# Finding cut-off point for utility function

OK, so apologies for the easy question, but I'm new to this! This is somewhere between elementary algebra, and beginner's game theory. The question comes from a paper I read here (see p. 193): http://home.uchicago.edu/~sashwort/valence.pdf

The following is a utility function for an individual comparing two alternatives (call them L and R). The individual, $i$, prefers L to R when:

$V_L - (x^* - x_L)^2 > V_R - (x^* - x_R)^2$

So far so good. The difficulty I'm having is figuring out how we can get from here to a cutoff rule, such that $i$ will prefer L if and only if:

$x^* < \hat{x}(x_L,x_R,v_L,v_R)$

The paper says that this can be accomplished via "straightforward algebra" to reach:

$\hat{x}(x_L,x_R,v_L,v_R) = \frac{1}{2}(x_R + x_L) + \frac{V_L - V_R}{2(X_R-X_L)}$

Sadly, for me, this algebra ain't so straightforward. If anyone could walk me through the steps to reach this point (or point out how I should approach this) that'd be great. Of course, in the SO tradition, anything more general that can help make this question more applicable to others is also very welcome.

Thanks!

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EDIT: posted this q this morning, and have had some views but no nibbles... anyone got any suggestions? Thanks so much!

• 1. Expand both sides. 2. Cancel the $(x^*)^2$ that appears on both sides. 3. Solve for $x^*$. 4. Simplify, remembering that $(x_L^2-x_R^2)=(x_L+x_R)(x_L-x_R)$. 5. Drop the "algebraic-geometry" tag! :) Commented Jul 22, 2014 at 16:49
• Thanks so much - that's really great! Commented Jul 22, 2014 at 17:20

Just to avoid cumbersome effects, use $x_*$ instead of $x^*$.

Then we have (step by step)

$V_R - (x_* - x_R)^2 < V_L - (x_* - x_L)^2$,
$V_R - x_{*}^2 - x_{R}^2 + 2x_* x_R < V_L - x_{*}^2 - x_{L}^2 + 2x_* x_L$,
$V_R - x_{R}^2 + 2x_* x_R < V_L - x_{L}^2 + 2x_* x_L$,
$2x_* x_R - 2x_* x_L + x_{L}^2 - x_{R}^2 < V_L - V_R$,
$2x_* ( x_R - x_L) < (V_L - V_R) + (x_{R}^2 - x_{L}^2)$,
$x_* < \frac{(V_L - V_R)}{2 ( x_R - x_L)} + \frac{1}{2}(x_{R} + x_{L})$.

As somebody suggested, drop the algebraic topology tag. ;)
I hope it helps!

• Many thanks for this! It's great. Just one thing: in the last step, when you divide $(V_R - V_R)$ by $2(X_R - X_L)$, why is the other term $(X_R^2 - X_L^2)$ not also divided by $(X_R - X_L)$? I understand the $\frac{1}{2}$ part but don't understand where the other bit goes! Apologies for confusion. Many thanks. Commented Jul 22, 2014 at 17:23
• Set $x_R = a$ and $x_L =b$. Then you have $\frac{a^2 - b^2}{2(a-b)}$. But this is nothing more than $\frac{(a - b)(a+b)}{2(a-b)}$, and you simplify to get $\frac{(a+b)}{2}$. Commented Jul 22, 2014 at 17:36
• Ah I see. In which case, I think the last term in the last line ought to be $\frac{1}{2}(x_R + x_L)$ i.e., without the square term on $x_R$ and $x_L$? Commented Jul 22, 2014 at 22:35
• Indeed, I corrected the typo. Commented Jul 23, 2014 at 6:23
• Great, many thanks again. Commented Jul 23, 2014 at 8:35