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?


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 $\displaystyle\beta = \frac{v(w) - v(b)}{u(w) - u(b)} > 0$.

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    $\begingroup$ An MSc/PhD micro coursework is being outsourced to you. $\endgroup$ – Giskard May 12 '17 at 19:44

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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