD micro coursework is being outsourced to you.  

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### Answer by username_3 on 2017-05-12. Score: 0

Well, the "if" part is certainly straightforward. Make v = a+bu where b>0. The a's cancel through subtraction in each numerator and denominator; once this is done, the b's cancel in numerator and denominator.

The "only if" part is a little harder.


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